The maximum speed of the bob.

Question

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Answer 1

Question

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

Chapter 12, Problem 35P

(a)

To determine

The maximum speed of the bob.

(a)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.820 m/s$ .

Explanation of Solution

Section 1:

To determine: The amplitude of the motion.

Answer: The maximum speed of the amplitude of the motion is $0.262 m$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate amplitude is,

$A=Lθ$

$L$ is the length of pendulum.
$θ$ is the displaced angle.

Substitute $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $A$ .

$A=(1.00 m)(15.0°(π180°))=0.262 m$

Section 2:

To determine: The angular frequency of the motion.

Answer: The angular frequency of the motion is $3.13 rad/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate angular frequency is,

$ω=gL$

$g$ is the acceleration due to gravity.

Substitute $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $ω$ .

$ω=9.8 m/s21.00 m=3.13 rad/s$

Section 3:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.820 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum speed is,

$vmax=Aω$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $vmax$ .

$vmax=(0.262 m)(3.13 rad/s)=0.820 m/s$

Conclusion:

Therefore, the maximum speed of the bob is $0.820 m/s$ .

(b)

To determine

The maximum acceleration of the bob.

(b)

Expert Solution

Answer to Problem 35P

The maximum acceleration of the bob is $2.57 rad/s2$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum acceleration of the bob is,

$amax=Aω2$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $amax$ .

$amax=(0.262 m)(3.13 rad/s)2=2.57 rad/s2$

Conclusion:

Therefore, the maximum acceleration of the bob is $2.57 rad/s2$ .

(c)

To determine

The maximum restoring force of the bob.

(c)

Expert Solution

Answer to Problem 35P

The maximum restoring force of the bob is $0.641 N$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum restoring force of the bob is,

$F=mamax$

$m$ is the mass of the pendulum.

Substitute $Aω2$ for $amax$ in above equation.

$F=mAω2$

Substitute $0.250 kg$ for $m$ , $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $F$ .

$F=(0.250 kg)(0.262 m)(3.13 rad/s)2=0.641 N$

Conclusion:

Therefore, the maximum restoring force of the bob is $0.641 N$ .

(d)

To determine

The maximum speed, angular acceleration and restoring force of the bob using the model introduced earlier chapter.

(d)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Explanation of Solution

Section 1:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.817 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

Consider the figure given below.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 35P

In triangle $ABC$ ,

$cosθ=ACABAC=ABcosθ=Lcosθ$

The height of the bob is,

$h=AD−AC=L−Lcosθ=L(1−cosθ)$

The law of conservation of energy is,

$mgh=12mvmax2$

Substitute $L(1−cosθ)$ for $h$ in above expression.

$mgL(1−cosθ)=12mvmax2gL(1−cosθ)=12vmax2$

Substitute $15.0°$ for $θ$ , $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $vmax$ .

$(9.8 m/s2)(1.00 m)(1−cos(15.0°))=12vmax2vmax2=2(0.333) m2/s2vmax=2(0.333) m2/s2=0.817 m/s$

Section 2:

To determine: The angular acceleration of the bob.

Answer: The angular acceleration of the bob is $2.54 rad/s2$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula for the moment of inertia of the pendulum is,

$I=mL2$

The equation for the conservation of energy is,

$Iα=mgLsinθ$

$α$ is the angular acceleration.

Substitute $mL2$ for $I$ in above expression and rearrange for $α$ .

$mL2α=mgLsinθα=mgLsinθmL2=gsinθL$

Substitute $9.8 m/s2$ for $g$ , $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $α$ .

$α=(9.8 m/s2)sin(15.0°)1.00 m=2.54 rad/s2$

Section 3:

To determine: The restoring force of the bob.

Answer: The restoring force of the bob is $0.634 N$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The force is maximum, when the angle is maximum.

The restoring force is calculated as,

$F=mgsinθ$

Substitute $15.0°$ for $θ$ , $0.250 kg$ for $m$ and $9.8 m/s2$ for $g$ in above equation to find $F$ .

$F=(0.250 kg)(9.8 m/s2)sin(15.0°)=0.634 N$

Conclusion:

Therefore, the maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

(e)

To determine

To compare: The answers of part (a), part (c) and part (d).

