Chapter 9, Problem 89A

(a)

To determine

The axis of spin in the second and final parts of the gymnast’s routine.

Introduction:

Center of mass is that virtual point of the body where its complete mass is assumed to be concentrated.

Explanation:

The center of mass of an average human body lies near about its navel.

Gymnast spins about the center of mass of her body (near about navel of her) in the second and final parts of her routine. In the second part, she rotates in the tuck posture (position at B). In the final part, she straightens out (position at C).

Conclusion:

Thus, gymnast spins about the center of mass of her body in the second and final parts of her routine.

(b)

To Rank: The moment of inertia in three given positions from the greatest to the least.

The order of moment of inertia is,

  $I_A>I_C>I_B$

Introduction:

The moment of inertia is defined as the property of body which resist the angular motion of the body. It is directly proportional to the product of mass and square of distance from the axis of rotation.

  $I\propto M$

  $I\propto R^2$

Where, $I$ = moment of inertia

  $R$ = distance of center of mass from distance from axis of rotation

  $M$ = center of mass

Explanation:

The initial position of the gymnast was at point $A$ and her position changed continuously from point $A$ to $C$ . During that time interval, her axis of rotation was also changed. When she was at A, her axis of rotation was fixed about the bar but as she moved to position $B$ and $C$ , her new axis of rotation was an imaginary axis which was passing through the center of mass.

In overall process, her mass did not change but distance of center of mass ( $M_{com}$ ) from the axis of rotation kept changing. The distance of $M_{com}$ from axis of rotation is maximum at point $A$ and minimum at point $B$ . The order of the distance from axis of rotation is,

  $R_A>R_C>R_B$

If mass of object does not change then moment of inertia is directly proportional to the square of the distance from axis of rotation.

Therefore, the order of moment of inertia in decreasing order is:

  $I_A>I_C>I_B$

Conclusion:

The moment of inertia was maximum at position $A$ and least at position $B$ .

(c)

The order of angular velocity in three given positions from the greatest to the least.

  ${\omega }_B>{\omega }_C>{\omega }_A$

Introduction:

The moment of inertia is defined as the property of body which resist the angular motion of the body. It is directly proportional to the product of mass and square of distance from the axis of rotation.

  $I\propto M$

  $I\propto R^2$

Where, $I$ = moment of inertia

  $R$ = distance of center of mass from distance from axis of rotation

  $M$ = center of mass

The angular momentum for a rigid body rotating about an axis is defined as the product of moment of inertia and angular velocity ( $\omega$ ) and it is denoted by $L$ .

  $L=I\omega$

Where, $I$ = moment of inertia

  $\omega$ = angular velocity

According to conservation of momentum when no external torque acting on the system then the angular momentum of the rigid body will conserve.

Therefore,

  $L_{final}=L_{initial}$ = constant

Explanation:

The initial position of the gymnast was at point $A$ and her position changed continuously from point $A$ to $C$ . During that time interval her axis of rotation was also changed. When she was at A, her axis of rotation was a fixed bar but as she moved to position $B$ and $C$ , her new axis of rotation was an imaginary axis which was passing through the center of mass.

In overall process her mass did not change but distance of center of mass ( $M_{com}$ ) from the axis of rotation kept changing. The distance of $M_{com}$ from axis of rotation is maximum at point $A$ and minimum at point $B$ . The order of the distance from axis of rotation is,

  $R_A>R_C>R_B$

If mass of object does not change, then moment of inertia is directly proportional to the square of the distance from axis of rotation.

Therefore, the order of moment of inertia in decreasing order is:

  $I_A>I_C>I_B$

And according to conservation of momentum, $I\omega$ = constant.

When moment of inertia decreases, the angular velocity of rigid body should increase to make the product constant.

Thus, the order of angular velocity of gymnast at different points in decreasing order is:

  ${\omega }_B>{\omega }_C>{\omega }_A$

Conclusion:

The angular velocity is maximum at point $B$ and minimum at point $A$ .

