The moment of inertia of combination of platform-drum.

Question

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Answer 1

Question

Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

(a)

Expert Solution

Answer to Problem 82P

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance through which the object falls, $D=1.80 m$

Time, $t1=4.2 s$

Formula used:

Torque, $τ=Iα$

Where, I is the moment of inertia and $α$ is the angular acceleration.

From Newton’s second law of motion:

$∑Fi=miai$

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

$s=v0t+12at2$

Where, s is the displacement, t is the time, a is the acceleration and $v0$ is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

$∑τ=TR =I0α .......(1)$

Apply the Newton’s second law of motion to the weight:

$∑Fy=Mg−T =Ma .......(2)$

Relation between angular acceleration and acceleration is $α=aR$

Substituting for $α$ in equation $(1)$ then the tension in the string is

$T=I0aR2$

Now, substituting for the tension in equation $(2)$ , and solve for the moment of inertia,

$Mg−I0aR2=Ma I0=MR2( g−a)a=MR2(ga−1) ..........(3)$

The relation among the distance, acceleration and time is given by

$s=v0t+12at2$

The intial velocity $(v0)$ of the object is zero.

Substitute for the initial velocity $(v0)$ of the object and distance in the above equation:

$D=12at2a=2Dt2$

The moment of inertia of combinaiton of platform-drum is: $I0=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 4.20 s ) 2 2( 1.80 m )−1) =1.18 kg⋅m2 ≈1.2 kg⋅m2$

Conclusion:

The moment of inertia of combination of platform-drum is $1.2 kg/m2$ .

(b)

To determine

To Calculate: The total moment of inertia.

(b)

Expert Solution

Answer to Problem 82P

The total moment of inertia is $3.1 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance fall by the object, $D=1.80 m$

Time, $t2=6.8 s$

Formula used:

From the previous part:

$Itotal=MR2(gt22D−1)$

Calculation:

Substitute the values and solve for total moment of inertia:

$Itotal=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 6.80 s ) 2 2( 1.80 m )−1) =3.1 kg⋅m2$

Conclusion:

Total moment of inertia is $3.1 kg⋅m2$ .

(c)

To determine

To Calculate: The moment of inertia of the object.

(c)

Expert Solution

Answer to Problem 82P

The moment of inertia of the object is $1.9 kg⋅m2$ .

Explanation of Solution

Given data:

The total moment of inertia is $3.1 kg⋅m2$ .

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

$I=Itotal−I0$

Calculation:

Moment of inertia of the object, $I=Itotal−I0$

$=3.12 kg⋅m2−1.18 kg⋅m2=1.9 kg⋅m2$

Conclusion:

Moment of inertia of the object is $1.9 kg⋅m2$ .

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Answer 2

Question

Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

(a)

Expert Solution

Answer to Problem 82P

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance through which the object falls, $D=1.80 m$

Time, $t1=4.2 s$

Formula used:

Torque, $τ=Iα$

Where, I is the moment of inertia and $α$ is the angular acceleration.

From Newton’s second law of motion:

$∑Fi=miai$

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

$s=v0t+12at2$

Where, s is the displacement, t is the time, a is the acceleration and $v0$ is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

$∑τ=TR =I0α .......(1)$

Apply the Newton’s second law of motion to the weight:

$∑Fy=Mg−T =Ma .......(2)$

Relation between angular acceleration and acceleration is $α=aR$

Substituting for $α$ in equation $(1)$ then the tension in the string is

$T=I0aR2$

Now, substituting for the tension in equation $(2)$ , and solve for the moment of inertia,

$Mg−I0aR2=Ma I0=MR2( g−a)a=MR2(ga−1) ..........(3)$

The relation among the distance, acceleration and time is given by

$s=v0t+12at2$

The intial velocity $(v0)$ of the object is zero.

