A single bead can slide with negligible friction on a stiff wire that has been bent into a circular loop of radius 15.0 cm as shown in Figure P6.48. The circle is always in a vertical plane and rotates steadily about its vertical diameter with a period of 0.450 s. The position of the bead is described by the angle θ that the radial line, from the center of the loop to the bead, makes with the vertical. (a) At what angle up from the bottom of the circle can the bead slay motionless relative to the turning circle? (b) What If? Repeat the problem, this time taking the period of the circle’s rotation as 0.850 s. (c) Describe how the solution to part (b) is different from the solution to part (a). (d) For any period or loop size, is there always an angle at which the bead can stand still relative to the loop? (e) Are there ever more than two angles? Arnold Arons suggested the idea for this problem. Figure P6.48

Question

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Answer 1

Textbook Question

Chapter 6, Problem 68CP

A single bead can slide with negligible friction on a stiff wire that has been bent into a circular loop of radius 15.0 cm as shown in Figure P6.48. The circle is always in a vertical plane and rotates steadily about its vertical diameter with a period of 0.450 s. The position of the bead is described by the angle θ that the radial line, from the center of the loop to the bead, makes with the vertical. (a) At what angle up from the bottom of the circle can the bead slay motionless relative to the turning circle? (b) What If? Repeat the problem, this time taking the period of the circle’s rotation as 0.850 s. (c) Describe how the solution to part (b) is different from the solution to part (a). (d) For any period or loop size, is there always an angle at which the bead can stand still relative to the loop? (e) Are there ever more than two angles? Arnold Arons suggested the idea for this problem.

Figure P6.48

Chapter 6, Problem 68CP, A single bead can slide with negligible friction on a stiff wire that has been bent into a circular

(a)

Expert Solution

To determine

The angle from the bottom of the circle for which the bead can stay motionless.

Answer to Problem 68CP

The angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

The rough sketch of the force body diagram of the situation is shown below,

Physics for Scientists and Engineers With Modern Physics, Chapter 6, Problem 68CP

Figure (1)

The bead moves in a circle,

$r=Rsinθ$

Here,

$R$ is the radius of the hoop.

$r$ is the radius of the circle in which the bead moves.

$θ$ is the angle between the vertical and the radial line from the centre of the loop.

The speed of the bead is,

$v=2πrT$

Here,

$T$ is the time period of the rotation.

From the figure (1) the net force in $y$ direction is,

$Ncosθ−mg=0N=mgcosθ$

Here,

$N$ is the normal force acting on the bead.

$m$ is the amass of the bead.

$g$ is the acceleration due to gravity.

The net force in $x$ direction is,

$Nsinθ=mv2r$

Substitute $2πrT$ for $v$ in the above equation.

$Nsinθ=m[(2πrT)2r]$

Substitute $Rsinθ$ for $r$ and $mgcosθ$ for $N$ in the above equation.

$mgcosθsinθ=m[(2πRsinθT)2Rsinθ]gsinθcosθ−4π2RsinθT2=0sinθ(gcosθ−4π2RT2)=0$ . (1)

The two possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ−4π2RT2=0gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.450 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.450 s)24π2(15 cm×1 m100 cm)=0.335θ=cos-1(0.335)=70.42°$

Conclusion:

Therefore, the angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

(b)

Expert Solution

To determine

The angles from the bottom of the circle for which the bead can stay motionless for the time period as $0.850 s$ .

Answer to Problem 68CP

The only one possible angle from the bottom of the circle for which the bead can stay motionless for time period $0.850 s$ is $0.0°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

Form the part (a) equation (1).

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.850 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.850 s)24π2(15 cm×1 m100 cm)=1.19θ=cos−1(1.19)$

The above value is not possible.

Conclusion:

Therefore, the only one possible angle from the bottom of the circle for which the bead can stay motionless at time period $0.850 s$ is $0.0°$ .

(c)

Expert Solution

To determine

The difference between the solution of the part (a) and part (b).

Answer to Problem 68CP

The part (b) has only one solution as the time period is very large.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2Rcosθ∝T2$

As the second solution depends directly on the square of the amplitude of the time period so, as the time period increases the value of $cosθ$ increases to value greater than $1$ which is not possible as the range of the cosine function is $−1≤cosθ≤+1$ .

Conclusion:

Therefore, the part (b) has only one solution as the time period is very large.