(e)

Expert Solution

Explanation of Solution

Introduction: The restoring force is defined as the force or torque that tends to restore a system to equilibrium after displacement.

The answers are closest but not exactly the same. The angular amplitude of $15.0°$ is not small, so the simple harmonic oscillation is not accurate. The answers computed from conservation of the energy and from Newton’s second law are more accurate.

Conclusion:

Therefore, the answers are closest but not exactly the same.

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Answer 2

Question

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

Chapter 12, Problem 35P

(a)

To determine

The maximum speed of the bob.

(a)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.820 m/s$ .

Explanation of Solution

Section 1:

To determine: The amplitude of the motion.

Answer: The maximum speed of the amplitude of the motion is $0.262 m$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate amplitude is,

$A=Lθ$

$L$ is the length of pendulum.
$θ$ is the displaced angle.

Substitute $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $A$ .

$A=(1.00 m)(15.0°(π180°))=0.262 m$

Section 2:

To determine: The angular frequency of the motion.

Answer: The angular frequency of the motion is $3.13 rad/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate angular frequency is,

$ω=gL$

$g$ is the acceleration due to gravity.

Substitute $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $ω$ .

$ω=9.8 m/s21.00 m=3.13 rad/s$

Section 3:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.820 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum speed is,

$vmax=Aω$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $vmax$ .

$vmax=(0.262 m)(3.13 rad/s)=0.820 m/s$

Conclusion:

Therefore, the maximum speed of the bob is $0.820 m/s$ .

(b)

To determine

The maximum acceleration of the bob.

(b)

Expert Solution

Answer to Problem 35P

The maximum acceleration of the bob is $2.57 rad/s2$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum acceleration of the bob is,

$amax=Aω2$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $amax$ .

$amax=(0.262 m)(3.13 rad/s)2=2.57 rad/s2$

Conclusion:

Therefore, the maximum acceleration of the bob is $2.57 rad/s2$ .

(c)

To determine

The maximum restoring force of the bob.

(c)

Expert Solution

Answer to Problem 35P

The maximum restoring force of the bob is $0.641 N$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum restoring force of the bob is,

$F=mamax$

$m$ is the mass of the pendulum.

Substitute $Aω2$ for $amax$ in above equation.

$F=mAω2$

Substitute $0.250 kg$ for $m$ , $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $F$ .

$F=(0.250 kg)(0.262 m)(3.13 rad/s)2=0.641 N$

Conclusion:

Therefore, the maximum restoring force of the bob is $0.641 N$ .

(d)

To determine

The maximum speed, angular acceleration and restoring force of the bob using the model introduced earlier chapter.

(d)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Explanation of Solution

Section 1:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.817 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

Consider the figure given below.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 35P

In triangle $ABC$ ,

$cosθ=ACABAC=ABcosθ=Lcosθ$

The height of the bob is,

$h=AD−AC=L−Lcosθ=L(1−cosθ)$

The law of conservation of energy is,

$mgh=12mvmax2$

Substitute $L(1−cosθ)$ for $h$ in above expression.

$mgL(1−cosθ)=12mvmax2gL(1−cosθ)=12vmax2$

Substitute $15.0°$ for $θ$ , $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $vmax$ .

$(9.8 m/s2)(1.00 m)(1−cos(15.0°))=12vmax2vmax2=2(0.333) m2/s2vmax=2(0.333) m2/s2=0.817 m/s$

Section 2:

To determine: The angular acceleration of the bob.

Answer: The angular acceleration of the bob is $2.54 rad/s2$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula for the moment of inertia of the pendulum is,

$I=mL2$

The equation for the conservation of energy is,

$Iα=mgLsinθ$

$α$ is the angular acceleration.

Substitute $mL2$ for $I$ in above expression and rearrange for $α$ .

$mL2α=mgLsinθα=mgLsinθmL2=gsinθL$

Substitute $9.8 m/s2$ for $g$ , $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $α$ .

$α=(9.8 m/s2)sin(15.0°)1.00 m=2.54 rad/s2$

Section 3:

To determine: The restoring force of the bob.

