Question

A girl moves quickly to the center of a spinning merry-go-round, traveling along the radius of the merry-go-round. Which of the following statements are true?

Check all that apply.

Check all that apply.

	The angular speed of the system increases.
	The moment of inertia of the system remains constant.
	The angular speed of the system decreases.
	The moment of inertia of the system increases.
	The moment of inertia of the system decreases.
	The angular speed of the system remains constant.

Answer 1

Concepts and reason

The concepts used in this question are moment of inertia and angular momentum.

Firstly, write the expression for the moment of inertia of the entire system and then find the effect of the movement of the girl along the radius on the moment of inertia.

Finally, use the conservation of angular momentum and find out the change in angular speed as the moment of inertia changes using the angular momentum expression.

Fundamentals

The moment of inertia of an object of mass m rotating about an axis at a radius r is as follows:

$I = m{r^2}$

The angular momentum of a system is given as follows:

$L = I\omega$

Here, I is the moment of inertia of the system and $\omega$ is the angular speed of the system.

The moment of inertia of the girl is as follows:

$I = m{r^2}$

The moment of inertia of the merry-go-round ride is as follows:

$I = {I_{{\rm{ride}}}}$

The total moment of inertia of the system is the sum of the moment of inertia of the ride and the moment of inertia of the girl.

${I_{\rm{t}}} = {I_{{\rm{ride}}}} + m{r^2}$

The total moment of inertia of the system depends upon the distance of the girl from the center. As the distance decreases, the total moment of inertia also decreases.

When the girl reaches the center, the total moment of inertia does not increase or does not remain constant.

The angular momentum of the system is given as follows:

$L = {I_{\rm{t}}}\omega$

The angular momentum is directly proportional to the moment of inertia of the system as well as the angular speed of the system.

The moment of inertia of the system is conserved so that its value remains constant at every point.

The moment of inertia of the system is decreasing so that some other quantity must increase so as to balance its effect. For the angular momentum to remain constant, the angular speed must increase.

Thus, the angular speed cannot decrease or does not remain constant.

1)

The angular momentum of the system is given as follows:

$L = {I_{\rm{t}}}\omega$

Substitute ${I_{{\rm{ride}}}} + m{r^2}$ for I_t in the above expression.

$L = \left( {{I_{{\rm{ride}}}} + m{r^2}} \right)\omega$

The angular momentum is directly proportional to the moment of inertia of the system as well as the angular speed of the system.

The moment of inertia of the system is conserved so that its value remains constant at every point.

The moment of inertia of the system is decreasing so that some other quantity must increase so as to balance its effect. For the angular momentum to remain constant, the angular speed must increase.

5)

The moment of inertia of the merry-go-round ride is as follows:

$I = {I_{{\rm{ride}}}}$

The total moment of inertia of the system is the sum of the moment of inertia of the ride and the moment of inertia of the girl.

${I_{\rm{t}}} = {I_{{\rm{ride}}}} + m{r^2}$

The total moment of inertia of the system depends upon the distance of the girl from the center. As the distance decreases, the total moment of inertia also decreases.

Ans: Part 1

The angular speed of the system increases.