(d)

Expert Solution

To determine

The angle at which the bead can stand still relative to the loop.

Answer to Problem 68CP

The angle for which condition $gT2<4π2R$ and $θ=0°$ is followed the angles are possible for the beads to stand still.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

As the range of the cosine function is $−1≤cosθ≤+1$ , the solution to occur the condition is to be followed is,

$gT24π2R<1gT2<4π2R$

So, the value of the cosine of the angle is always less than $1$ .

Conclusion:

Therefore, for the condition $gT2<4π2R$ and $θ=0°$ the angles are possible for the beads to stand still.

(e)

Expert Solution

To determine

Whether there are more than two angles.

Answer to Problem 68CP

The number of possible angles are $2$ for a given time period.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Form the above expression the second solution depends on time period, acceleration due to gravity and the radius of the hoop and for a given case the parameters has only one value so the second solution has only one answer.

Conclusion:

Therefore, the number of possible angles are $2$ for a given time period.

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Answer 2

Textbook Question

Chapter 6, Problem 68CP

A single bead can slide with negligible friction on a stiff wire that has been bent into a circular loop of radius 15.0 cm as shown in Figure P6.48. The circle is always in a vertical plane and rotates steadily about its vertical diameter with a period of 0.450 s. The position of the bead is described by the angle θ that the radial line, from the center of the loop to the bead, makes with the vertical. (a) At what angle up from the bottom of the circle can the bead slay motionless relative to the turning circle? (b) What If? Repeat the problem, this time taking the period of the circle’s rotation as 0.850 s. (c) Describe how the solution to part (b) is different from the solution to part (a). (d) For any period or loop size, is there always an angle at which the bead can stand still relative to the loop? (e) Are there ever more than two angles? Arnold Arons suggested the idea for this problem.

Figure P6.48

Chapter 6, Problem 68CP, A single bead can slide with negligible friction on a stiff wire that has been bent into a circular

(a)

Expert Solution

To determine

The angle from the bottom of the circle for which the bead can stay motionless.

Answer to Problem 68CP

The angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

The rough sketch of the force body diagram of the situation is shown below,

Physics for Scientists and Engineers With Modern Physics, Chapter 6, Problem 68CP

Figure (1)

The bead moves in a circle,

$r=Rsinθ$

Here,

$R$ is the radius of the hoop.

$r$ is the radius of the circle in which the bead moves.

$θ$ is the angle between the vertical and the radial line from the centre of the loop.

The speed of the bead is,

$v=2πrT$

Here,

$T$ is the time period of the rotation.

From the figure (1) the net force in $y$ direction is,

$Ncosθ−mg=0N=mgcosθ$

Here,

$N$ is the normal force acting on the bead.

$m$ is the amass of the bead.

$g$ is the acceleration due to gravity.

The net force in $x$ direction is,

$Nsinθ=mv2r$

Substitute $2πrT$ for $v$ in the above equation.

$Nsinθ=m[(2πrT)2r]$

Substitute $Rsinθ$ for $r$ and $mgcosθ$ for $N$ in the above equation.

$mgcosθsinθ=m[(2πRsinθT)2Rsinθ]gsinθcosθ−4π2RsinθT2=0sinθ(gcosθ−4π2RT2)=0$ . (1)

The two possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ−4π2RT2=0gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.450 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.450 s)24π2(15 cm×1 m100 cm)=0.335θ=cos-1(0.335)=70.42°$

Conclusion:

Therefore, the angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

(b)

Expert Solution

To determine

The angles from the bottom of the circle for which the bead can stay motionless for the time period as $0.850 s$ .

Answer to Problem 68CP

The only one possible angle from the bottom of the circle for which the bead can stay motionless for time period $0.850 s$ is $0.0°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

Form the part (a) equation (1).

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.850 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.850 s)24π2(15 cm×1 m100 cm)=1.19θ=cos−1(1.19)$

The above value is not possible.

Conclusion:

Therefore, the only one possible angle from the bottom of the circle for which the bead can stay motionless at time period $0.850 s$ is $0.0°$ .

(c)

Expert Solution

To determine

The difference between the solution of the part (a) and part (b).

Answer to Problem 68CP

The part (b) has only one solution as the time period is very large.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2Rcosθ∝T2$

As the second solution depends directly on the square of the amplitude of the time period so, as the time period increases the value of $cosθ$ increases to value greater than $1$ which is not possible as the range of the cosine function is $−1≤cosθ≤+1$ .