Answer: The restoring force of the bob is $0.634 N$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The force is maximum, when the angle is maximum.

The restoring force is calculated as,

$F=mgsinθ$

Substitute $15.0°$ for $θ$ , $0.250 kg$ for $m$ and $9.8 m/s2$ for $g$ in above equation to find $F$ .

$F=(0.250 kg)(9.8 m/s2)sin(15.0°)=0.634 N$

Conclusion:

Therefore, the maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

(e)

To determine

To compare: The answers of part (a), part (c) and part (d).

(e)

Expert Solution

Explanation of Solution

Introduction: The restoring force is defined as the force or torque that tends to restore a system to equilibrium after displacement.

The answers are closest but not exactly the same. The angular amplitude of $15.0°$ is not small, so the simple harmonic oscillation is not accurate. The answers computed from conservation of the energy and from Newton’s second law are more accurate.

Conclusion:

Therefore, the answers are closest but not exactly the same.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

Chapter 12, Problem 35P

(a)

To determine

The maximum speed of the bob.

(a)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.820 m/s$ .

Explanation of Solution

Section 1:

To determine: The amplitude of the motion.

Answer: The maximum speed of the amplitude of the motion is $0.262 m$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate amplitude is,

$A=Lθ$

$L$ is the length of pendulum.
$θ$ is the displaced angle.

Substitute $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $A$ .

$A=(1.00 m)(15.0°(π180°))=0.262 m$

Section 2:

To determine: The angular frequency of the motion.

Answer: The angular frequency of the motion is $3.13 rad/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate angular frequency is,

$ω=gL$

$g$ is the acceleration due to gravity.

Substitute $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $ω$ .

$ω=9.8 m/s21.00 m=3.13 rad/s$

Section 3:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.820 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum speed is,

$vmax=Aω$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $vmax$ .

$vmax=(0.262 m)(3.13 rad/s)=0.820 m/s$

Conclusion:

Therefore, the maximum speed of the bob is $0.820 m/s$ .

(b)

To determine

The maximum acceleration of the bob.

(b)

Expert Solution

Answer to Problem 35P

The maximum acceleration of the bob is $2.57 rad/s2$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum acceleration of the bob is,

$amax=Aω2$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $amax$ .

$amax=(0.262 m)(3.13 rad/s)2=2.57 rad/s2$

Conclusion:

Therefore, the maximum acceleration of the bob is $2.57 rad/s2$ .

(c)

To determine

The maximum restoring force of the bob.

(c)

Expert Solution

Answer to Problem 35P

The maximum restoring force of the bob is $0.641 N$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum restoring force of the bob is,

$F=mamax$

$m$ is the mass of the pendulum.

Substitute $Aω2$ for $amax$ in above equation.

$F=mAω2$

Substitute $0.250 kg$ for $m$ , $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $F$ .

$F=(0.250 kg)(0.262 m)(3.13 rad/s)2=0.641 N$

Conclusion:

Therefore, the maximum restoring force of the bob is $0.641 N$ .

(d)

To determine

The maximum speed, angular acceleration and restoring force of the bob using the model introduced earlier chapter.

(d)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Explanation of Solution

Section 1:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.817 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

Consider the figure given below.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 35P

In triangle $ABC$ ,

$cosθ=ACABAC=ABcosθ=Lcosθ$

The height of the bob is,

$h=AD−AC=L−Lcosθ=L(1−cosθ)$

The law of conservation of energy is,

$mgh=12mvmax2$

Substitute $L(1−cosθ)$ for $h$ in above expression.

$mgL(1−cosθ)=12mvmax2gL(1−cosθ)=12vmax2$

Substitute $15.0°$ for $θ$ , $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $vmax$ .

$(9.8 m/s2)(1.00 m)(1−cos(15.0°))=12vmax2vmax2=2(0.333) m2/s2vmax=2(0.333) m2/s2=0.817 m/s$

Section 2:

To determine: The angular acceleration of the bob.

Answer: The angular acceleration of the bob is $2.54 rad/s2$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula for the moment of inertia of the pendulum is,

$I=mL2$

The equation for the conservation of energy is,

$Iα=mgLsinθ$

$α$ is the angular acceleration.