Substitute for the initial velocity $(v0)$ of the object and distance in the above equation:

$D=12at2a=2Dt2$

The moment of inertia of combinaiton of platform-drum is: $I0=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 4.20 s ) 2 2( 1.80 m )−1) =1.18 kg⋅m2 ≈1.2 kg⋅m2$

Conclusion:

The moment of inertia of combination of platform-drum is $1.2 kg/m2$ .

(b)

To determine

To Calculate: The total moment of inertia.

(b)

Expert Solution

Answer to Problem 82P

The total moment of inertia is $3.1 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance fall by the object, $D=1.80 m$

Time, $t2=6.8 s$

Formula used:

From the previous part:

$Itotal=MR2(gt22D−1)$

Calculation:

Substitute the values and solve for total moment of inertia:

$Itotal=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 6.80 s ) 2 2( 1.80 m )−1) =3.1 kg⋅m2$

Conclusion:

Total moment of inertia is $3.1 kg⋅m2$ .

(c)

To determine

To Calculate: The moment of inertia of the object.

(c)

Expert Solution

Answer to Problem 82P

The moment of inertia of the object is $1.9 kg⋅m2$ .

Explanation of Solution

Given data:

The total moment of inertia is $3.1 kg⋅m2$ .

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

$I=Itotal−I0$

Calculation:

Moment of inertia of the object, $I=Itotal−I0$

$=3.12 kg⋅m2−1.18 kg⋅m2=1.9 kg⋅m2$

Conclusion:

Moment of inertia of the object is $1.9 kg⋅m2$ .

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Answer 3

Question

Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

(a)

Expert Solution

Answer to Problem 82P

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance through which the object falls, $D=1.80 m$

Time, $t1=4.2 s$

Formula used:

Torque, $τ=Iα$

Where, I is the moment of inertia and $α$ is the angular acceleration.

From Newton’s second law of motion:

$∑Fi=miai$

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

$s=v0t+12at2$

Where, s is the displacement, t is the time, a is the acceleration and $v0$ is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

$∑τ=TR =I0α .......(1)$

Apply the Newton’s second law of motion to the weight:

$∑Fy=Mg−T =Ma .......(2)$

Relation between angular acceleration and acceleration is $α=aR$

Substituting for $α$ in equation $(1)$ then the tension in the string is

$T=I0aR2$

Now, substituting for the tension in equation $(2)$ , and solve for the moment of inertia,

$Mg−I0aR2=Ma I0=MR2( g−a)a=MR2(ga−1) ..........(3)$

The relation among the distance, acceleration and time is given by

$s=v0t+12at2$

The intial velocity $(v0)$ of the object is zero.

Substitute for the initial velocity $(v0)$ of the object and distance in the above equation:

$D=12at2a=2Dt2$

The moment of inertia of combinaiton of platform-drum is: $I0=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 4.20 s ) 2 2( 1.80 m )−1) =1.18 kg⋅m2 ≈1.2 kg⋅m2$

Conclusion:

The moment of inertia of combination of platform-drum is $1.2 kg/m2$ .

(b)

To determine

To Calculate: The total moment of inertia.

(b)

Expert Solution

Answer to Problem 82P

The total moment of inertia is $3.1 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance fall by the object, $D=1.80 m$

Time, $t2=6.8 s$

Formula used:

From the previous part:

$Itotal=MR2(gt22D−1)$

Calculation:

Substitute the values and solve for total moment of inertia:

$Itotal=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 6.80 s ) 2 2( 1.80 m )−1) =3.1 kg⋅m2$

Conclusion:

Total moment of inertia is $3.1 kg⋅m2$ .

(c)

To determine

To Calculate: The moment of inertia of the object.

(c)

Expert Solution

Answer to Problem 82P

The moment of inertia of the object is $1.9 kg⋅m2$ .

Explanation of Solution

Given data:

The total moment of inertia is $3.1 kg⋅m2$ .

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

$I=Itotal−I0$

Calculation:

Moment of inertia of the object, $I=Itotal−I0$

$=3.12 kg⋅m2−1.18 kg⋅m2=1.9 kg⋅m2$

Conclusion:

Moment of inertia of the object is $1.9 kg⋅m2$ .