Conclusion:

Therefore, the part (b) has only one solution as the time period is very large.

(d)

Expert Solution

To determine

The angle at which the bead can stand still relative to the loop.

Answer to Problem 68CP

The angle for which condition $gT2<4π2R$ and $θ=0°$ is followed the angles are possible for the beads to stand still.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

As the range of the cosine function is $−1≤cosθ≤+1$ , the solution to occur the condition is to be followed is,

$gT24π2R<1gT2<4π2R$

So, the value of the cosine of the angle is always less than $1$ .

Conclusion:

Therefore, for the condition $gT2<4π2R$ and $θ=0°$ the angles are possible for the beads to stand still.

(e)

Expert Solution

To determine

Whether there are more than two angles.

Answer to Problem 68CP

The number of possible angles are $2$ for a given time period.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Form the above expression the second solution depends on time period, acceleration due to gravity and the radius of the hoop and for a given case the parameters has only one value so the second solution has only one answer.

Conclusion:

Therefore, the number of possible angles are $2$ for a given time period.

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Answer 3

Textbook Question

Chapter 6, Problem 68CP

A single bead can slide with negligible friction on a stiff wire that has been bent into a circular loop of radius 15.0 cm as shown in Figure P6.48. The circle is always in a vertical plane and rotates steadily about its vertical diameter with a period of 0.450 s. The position of the bead is described by the angle θ that the radial line, from the center of the loop to the bead, makes with the vertical. (a) At what angle up from the bottom of the circle can the bead slay motionless relative to the turning circle? (b) What If? Repeat the problem, this time taking the period of the circle’s rotation as 0.850 s. (c) Describe how the solution to part (b) is different from the solution to part (a). (d) For any period or loop size, is there always an angle at which the bead can stand still relative to the loop? (e) Are there ever more than two angles? Arnold Arons suggested the idea for this problem.

Figure P6.48

Chapter 6, Problem 68CP, A single bead can slide with negligible friction on a stiff wire that has been bent into a circular

(a)

Expert Solution

To determine

The angle from the bottom of the circle for which the bead can stay motionless.

Answer to Problem 68CP

The angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

The rough sketch of the force body diagram of the situation is shown below,

Physics for Scientists and Engineers With Modern Physics, Chapter 6, Problem 68CP

Figure (1)

The bead moves in a circle,

$r=Rsinθ$

Here,

$R$ is the radius of the hoop.

$r$ is the radius of the circle in which the bead moves.

$θ$ is the angle between the vertical and the radial line from the centre of the loop.

The speed of the bead is,

$v=2πrT$

Here,

$T$ is the time period of the rotation.

From the figure (1) the net force in $y$ direction is,

$Ncosθ−mg=0N=mgcosθ$

Here,

$N$ is the normal force acting on the bead.

$m$ is the amass of the bead.

$g$ is the acceleration due to gravity.

The net force in $x$ direction is,

$Nsinθ=mv2r$

Substitute $2πrT$ for $v$ in the above equation.

$Nsinθ=m[(2πrT)2r]$

Substitute $Rsinθ$ for $r$ and $mgcosθ$ for $N$ in the above equation.

$mgcosθsinθ=m[(2πRsinθT)2Rsinθ]gsinθcosθ−4π2RsinθT2=0sinθ(gcosθ−4π2RT2)=0$ . (1)

The two possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ−4π2RT2=0gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.450 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.450 s)24π2(15 cm×1 m100 cm)=0.335θ=cos-1(0.335)=70.42°$

Conclusion:

Therefore, the angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

(b)

Expert Solution

To determine

The angles from the bottom of the circle for which the bead can stay motionless for the time period as $0.850 s$ .

Answer to Problem 68CP

The only one possible angle from the bottom of the circle for which the bead can stay motionless for time period $0.850 s$ is $0.0°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

Form the part (a) equation (1).

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.850 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.850 s)24π2(15 cm×1 m100 cm)=1.19θ=cos−1(1.19)$

The above value is not possible.

Conclusion:

Therefore, the only one possible angle from the bottom of the circle for which the bead can stay motionless at time period $0.850 s$ is $0.0°$ .

(c)

Expert Solution

To determine

The difference between the solution of the part (a) and part (b).