Substitute $mL2$ for $I$ in above expression and rearrange for $α$ .

$mL2α=mgLsinθα=mgLsinθmL2=gsinθL$

Substitute $9.8 m/s2$ for $g$ , $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $α$ .

$α=(9.8 m/s2)sin(15.0°)1.00 m=2.54 rad/s2$

Section 3:

To determine: The restoring force of the bob.

Answer: The restoring force of the bob is $0.634 N$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The force is maximum, when the angle is maximum.

The restoring force is calculated as,

$F=mgsinθ$

Substitute $15.0°$ for $θ$ , $0.250 kg$ for $m$ and $9.8 m/s2$ for $g$ in above equation to find $F$ .

$F=(0.250 kg)(9.8 m/s2)sin(15.0°)=0.634 N$

Conclusion:

Therefore, the maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

(e)

To determine

To compare: The answers of part (a), part (c) and part (d).

(e)

Expert Solution

Explanation of Solution

Introduction: The restoring force is defined as the force or torque that tends to restore a system to equilibrium after displacement.

The answers are closest but not exactly the same. The angular amplitude of $15.0°$ is not small, so the simple harmonic oscillation is not accurate. The answers computed from conservation of the energy and from Newton’s second law are more accurate.

Conclusion:

Therefore, the answers are closest but not exactly the same.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

Chapter 12, Problem 35P

(a)

To determine

The maximum speed of the bob.

(a)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.820 m/s$ .

Explanation of Solution

Section 1:

To determine: The amplitude of the motion.

Answer: The maximum speed of the amplitude of the motion is $0.262 m$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate amplitude is,

$A=Lθ$

$L$ is the length of pendulum.
$θ$ is the displaced angle.

Substitute $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $A$ .

$A=(1.00 m)(15.0°(π180°))=0.262 m$

Section 2:

To determine: The angular frequency of the motion.

Answer: The angular frequency of the motion is $3.13 rad/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate angular frequency is,

$ω=gL$

$g$ is the acceleration due to gravity.

Substitute $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $ω$ .

$ω=9.8 m/s21.00 m=3.13 rad/s$

Section 3:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.820 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum speed is,

$vmax=Aω$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $vmax$ .

$vmax=(0.262 m)(3.13 rad/s)=0.820 m/s$

Conclusion:

Therefore, the maximum speed of the bob is $0.820 m/s$ .

(b)

To determine

The maximum acceleration of the bob.

(b)

Expert Solution

Answer to Problem 35P

The maximum acceleration of the bob is $2.57 rad/s2$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum acceleration of the bob is,

$amax=Aω2$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $amax$ .

$amax=(0.262 m)(3.13 rad/s)2=2.57 rad/s2$

Conclusion:

Therefore, the maximum acceleration of the bob is $2.57 rad/s2$ .

(c)

To determine

The maximum restoring force of the bob.

(c)

Expert Solution

Answer to Problem 35P

The maximum restoring force of the bob is $0.641 N$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum restoring force of the bob is,

$F=mamax$

$m$ is the mass of the pendulum.

Substitute $Aω2$ for $amax$ in above equation.

$F=mAω2$

Substitute $0.250 kg$ for $m$ , $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $F$ .

$F=(0.250 kg)(0.262 m)(3.13 rad/s)2=0.641 N$

Conclusion:

Therefore, the maximum restoring force of the bob is $0.641 N$ .

(d)

To determine

The maximum speed, angular acceleration and restoring force of the bob using the model introduced earlier chapter.

(d)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Explanation of Solution

Section 1:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.817 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

Consider the figure given below.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 35P

In triangle $ABC$ ,

$cosθ=ACABAC=ABcosθ=Lcosθ$

The height of the bob is,

$h=AD−AC=L−Lcosθ=L(1−cosθ)$

The law of conservation of energy is,

$mgh=12mvmax2$

Substitute $L(1−cosθ)$ for $h$ in above expression.

$mgL(1−cosθ)=12mvmax2gL(1−cosθ)=12vmax2$

Substitute $15.0°$ for $θ$ , $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $vmax$ .

