Physics: A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torque of 5.0 N•m is applied, and continues for 4.0 s. What is the change in angular momentum of the wheel?

it could be either or because it doesnt just depend on the height but it also depends on the pressure but then again the question didnt ask anything about the pressure so the answer should be trueExplanation:...

A spinning wheel has a rotational inertia of

Ответ:

zavej400

22.01.2022

Physics

-20.0 kg m^2/s

Explanation:

The angular momentum of an object in rotation is given by

$L=I \omega$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, initially we have

I=2 kg m^2 is the moment of inertia of the wheel

$\omega_i = 6.0 rad/s$ is the initial angular speed

So the initial angular momentum is

$L_i = I\omega_i = (2)(6.0)=12 kg m^2/s$

Later, a counterclockwise torque of

$\tau=-5.0 Nm$ is applied

So the angular acceleration of the wheel is:

$\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2$ in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

$\omega_f = \omega_i + \alpha t$

where

t = 4.0 is the time interval

Solving,

$\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s$

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

$L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s$

So, the change in angular momentum is:

$\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2$

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A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torque of 5.0 N•m is applied, and continues for 4.0 s. What is the change in angular momentum of the wheel?

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