Answer to Problem 68CP

The part (b) has only one solution as the time period is very large.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2Rcosθ∝T2$

As the second solution depends directly on the square of the amplitude of the time period so, as the time period increases the value of $cosθ$ increases to value greater than $1$ which is not possible as the range of the cosine function is $−1≤cosθ≤+1$ .

Conclusion:

Therefore, the part (b) has only one solution as the time period is very large.

(d)

Expert Solution

To determine

The angle at which the bead can stand still relative to the loop.

Answer to Problem 68CP

The angle for which condition $gT2<4π2R$ and $θ=0°$ is followed the angles are possible for the beads to stand still.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

As the range of the cosine function is $−1≤cosθ≤+1$ , the solution to occur the condition is to be followed is,

$gT24π2R<1gT2<4π2R$

So, the value of the cosine of the angle is always less than $1$ .

Conclusion:

Therefore, for the condition $gT2<4π2R$ and $θ=0°$ the angles are possible for the beads to stand still.

(e)

Expert Solution

To determine

Whether there are more than two angles.

Answer to Problem 68CP

The number of possible angles are $2$ for a given time period.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Form the above expression the second solution depends on time period, acceleration due to gravity and the radius of the hoop and for a given case the parameters has only one value so the second solution has only one answer.

Conclusion:

Therefore, the number of possible angles are $2$ for a given time period.

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Answer 4

Textbook Question

Chapter 6, Problem 68CP

A single bead can slide with negligible friction on a stiff wire that has been bent into a circular loop of radius 15.0 cm as shown in Figure P6.48. The circle is always in a vertical plane and rotates steadily about its vertical diameter with a period of 0.450 s. The position of the bead is described by the angle θ that the radial line, from the center of the loop to the bead, makes with the vertical. (a) At what angle up from the bottom of the circle can the bead slay motionless relative to the turning circle? (b) What If? Repeat the problem, this time taking the period of the circle’s rotation as 0.850 s. (c) Describe how the solution to part (b) is different from the solution to part (a). (d) For any period or loop size, is there always an angle at which the bead can stand still relative to the loop? (e) Are there ever more than two angles? Arnold Arons suggested the idea for this problem.

Figure P6.48

Chapter 6, Problem 68CP, A single bead can slide with negligible friction on a stiff wire that has been bent into a circular

(a)

Expert Solution

To determine

The angle from the bottom of the circle for which the bead can stay motionless.

Answer to Problem 68CP

The angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

The rough sketch of the force body diagram of the situation is shown below,

Physics for Scientists and Engineers With Modern Physics, Chapter 6, Problem 68CP

Figure (1)

The bead moves in a circle,

$r=Rsinθ$

Here,

$R$ is the radius of the hoop.

$r$ is the radius of the circle in which the bead moves.

$θ$ is the angle between the vertical and the radial line from the centre of the loop.

The speed of the bead is,

$v=2πrT$

Here,

$T$ is the time period of the rotation.

From the figure (1) the net force in $y$ direction is,

$Ncosθ−mg=0N=mgcosθ$

Here,

$N$ is the normal force acting on the bead.

$m$ is the amass of the bead.

$g$ is the acceleration due to gravity.

The net force in $x$ direction is,

$Nsinθ=mv2r$

Substitute $2πrT$ for $v$ in the above equation.

$Nsinθ=m[(2πrT)2r]$

Substitute $Rsinθ$ for $r$ and $mgcosθ$ for $N$ in the above equation.

$mgcosθsinθ=m[(2πRsinθT)2Rsinθ]gsinθcosθ−4π2RsinθT2=0sinθ(gcosθ−4π2RT2)=0$ . (1)

The two possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ−4π2RT2=0gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.450 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.450 s)24π2(15 cm×1 m100 cm)=0.335θ=cos-1(0.335)=70.42°$

Conclusion:

Therefore, the angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

(b)

Expert Solution

To determine

The angles from the bottom of the circle for which the bead can stay motionless for the time period as $0.850 s$ .

Answer to Problem 68CP

The only one possible angle from the bottom of the circle for which the bead can stay motionless for time period $0.850 s$ is $0.0°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

Form the part (a) equation (1).

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.850 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.850 s)24π2(15 cm×1 m100 cm)=1.19θ=cos−1(1.19)$

The above value is not possible.