$(9.8 m/s2)(1.00 m)(1−cos(15.0°))=12vmax2vmax2=2(0.333) m2/s2vmax=2(0.333) m2/s2=0.817 m/s$

Section 2:

To determine: The angular acceleration of the bob.

Answer: The angular acceleration of the bob is $2.54 rad/s2$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula for the moment of inertia of the pendulum is,

$I=mL2$

The equation for the conservation of energy is,

$Iα=mgLsinθ$

$α$ is the angular acceleration.

Substitute $mL2$ for $I$ in above expression and rearrange for $α$ .

$mL2α=mgLsinθα=mgLsinθmL2=gsinθL$

Substitute $9.8 m/s2$ for $g$ , $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $α$ .

$α=(9.8 m/s2)sin(15.0°)1.00 m=2.54 rad/s2$

Section 3:

To determine: The restoring force of the bob.

Answer: The restoring force of the bob is $0.634 N$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The force is maximum, when the angle is maximum.

The restoring force is calculated as,

$F=mgsinθ$

Substitute $15.0°$ for $θ$ , $0.250 kg$ for $m$ and $9.8 m/s2$ for $g$ in above equation to find $F$ .

$F=(0.250 kg)(9.8 m/s2)sin(15.0°)=0.634 N$

Conclusion:

Therefore, the maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

(e)

To determine

To compare: The answers of part (a), part (c) and part (d).

(e)

Expert Solution

Explanation of Solution

Introduction: The restoring force is defined as the force or torque that tends to restore a system to equilibrium after displacement.

The answers are closest but not exactly the same. The angular amplitude of $15.0°$ is not small, so the simple harmonic oscillation is not accurate. The answers computed from conservation of the energy and from Newton’s second law are more accurate.

Conclusion:

Therefore, the answers are closest but not exactly the same.

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Answer 5

Chapter 12, Problem 35P

(a)

To determine

The maximum speed of the bob.

(a)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.820 m/s$ .

Explanation of Solution

Section 1:

To determine: The amplitude of the motion.

Answer: The maximum speed of the amplitude of the motion is $0.262 m$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate amplitude is,

$A=Lθ$

$L$ is the length of pendulum.
$θ$ is the displaced angle.

Substitute $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $A$ .

$A=(1.00 m)(15.0°(π180°))=0.262 m$

Section 2:

To determine: The angular frequency of the motion.

Answer: The angular frequency of the motion is $3.13 rad/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate angular frequency is,

$ω=gL$

$g$ is the acceleration due to gravity.

Substitute $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $ω$ .

$ω=9.8 m/s21.00 m=3.13 rad/s$

Section 3:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.820 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum speed is,

$vmax=Aω$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $vmax$ .

$vmax=(0.262 m)(3.13 rad/s)=0.820 m/s$

Conclusion:

Therefore, the maximum speed of the bob is $0.820 m/s$ .

(b)

To determine

The maximum acceleration of the bob.

(b)

Expert Solution

Answer to Problem 35P

The maximum acceleration of the bob is $2.57 rad/s2$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum acceleration of the bob is,

$amax=Aω2$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $amax$ .

$amax=(0.262 m)(3.13 rad/s)2=2.57 rad/s2$

Conclusion:

Therefore, the maximum acceleration of the bob is $2.57 rad/s2$ .

(c)

To determine

The maximum restoring force of the bob.

(c)

Expert Solution

Answer to Problem 35P

The maximum restoring force of the bob is $0.641 N$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum restoring force of the bob is,

$F=mamax$

$m$ is the mass of the pendulum.

Substitute $Aω2$ for $amax$ in above equation.

$F=mAω2$

Substitute $0.250 kg$ for $m$ , $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $F$ .

$F=(0.250 kg)(0.262 m)(3.13 rad/s)2=0.641 N$

Conclusion:

Therefore, the maximum restoring force of the bob is $0.641 N$ .

(d)

To determine

The maximum speed, angular acceleration and restoring force of the bob using the model introduced earlier chapter.