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Answer 4

Question

Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

(a)

Expert Solution

Answer to Problem 82P

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance through which the object falls, $D=1.80 m$

Time, $t1=4.2 s$

Formula used:

Torque, $τ=Iα$

Where, I is the moment of inertia and $α$ is the angular acceleration.

From Newton’s second law of motion:

$∑Fi=miai$

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

$s=v0t+12at2$

Where, s is the displacement, t is the time, a is the acceleration and $v0$ is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

$∑τ=TR =I0α .......(1)$

Apply the Newton’s second law of motion to the weight:

$∑Fy=Mg−T =Ma .......(2)$

Relation between angular acceleration and acceleration is $α=aR$

Substituting for $α$ in equation $(1)$ then the tension in the string is

$T=I0aR2$

Now, substituting for the tension in equation $(2)$ , and solve for the moment of inertia,

$Mg−I0aR2=Ma I0=MR2( g−a)a=MR2(ga−1) ..........(3)$

The relation among the distance, acceleration and time is given by

$s=v0t+12at2$

The intial velocity $(v0)$ of the object is zero.

Substitute for the initial velocity $(v0)$ of the object and distance in the above equation:

$D=12at2a=2Dt2$

The moment of inertia of combinaiton of platform-drum is: $I0=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 4.20 s ) 2 2( 1.80 m )−1) =1.18 kg⋅m2 ≈1.2 kg⋅m2$

Conclusion:

The moment of inertia of combination of platform-drum is $1.2 kg/m2$ .

(b)

To determine

To Calculate: The total moment of inertia.

(b)

Expert Solution

Answer to Problem 82P

The total moment of inertia is $3.1 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance fall by the object, $D=1.80 m$

Time, $t2=6.8 s$

Formula used:

From the previous part:

$Itotal=MR2(gt22D−1)$

Calculation:

Substitute the values and solve for total moment of inertia:

$Itotal=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 6.80 s ) 2 2( 1.80 m )−1) =3.1 kg⋅m2$

Conclusion:

Total moment of inertia is $3.1 kg⋅m2$ .

(c)

To determine

To Calculate: The moment of inertia of the object.

(c)

Expert Solution

Answer to Problem 82P

The moment of inertia of the object is $1.9 kg⋅m2$ .

Explanation of Solution

Given data:

The total moment of inertia is $3.1 kg⋅m2$ .

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

$I=Itotal−I0$

Calculation:

Moment of inertia of the object, $I=Itotal−I0$

$=3.12 kg⋅m2−1.18 kg⋅m2=1.9 kg⋅m2$

Conclusion:

Moment of inertia of the object is $1.9 kg⋅m2$ .

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Answer 5

Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

(a)

Expert Solution

Answer to Problem 82P

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance through which the object falls, $D=1.80 m$

Time, $t1=4.2 s$

Formula used:

Torque, $τ=Iα$

Where, I is the moment of inertia and $α$ is the angular acceleration.

From Newton’s second law of motion:

$∑Fi=miai$

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

$s=v0t+12at2$

Where, s is the displacement, t is the time, a is the acceleration and $v0$ is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

$∑τ=TR =I0α .......(1)$

Apply the Newton’s second law of motion to the weight:

$∑Fy=Mg−T =Ma .......(2)$

Relation between angular acceleration and acceleration is $α=aR$

Substituting for $α$ in equation $(1)$ then the tension in the string is

$T=I0aR2$

Now, substituting for the tension in equation $(2)$ , and solve for the moment of inertia,

$Mg−I0aR2=Ma I0=MR2( g−a)a=MR2(ga−1) ..........(3)$

The relation among the distance, acceleration and time is given by

$s=v0t+12at2$

The intial velocity $(v0)$ of the object is zero.