Conclusion:

Therefore, the only one possible angle from the bottom of the circle for which the bead can stay motionless at time period $0.850 s$ is $0.0°$ .

(c)

Expert Solution

To determine

The difference between the solution of the part (a) and part (b).

Answer to Problem 68CP

The part (b) has only one solution as the time period is very large.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2Rcosθ∝T2$

As the second solution depends directly on the square of the amplitude of the time period so, as the time period increases the value of $cosθ$ increases to value greater than $1$ which is not possible as the range of the cosine function is $−1≤cosθ≤+1$ .

Conclusion:

Therefore, the part (b) has only one solution as the time period is very large.

(d)

Expert Solution

To determine

The angle at which the bead can stand still relative to the loop.

Answer to Problem 68CP

The angle for which condition $gT2<4π2R$ and $θ=0°$ is followed the angles are possible for the beads to stand still.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

As the range of the cosine function is $−1≤cosθ≤+1$ , the solution to occur the condition is to be followed is,

$gT24π2R<1gT2<4π2R$

So, the value of the cosine of the angle is always less than $1$ .

Conclusion:

Therefore, for the condition $gT2<4π2R$ and $θ=0°$ the angles are possible for the beads to stand still.

(e)

Expert Solution

To determine

Whether there are more than two angles.

Answer to Problem 68CP

The number of possible angles are $2$ for a given time period.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Form the above expression the second solution depends on time period, acceleration due to gravity and the radius of the hoop and for a given case the parameters has only one value so the second solution has only one answer.

Conclusion:

Therefore, the number of possible angles are $2$ for a given time period.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The angle from the bottom of the circle for which the bead can stay motionless.

Answer to Problem 68CP

The angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

The rough sketch of the force body diagram of the situation is shown below,

Physics for Scientists and Engineers With Modern Physics, Chapter 6, Problem 68CP

Figure (1)

The bead moves in a circle,

$r=Rsinθ$

Here,

$R$ is the radius of the hoop.

$r$ is the radius of the circle in which the bead moves.

$θ$ is the angle between the vertical and the radial line from the centre of the loop.

The speed of the bead is,

$v=2πrT$

Here,

$T$ is the time period of the rotation.

From the figure (1) the net force in $y$ direction is,

$Ncosθ−mg=0N=mgcosθ$

Here,

$N$ is the normal force acting on the bead.

$m$ is the amass of the bead.

$g$ is the acceleration due to gravity.

The net force in $x$ direction is,

$Nsinθ=mv2r$

Substitute $2πrT$ for $v$ in the above equation.

$Nsinθ=m[(2πrT)2r]$

Substitute $Rsinθ$ for $r$ and $mgcosθ$ for $N$ in the above equation.

$mgcosθsinθ=m[(2πRsinθT)2Rsinθ]gsinθcosθ−4π2RsinθT2=0sinθ(gcosθ−4π2RT2)=0$ . (1)

The two possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ−4π2RT2=0gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.450 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.450 s)24π2(15 cm×1 m100 cm)=0.335θ=cos-1(0.335)=70.42°$

Conclusion:

Therefore, the angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

(b)

Expert Solution

To determine

The angles from the bottom of the circle for which the bead can stay motionless for the time period as $0.850 s$ .

Answer to Problem 68CP

The only one possible angle from the bottom of the circle for which the bead can stay motionless for time period $0.850 s$ is $0.0°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

Form the part (a) equation (1).

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.850 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.850 s)24π2(15 cm×1 m100 cm)=1.19θ=cos−1(1.19)$

The above value is not possible.

Conclusion:

Therefore, the only one possible angle from the bottom of the circle for which the bead can stay motionless at time period $0.850 s$ is $0.0°$ .

(c)

Expert Solution

To determine

The difference between the solution of the part (a) and part (b).

Answer to Problem 68CP

The part (b) has only one solution as the time period is very large.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2Rcosθ∝T2$

As the second solution depends directly on the square of the amplitude of the time period so, as the time period increases the value of $cosθ$ increases to value greater than $1$ which is not possible as the range of the cosine function is $−1≤cosθ≤+1$ .

Conclusion:

Therefore, the part (b) has only one solution as the time period is very large.

(d)

Expert Solution

To determine

The angle at which the bead can stand still relative to the loop.