(d)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Explanation of Solution

Section 1:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.817 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

Consider the figure given below.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 35P

In triangle $ABC$ ,

$cosθ=ACABAC=ABcosθ=Lcosθ$

The height of the bob is,

$h=AD−AC=L−Lcosθ=L(1−cosθ)$

The law of conservation of energy is,

$mgh=12mvmax2$

Substitute $L(1−cosθ)$ for $h$ in above expression.

$mgL(1−cosθ)=12mvmax2gL(1−cosθ)=12vmax2$

Substitute $15.0°$ for $θ$ , $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $vmax$ .

$(9.8 m/s2)(1.00 m)(1−cos(15.0°))=12vmax2vmax2=2(0.333) m2/s2vmax=2(0.333) m2/s2=0.817 m/s$

Section 2:

To determine: The angular acceleration of the bob.

Answer: The angular acceleration of the bob is $2.54 rad/s2$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula for the moment of inertia of the pendulum is,

$I=mL2$

The equation for the conservation of energy is,

$Iα=mgLsinθ$

$α$ is the angular acceleration.

Substitute $mL2$ for $I$ in above expression and rearrange for $α$ .

$mL2α=mgLsinθα=mgLsinθmL2=gsinθL$

Substitute $9.8 m/s2$ for $g$ , $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $α$ .

$α=(9.8 m/s2)sin(15.0°)1.00 m=2.54 rad/s2$

Section 3:

To determine: The restoring force of the bob.

Answer: The restoring force of the bob is $0.634 N$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The force is maximum, when the angle is maximum.

The restoring force is calculated as,

$F=mgsinθ$

Substitute $15.0°$ for $θ$ , $0.250 kg$ for $m$ and $9.8 m/s2$ for $g$ in above equation to find $F$ .

$F=(0.250 kg)(9.8 m/s2)sin(15.0°)=0.634 N$

Conclusion:

Therefore, the maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

(e)

To determine

To compare: The answers of part (a), part (c) and part (d).

(e)

Expert Solution

Explanation of Solution

Introduction: The restoring force is defined as the force or torque that tends to restore a system to equilibrium after displacement.

The answers are closest but not exactly the same. The angular amplitude of $15.0°$ is not small, so the simple harmonic oscillation is not accurate. The answers computed from conservation of the energy and from Newton’s second law are more accurate.

Conclusion:

Therefore, the answers are closest but not exactly the same.

Answer 6

Chapter 12, Problem 35P

(a)

To determine

The maximum speed of the bob.

(a)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.820 m/s$ .

Explanation of Solution

Section 1:

To determine: The amplitude of the motion.

Answer: The maximum speed of the amplitude of the motion is $0.262 m$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate amplitude is,

$A=Lθ$

$L$ is the length of pendulum.
$θ$ is the displaced angle.

Substitute $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $A$ .

$A=(1.00 m)(15.0°(π180°))=0.262 m$

Section 2:

To determine: The angular frequency of the motion.

Answer: The angular frequency of the motion is $3.13 rad/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate angular frequency is,

$ω=gL$

$g$ is the acceleration due to gravity.

Substitute $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $ω$ .

$ω=9.8 m/s21.00 m=3.13 rad/s$

Section 3:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.820 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum speed is,

$vmax=Aω$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $vmax$ .

$vmax=(0.262 m)(3.13 rad/s)=0.820 m/s$

Conclusion:

Therefore, the maximum speed of the bob is $0.820 m/s$ .

Answer 7

Chapter 12, Problem 35P

(a)

To determine

The maximum speed of the bob.

Answer 8

(a)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.820 m/s$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Section 1:

To determine: The amplitude of the motion.

Answer: The maximum speed of the amplitude of the motion is $0.262 m$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate amplitude is,

$A=Lθ$

$L$ is the length of pendulum.
$θ$ is the displaced angle.

Substitute $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $A$ .

$A=(1.00 m)(15.0°(π180°))=0.262 m$

Section 2:

To determine: The angular frequency of the motion.

Answer: The angular frequency of the motion is $3.13 rad/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate angular frequency is,

$ω=gL$

$g$ is the acceleration due to gravity.

Substitute $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $ω$ .

