A student is at rest on a stool that may freely spin about its central axis of rotation. As the stool spins, the student holds onto two dumbbells as the stool spins at an angular speed of 1.2 rad/s with the student’s arms completely stretched out from the student’s body. At this instant, the student-dumbbell system has rotational inertia of 6 kg⋅m^2. The student then brings their arms close to their body, and rotational inertia of the student-dumbbell system is changed to 2 kg⋅m^2.
What is the new angular speed of the student?

A) 0.4 rad/s B) 1.2 rad/s C) 3.6 rad/s D) 7.2 rad/s

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aristocles aristocles · Answer 1 · 2020-03-25T07:23:06+01:00

Answer:

the final speed of the stool is 3.6 rad/s

Explanation:

As we know that there is no external torque on the system

So we can use concept of angular momentum conservation

So we will have

[tex]I_1\omega_1 = I_2\omega_2[/tex]

now we will have

[tex]6\times 1.2 = 2\times \omega[/tex]

[tex]\omega = \frac{6 \times 1.2}{2}[/tex]

[tex]\omega = 3.6 rad/s[/tex]

So the final speed of the stool is 3.6 rad/s

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priyankatutor priyankatutor · Answer 2 · 2022-01-11T07:01:17+01:00

The new angular speed of the student is 3.6 rad/s when the rotational inertia is [tex]\bold{ 2\ kgm^2}[/tex].

It can be defined as the velocity of an object on an axis. It is equal to the product of inertia and angular speed.

[tex]I_1\omega _1 = I_2\omega _2[/tex]

Where,

[tex]I _1\\[/tex]- Initial rotational inertia = [tex]\bold{6\ kgm^2 }[/tex]

[tex]\omega _1[/tex] - initial angular speed = 1.2 rad/s

[tex]I_2[/tex] - final rotational inertia = [tex]\bold{2\ kgm^2 }[/tex]

[tex]\omega _2[/tex] - final angular momentum =?

Put the values in the equation,

[tex]6 \times 1.2 = 2\times \omega_2 \\\\\omega_2 = 3.6 \rm \ rad/s[/tex]

Therefore, the new angular speed of the student is 3.6 rad/s.

To know more about angular momentum,

https://brainly.com/question/6672895