Answer to Problem 89A

The order of moment of inertia is,

  $I_A>I_C>I_B$

Explanation of Solution

Introduction:

Center of mass is that virtual point of the body where its complete mass is assumed to be concentrated.

The center of mass of an average human body lies near about its navel.

Gymnast spins about the center of mass of her body (near about navel of her) in the second and final parts of her routine. In the second part, she rotates in the tuck posture (position at B). In the final part, she straightens out (position at C).

Conclusion:

Thus, gymnast spins about the center of mass of her body in the second and final parts of her routine.

(b)

To determine

To Rank: The moment of inertia in three given positions from the greatest to the least.

Answer to Problem 89A

The order of moment of inertia is,

  $I_A>I_C>I_B$

Explanation of Solution

Introduction:

The moment of inertia is defined as the property of body which resist the angular motion of the body. It is directly proportional to the product of mass and square of distance from the axis of rotation.

  $I\propto M$

  $I\propto R^2$

Where, $I$ = moment of inertia

  $R$ = distance of center of mass from distance from axis of rotation

  $M$ = center of mass

The initial position of the gymnast was at point $A$ and her position changed continuously from point $A$ to $C$ . During that time interval, her axis of rotation was also changed. When she was at A, her axis of rotation was fixed about the bar but as she moved to position $B$ and $C$ , her new axis of rotation was an imaginary axis which was passing through the center of mass.

In overall process, her mass did not change but distance of center of mass ( $M_{com}$ ) from the axis of rotation kept changing. The distance of $M_{com}$ from axis of rotation is maximum at point $A$ and minimum at point $B$ . The order of the distance from axis of rotation is,

  $R_A>R_C>R_B$

If mass of object does not change then moment of inertia is directly proportional to the square of the distance from axis of rotation.

Therefore, the order of moment of inertia in decreasing order is:

  $I_A>I_C>I_B$

Conclusion:

The moment of inertia was maximum at position $A$ and least at position $B$ .

(c)

To determine

The order of angular velocity in three given positions from the greatest to the least.

Answer to Problem 89A

  ${\omega }_B>{\omega }_C>{\omega }_A$

Explanation of Solution

Introduction:

The moment of inertia is defined as the property of body which resist the angular motion of the body. It is directly proportional to the product of mass and square of distance from the axis of rotation.

  $I\propto M$

  $I\propto R^2$

Where, $I$ = moment of inertia

  $R$ = distance of center of mass from distance from axis of rotation

  $M$ = center of mass

The angular momentum for a rigid body rotating about an axis is defined as the product of moment of inertia and angular velocity ( $\omega$ ) and it is denoted by $L$ .

  $L=I\omega$

Where, $I$ = moment of inertia

  $\omega$ = angular velocity

According to conservation of momentum when no external torque acting on the system then the angular momentum of the rigid body will conserve.

Therefore,

  $L_{final}=L_{initial}$ = constant

The initial position of the gymnast was at point $A$ and her position changed continuously from point $A$ to $C$ . During that time interval her axis of rotation was also changed. When she was at A, her axis of rotation was a fixed bar but as she moved to position $B$ and $C$ , her new axis of rotation was an imaginary axis which was passing through the center of mass.

In overall process her mass did not change but distance of center of mass ( $M_{com}$ ) from the axis of rotation kept changing. The distance of $M_{com}$ from axis of rotation is maximum at point $A$ and minimum at point $B$ . The order of the distance from axis of rotation is,

  $R_A>R_C>R_B$

If mass of object does not change, then moment of inertia is directly proportional to the square of the distance from axis of rotation.

Therefore, the order of moment of inertia in decreasing order is:

  $I_A>I_C>I_B$

And according to conservation of momentum, $I\omega$ = constant.

When moment of inertia decreases, the angular velocity of rigid body should increase to make the product constant.

Thus, the order of angular velocity of gymnast at different points in decreasing order is:

  ${\omega }_B>{\omega }_C>{\omega }_A$

Conclusion:

The angular velocity is maximum at point $B$ and minimum at point $A$ .