Answer to Problem 68CP

The angle for which condition $gT2<4π2R$ and $θ=0°$ is followed the angles are possible for the beads to stand still.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

As the range of the cosine function is $−1≤cosθ≤+1$ , the solution to occur the condition is to be followed is,

$gT24π2R<1gT2<4π2R$

So, the value of the cosine of the angle is always less than $1$ .

Conclusion:

Therefore, for the condition $gT2<4π2R$ and $θ=0°$ the angles are possible for the beads to stand still.

(e)

Expert Solution

To determine

Whether there are more than two angles.

Answer to Problem 68CP

The number of possible angles are $2$ for a given time period.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Form the above expression the second solution depends on time period, acceleration due to gravity and the radius of the hoop and for a given case the parameters has only one value so the second solution has only one answer.

Conclusion:

Therefore, the number of possible angles are $2$ for a given time period.

Answer 8

(a)

Expert Solution

To determine

The angle from the bottom of the circle for which the bead can stay motionless.

Answer to Problem 68CP

The angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

The rough sketch of the force body diagram of the situation is shown below,

Physics for Scientists and Engineers With Modern Physics, Chapter 6, Problem 68CP

Figure (1)

The bead moves in a circle,

$r=Rsinθ$

Here,

$R$ is the radius of the hoop.

$r$ is the radius of the circle in which the bead moves.

$θ$ is the angle between the vertical and the radial line from the centre of the loop.

The speed of the bead is,

$v=2πrT$

Here,

$T$ is the time period of the rotation.

From the figure (1) the net force in $y$ direction is,

$Ncosθ−mg=0N=mgcosθ$

Here,

$N$ is the normal force acting on the bead.

$m$ is the amass of the bead.

$g$ is the acceleration due to gravity.

The net force in $x$ direction is,

$Nsinθ=mv2r$

Substitute $2πrT$ for $v$ in the above equation.

$Nsinθ=m[(2πrT)2r]$

Substitute $Rsinθ$ for $r$ and $mgcosθ$ for $N$ in the above equation.

$mgcosθsinθ=m[(2πRsinθT)2Rsinθ]gsinθcosθ−4π2RsinθT2=0sinθ(gcosθ−4π2RT2)=0$ . (1)

The two possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ−4π2RT2=0gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.450 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.450 s)24π2(15 cm×1 m100 cm)=0.335θ=cos-1(0.335)=70.42°$

Conclusion:

Therefore, the angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The angle from the bottom of the circle for which the bead can stay motionless.

Answer 13

Answer to Problem 68CP

The angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Answer 14

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

The rough sketch of the force body diagram of the situation is shown below,

Physics for Scientists and Engineers With Modern Physics, Chapter 6, Problem 68CP

Figure (1)

The bead moves in a circle,

$r=Rsinθ$

Here,

$R$ is the radius of the hoop.

$r$ is the radius of the circle in which the bead moves.

$θ$ is the angle between the vertical and the radial line from the centre of the loop.

The speed of the bead is,

$v=2πrT$

Here,

$T$ is the time period of the rotation.

From the figure (1) the net force in $y$ direction is,

$Ncosθ−mg=0N=mgcosθ$

Here,

$N$ is the normal force acting on the bead.

$m$ is the amass of the bead.

$g$ is the acceleration due to gravity.

The net force in $x$ direction is,

$Nsinθ=mv2r$

Substitute $2πrT$ for $v$ in the above equation.

$Nsinθ=m[(2πrT)2r]$

Substitute $Rsinθ$ for $r$ and $mgcosθ$ for $N$ in the above equation.

$mgcosθsinθ=m[(2πRsinθT)2Rsinθ]gsinθcosθ−4π2RsinθT2=0sinθ(gcosθ−4π2RT2)=0$ . (1)

The two possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ−4π2RT2=0gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.450 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.450 s)24π2(15 cm×1 m100 cm)=0.335θ=cos-1(0.335)=70.42°$

Conclusion:

Therefore, the angles from the bottom of the circle for which the bead can stay motionless are $0.0°$ and $70.4°$ .

Answer 15

(b)

Expert Solution

To determine

The angles from the bottom of the circle for which the bead can stay motionless for the time period as $0.850 s$ .

Answer to Problem 68CP

The only one possible angle from the bottom of the circle for which the bead can stay motionless for time period $0.850 s$ is $0.0°$ .

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

Form the part (a) equation (1).