$ω=9.8 m/s21.00 m=3.13 rad/s$

Section 3:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.820 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum speed is,

$vmax=Aω$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $vmax$ .

$vmax=(0.262 m)(3.13 rad/s)=0.820 m/s$

Conclusion:

Therefore, the maximum speed of the bob is $0.820 m/s$ .

Answer 13

(b)

To determine

The maximum acceleration of the bob.

(b)

Expert Solution

Answer to Problem 35P

The maximum acceleration of the bob is $2.57 rad/s2$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum acceleration of the bob is,

$amax=Aω2$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $amax$ .

$amax=(0.262 m)(3.13 rad/s)2=2.57 rad/s2$

Conclusion:

Therefore, the maximum acceleration of the bob is $2.57 rad/s2$ .

Answer 14

(b)

To determine

The maximum acceleration of the bob.

Answer 15

(b)

Expert Solution

Answer to Problem 35P

The maximum acceleration of the bob is $2.57 rad/s2$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum acceleration of the bob is,

$amax=Aω2$

Substitute $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $amax$ .

$amax=(0.262 m)(3.13 rad/s)2=2.57 rad/s2$

Conclusion:

Therefore, the maximum acceleration of the bob is $2.57 rad/s2$ .

Answer 20

(c)

To determine

The maximum restoring force of the bob.

(c)

Expert Solution

Answer to Problem 35P

The maximum restoring force of the bob is $0.641 N$ .

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum restoring force of the bob is,

$F=mamax$

$m$ is the mass of the pendulum.

Substitute $Aω2$ for $amax$ in above equation.

$F=mAω2$

Substitute $0.250 kg$ for $m$ , $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $F$ .

$F=(0.250 kg)(0.262 m)(3.13 rad/s)2=0.641 N$

Conclusion:

Therefore, the maximum restoring force of the bob is $0.641 N$ .

Answer 21

(c)

To determine

The maximum restoring force of the bob.

Answer 22

(c)

Expert Solution

Answer to Problem 35P

The maximum restoring force of the bob is $0.641 N$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula to calculate maximum restoring force of the bob is,

$F=mamax$

$m$ is the mass of the pendulum.

Substitute $Aω2$ for $amax$ in above equation.

$F=mAω2$

Substitute $0.250 kg$ for $m$ , $0.262 m$ for $A$ and $3.13 rad/s$ for $ω$ in above equation to find $F$ .

$F=(0.250 kg)(0.262 m)(3.13 rad/s)2=0.641 N$

Conclusion:

Therefore, the maximum restoring force of the bob is $0.641 N$ .

Answer 27

(d)

To determine

The maximum speed, angular acceleration and restoring force of the bob using the model introduced earlier chapter.

(d)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Explanation of Solution

Section 1:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.817 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

Consider the figure given below.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 35P

In triangle $ABC$ ,

$cosθ=ACABAC=ABcosθ=Lcosθ$

The height of the bob is,

$h=AD−AC=L−Lcosθ=L(1−cosθ)$

The law of conservation of energy is,

$mgh=12mvmax2$

Substitute $L(1−cosθ)$ for $h$ in above expression.

$mgL(1−cosθ)=12mvmax2gL(1−cosθ)=12vmax2$

Substitute $15.0°$ for $θ$ , $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $vmax$ .

$(9.8 m/s2)(1.00 m)(1−cos(15.0°))=12vmax2vmax2=2(0.333) m2/s2vmax=2(0.333) m2/s2=0.817 m/s$

Section 2:

To determine: The angular acceleration of the bob.

Answer: The angular acceleration of the bob is $2.54 rad/s2$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula for the moment of inertia of the pendulum is,

$I=mL2$

The equation for the conservation of energy is,

$Iα=mgLsinθ$

$α$ is the angular acceleration.

Substitute $mL2$ for $I$ in above expression and rearrange for $α$ .

$mL2α=mgLsinθα=mgLsinθmL2=gsinθL$

Substitute $9.8 m/s2$ for $g$ , $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $α$ .

$α=(9.8 m/s2)sin(15.0°)1.00 m=2.54 rad/s2$

Section 3:

To determine: The restoring force of the bob.

