Question

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a radius of 3.22 m and a rotational inertia of 275 kg·m2 about the axis of rotation. A 62.3 kg student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.33 rad/s when the student starts at the rim, what is the angular speed when she is 0.861 m from the center?

Answer 1

Answer:

The angular speed is $\omega_2= 3.81s^{-1}$.

Explanation:

The law of conservation of angular momentum says that for an isolated system

$I_1\omega_1 = I_2 \omega_2$

Now, when the student is at the rim of the platform the moment of inertia of the system is

$I_1 = mr_1^2+275kg\cdot m^2$

$I _1 = (62.3kg)(3.22m)^2+275kg\cdot m^2$

$I_1 = 920.95kg\cdot m^2$,

and the angular speed is

$\omega_1 = 1.33s^{-1}$.

When the student is $r_2 = 0.861m$ from the center,the moment of inertia of the system becomes

$I_2 =mr_2^2+275kg\cdot m^2$

$I_2 =(62.3kg)(0.861m)^2+275kg\cdot m^2$

$I_2= 321.18kg\cdot m^2$

Thus, from conservation of angular momentum

$(920.95kg\cdot m^2)(1.33s^{-1})= (321.18kg\cdot m^2)\omega_2$

$\omega_2=\dfrac{ (920.95kg\cdot m^2)(1.33s^{-1})}{(321.18kg\cdot m^2)}$

$\boxed{\omega_2= 3.81s^{-1}}$

which is the angular speed when the student is 0.861 meters from the center.