Substitute for the initial velocity $(v0)$ of the object and distance in the above equation:

$D=12at2a=2Dt2$

The moment of inertia of combinaiton of platform-drum is: $I0=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 4.20 s ) 2 2( 1.80 m )−1) =1.18 kg⋅m2 ≈1.2 kg⋅m2$

Conclusion:

The moment of inertia of combination of platform-drum is $1.2 kg/m2$ .

(b)

To determine

To Calculate: The total moment of inertia.

(b)

Expert Solution

Answer to Problem 82P

The total moment of inertia is $3.1 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance fall by the object, $D=1.80 m$

Time, $t2=6.8 s$

Formula used:

From the previous part:

$Itotal=MR2(gt22D−1)$

Calculation:

Substitute the values and solve for total moment of inertia:

$Itotal=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 6.80 s ) 2 2( 1.80 m )−1) =3.1 kg⋅m2$

Conclusion:

Total moment of inertia is $3.1 kg⋅m2$ .

(c)

To determine

To Calculate: The moment of inertia of the object.

(c)

Expert Solution

Answer to Problem 82P

The moment of inertia of the object is $1.9 kg⋅m2$ .

Explanation of Solution

Given data:

The total moment of inertia is $3.1 kg⋅m2$ .

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

$I=Itotal−I0$

Calculation:

Moment of inertia of the object, $I=Itotal−I0$

$=3.12 kg⋅m2−1.18 kg⋅m2=1.9 kg⋅m2$

Conclusion:

Moment of inertia of the object is $1.9 kg⋅m2$ .

Answer 6

Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

(a)

Expert Solution

Answer to Problem 82P

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance through which the object falls, $D=1.80 m$

Time, $t1=4.2 s$

Formula used:

Torque, $τ=Iα$

Where, I is the moment of inertia and $α$ is the angular acceleration.

From Newton’s second law of motion:

$∑Fi=miai$

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

$s=v0t+12at2$

Where, s is the displacement, t is the time, a is the acceleration and $v0$ is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

$∑τ=TR =I0α .......(1)$

Apply the Newton’s second law of motion to the weight:

$∑Fy=Mg−T =Ma .......(2)$

Relation between angular acceleration and acceleration is $α=aR$

Substituting for $α$ in equation $(1)$ then the tension in the string is

$T=I0aR2$

Now, substituting for the tension in equation $(2)$ , and solve for the moment of inertia,

$Mg−I0aR2=Ma I0=MR2( g−a)a=MR2(ga−1) ..........(3)$

The relation among the distance, acceleration and time is given by

$s=v0t+12at2$

The intial velocity $(v0)$ of the object is zero.

Substitute for the initial velocity $(v0)$ of the object and distance in the above equation:

$D=12at2a=2Dt2$

The moment of inertia of combinaiton of platform-drum is: $I0=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 4.20 s ) 2 2( 1.80 m )−1) =1.18 kg⋅m2 ≈1.2 kg⋅m2$

Conclusion:

The moment of inertia of combination of platform-drum is $1.2 kg/m2$ .

Answer 7

Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

Answer 8

(a)

Expert Solution

Answer to Problem 82P

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance through which the object falls, $D=1.80 m$

Time, $t1=4.2 s$

Formula used:

Torque, $τ=Iα$

Where, I is the moment of inertia and $α$ is the angular acceleration.

From Newton’s second law of motion:

$∑Fi=miai$

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

$s=v0t+12at2$

Where, s is the displacement, t is the time, a is the acceleration and $v0$ is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

$∑τ=TR =I0α .......(1)$

Apply the Newton’s second law of motion to the weight:

$∑Fy=Mg−T =Ma .......(2)$

Relation between angular acceleration and acceleration is $α=aR$

Substituting for $α$ in equation $(1)$ then the tension in the string is

$T=I0aR2$

Now, substituting for the tension in equation $(2)$ , and solve for the moment of inertia,

$Mg−I0aR2=Ma I0=MR2( g−a)a=MR2(ga−1) ..........(3)$

The relation among the distance, acceleration and time is given by

$s=v0t+12at2$

The intial velocity $(v0)$ of the object is zero.