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.850 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.850 s)24π2(15 cm×1 m100 cm)=1.19θ=cos−1(1.19)$

The above value is not possible.

Conclusion:

Therefore, the only one possible angle from the bottom of the circle for which the bead can stay motionless at time period $0.850 s$ is $0.0°$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The angles from the bottom of the circle for which the bead can stay motionless for the time period as $0.850 s$ .

Answer 20

Answer to Problem 68CP

The only one possible angle from the bottom of the circle for which the bead can stay motionless for time period $0.850 s$ is $0.0°$ .

Answer 21

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

The acceleration due to gravity is $9.8 m/s2$ .

Form the part (a) equation (1).

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Substitute $15.0 cm$ for $R$ , $0.850 s$ and $9.8 m/s2$ for $g$ in the above equation.

$cosθ=(9.8 m/s2)(0.850 s)24π2(15 cm×1 m100 cm)=1.19θ=cos−1(1.19)$

The above value is not possible.

Conclusion:

Therefore, the only one possible angle from the bottom of the circle for which the bead can stay motionless at time period $0.850 s$ is $0.0°$ .

Answer 22

(c)

Expert Solution

To determine

The difference between the solution of the part (a) and part (b).

Answer to Problem 68CP

The part (b) has only one solution as the time period is very large.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2Rcosθ∝T2$

As the second solution depends directly on the square of the amplitude of the time period so, as the time period increases the value of $cosθ$ increases to value greater than $1$ which is not possible as the range of the cosine function is $−1≤cosθ≤+1$ .

Conclusion:

Therefore, the part (b) has only one solution as the time period is very large.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The difference between the solution of the part (a) and part (b).

Answer 27

Answer to Problem 68CP

The part (b) has only one solution as the time period is very large.

Answer 28

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2Rcosθ∝T2$

As the second solution depends directly on the square of the amplitude of the time period so, as the time period increases the value of $cosθ$ increases to value greater than $1$ which is not possible as the range of the cosine function is $−1≤cosθ≤+1$ .

Conclusion:

Therefore, the part (b) has only one solution as the time period is very large.

Answer 29

(d)

Expert Solution

To determine

The angle at which the bead can stand still relative to the loop.

Answer to Problem 68CP

The angle for which condition $gT2<4π2R$ and $θ=0°$ is followed the angles are possible for the beads to stand still.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

As the range of the cosine function is $−1≤cosθ≤+1$ , the solution to occur the condition is to be followed is,

$gT24π2R<1gT2<4π2R$

So, the value of the cosine of the angle is always less than $1$ .

Conclusion:

Therefore, for the condition $gT2<4π2R$ and $θ=0°$ the angles are possible for the beads to stand still.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The angle at which the bead can stand still relative to the loop.

Answer 34

Answer to Problem 68CP

The angle for which condition $gT2<4π2R$ and $θ=0°$ is followed the angles are possible for the beads to stand still.

Answer 35

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

As the range of the cosine function is $−1≤cosθ≤+1$ , the solution to occur the condition is to be followed is,

$gT24π2R<1gT2<4π2R$

So, the value of the cosine of the angle is always less than $1$ .

Conclusion:

Therefore, for the condition $gT2<4π2R$ and $θ=0°$ the angles are possible for the beads to stand still.

Answer 36

(e)

Expert Solution

To determine

Whether there are more than two angles.

Answer to Problem 68CP

The number of possible angles are $2$ for a given time period.

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Form the above expression the second solution depends on time period, acceleration due to gravity and the radius of the hoop and for a given case the parameters has only one value so the second solution has only one answer.

Conclusion:

Therefore, the number of possible angles are $2$ for a given time period.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

Whether there are more than two angles.

Answer 41

Answer to Problem 68CP

The number of possible angles are $2$ for a given time period.

Answer 42

Explanation of Solution

Given info: The radius of the circular loop is $15.0 cm$ , the period of the rotation is $0.450 s$ , the angle between the vertical and the radial line from the centre of the loop.

Form equation (1)

$sinθ(gcosθ−4π2RT2)=0$

The possible solutions are,

$sinθ=0θ=0°$

and,

$gcosθ=4π2RT2cosθ=gT24π2R$

Form the above expression the second solution depends on time period, acceleration due to gravity and the radius of the hoop and for a given case the parameters has only one value so the second solution has only one answer.

Conclusion:

Therefore, the number of possible angles are $2$ for a given time period.