Answer: The restoring force of the bob is $0.634 N$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The force is maximum, when the angle is maximum.

The restoring force is calculated as,

$F=mgsinθ$

Substitute $15.0°$ for $θ$ , $0.250 kg$ for $m$ and $9.8 m/s2$ for $g$ in above equation to find $F$ .

$F=(0.250 kg)(9.8 m/s2)sin(15.0°)=0.634 N$

Conclusion:

Therefore, the maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Answer 28

(d)

To determine

The maximum speed, angular acceleration and restoring force of the bob using the model introduced earlier chapter.

Answer 29

(d)

Expert Solution

Answer to Problem 35P

The maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Section 1:

To determine: The maximum speed of the bob.

Answer: The maximum speed of the bob is $0.817 m/s$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

Consider the figure given below.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 35P

In triangle $ABC$ ,

$cosθ=ACABAC=ABcosθ=Lcosθ$

The height of the bob is,

$h=AD−AC=L−Lcosθ=L(1−cosθ)$

The law of conservation of energy is,

$mgh=12mvmax2$

Substitute $L(1−cosθ)$ for $h$ in above expression.

$mgL(1−cosθ)=12mvmax2gL(1−cosθ)=12vmax2$

Substitute $15.0°$ for $θ$ , $1.00 m$ for $L$ and $9.8 m/s2$ for $g$ in above equation to find $vmax$ .

$(9.8 m/s2)(1.00 m)(1−cos(15.0°))=12vmax2vmax2=2(0.333) m2/s2vmax=2(0.333) m2/s2=0.817 m/s$

Section 2:

To determine: The angular acceleration of the bob.

Answer: The angular acceleration of the bob is $2.54 rad/s2$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The formula for the moment of inertia of the pendulum is,

$I=mL2$

The equation for the conservation of energy is,

$Iα=mgLsinθ$

$α$ is the angular acceleration.

Substitute $mL2$ for $I$ in above expression and rearrange for $α$ .

$mL2α=mgLsinθα=mgLsinθmL2=gsinθL$

Substitute $9.8 m/s2$ for $g$ , $1.00 m$ for $L$ and $15.0°$ for $θ$ in above equation to find $α$ .

$α=(9.8 m/s2)sin(15.0°)1.00 m=2.54 rad/s2$

Section 3:

To determine: The restoring force of the bob.

Answer: The restoring force of the bob is $0.634 N$ .

Given information:

The mass of pendulum is $0.250 kg$ , the length of the pendulum is $1.00 m$ and the displace angle is $15.0°$ .

The force is maximum, when the angle is maximum.

The restoring force is calculated as,

$F=mgsinθ$

Substitute $15.0°$ for $θ$ , $0.250 kg$ for $m$ and $9.8 m/s2$ for $g$ in above equation to find $F$ .

$F=(0.250 kg)(9.8 m/s2)sin(15.0°)=0.634 N$

Conclusion:

Therefore, the maximum speed of the bob is $0.817 m/s$ , the angular acceleration of the bob is $2.54 rad/s2$ and the restoring force of the bob is $0.634 N$ .

Answer 34

(e)

To determine

To compare: The answers of part (a), part (c) and part (d).

(e)

Expert Solution

Explanation of Solution

Introduction: The restoring force is defined as the force or torque that tends to restore a system to equilibrium after displacement.

The answers are closest but not exactly the same. The angular amplitude of $15.0°$ is not small, so the simple harmonic oscillation is not accurate. The answers computed from conservation of the energy and from Newton’s second law are more accurate.

Conclusion:

Therefore, the answers are closest but not exactly the same.

Answer 35

(e)

To determine

To compare: The answers of part (a), part (c) and part (d).

Answer 36

(e)

Expert Solution

Explanation of Solution

Introduction: The restoring force is defined as the force or torque that tends to restore a system to equilibrium after displacement.

The answers are closest but not exactly the same. The angular amplitude of $15.0°$ is not small, so the simple harmonic oscillation is not accurate. The answers computed from conservation of the energy and from Newton’s second law are more accurate.

Conclusion:

Therefore, the answers are closest but not exactly the same.