Substitute for the initial velocity $(v0)$ of the object and distance in the above equation:

$D=12at2a=2Dt2$

The moment of inertia of combinaiton of platform-drum is: $I0=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 4.20 s ) 2 2( 1.80 m )−1) =1.18 kg⋅m2 ≈1.2 kg⋅m2$

Conclusion:

The moment of inertia of combination of platform-drum is $1.2 kg/m2$ .

Answer 13

(b)

To determine

To Calculate: The total moment of inertia.

(b)

Expert Solution

Answer to Problem 82P

The total moment of inertia is $3.1 kg⋅m2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance fall by the object, $D=1.80 m$

Time, $t2=6.8 s$

Formula used:

From the previous part:

$Itotal=MR2(gt22D−1)$

Calculation:

Substitute the values and solve for total moment of inertia:

$Itotal=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 6.80 s ) 2 2( 1.80 m )−1) =3.1 kg⋅m2$

Conclusion:

Total moment of inertia is $3.1 kg⋅m2$ .

Answer 14

(b)

To determine

To Calculate: The total moment of inertia.

Answer 15

(b)

Expert Solution

Answer to Problem 82P

The total moment of inertia is $3.1 kg⋅m2$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10 cm$

$=(10 cm)( 1 m 100 cm)=0.100 m$

Mass of the hanging object, $M=2.50 kg$

Distance fall by the object, $D=1.80 m$

Time, $t2=6.8 s$

Formula used:

From the previous part:

$Itotal=MR2(gt22D−1)$

Calculation:

Substitute the values and solve for total moment of inertia:

$Itotal=MR2( g t 2 2D−1) =(2.50 kg)(0.100 m)2( ( 9.80 m/s 2 ) ( 6.80 s ) 2 2( 1.80 m )−1) =3.1 kg⋅m2$

Conclusion:

Total moment of inertia is $3.1 kg⋅m2$ .

Answer 20

(c)

To determine

To Calculate: The moment of inertia of the object.

(c)

Expert Solution

Answer to Problem 82P

The moment of inertia of the object is $1.9 kg⋅m2$ .

Explanation of Solution

Given data:

The total moment of inertia is $3.1 kg⋅m2$ .

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

$I=Itotal−I0$

Calculation:

Moment of inertia of the object, $I=Itotal−I0$

$=3.12 kg⋅m2−1.18 kg⋅m2=1.9 kg⋅m2$

Conclusion:

Moment of inertia of the object is $1.9 kg⋅m2$ .

Answer 21

(c)

To determine

To Calculate: The moment of inertia of the object.

Answer 22

(c)

Expert Solution

Answer to Problem 82P

The moment of inertia of the object is $1.9 kg⋅m2$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given data:

The total moment of inertia is $3.1 kg⋅m2$ .

The moment of inertia of combination of platform-drum is $1.2 kg⋅m2$ .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

$I=Itotal−I0$

Calculation:

Moment of inertia of the object, $I=Itotal−I0$

$=3.12 kg⋅m2−1.18 kg⋅m2=1.9 kg⋅m2$

Conclusion:

Moment of inertia of the object is $1.9 kg⋅m2$ .

Answer 27

Ch. 9 - Prob. 1P Ch. 9 - Prob. 2P Ch. 9 - Prob. 3P Ch. 9 - Prob. 4P Ch. 9 - Prob. 5P Ch. 9 - Prob. 6P Ch. 9 - Prob. 7P Ch. 9 - Prob. 8P Ch. 9 - Prob. 9P Ch. 9 - Prob. 10P

The moment of inertia of combination of platform-drum.

Concept explainers

Videos

To Calculate: The moment of inertia of combination of platform-drum.

Answer to Problem 82P

Explanation of Solution

To Calculate: The total moment of inertia.

Answer to Problem 82P

Explanation of Solution

To Calculate: The moment of inertia of the object.

Answer to Problem 82P

Explanation of Solution

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