contestada

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 3.70 m and a rotational inertia of 274 kg·m2 about the axis of rotation. A 67.8 kg student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 2.62 rad/s when the student starts at the rim, what is the angular speed when she is 0.456 m from the center?

Respuesta :

MechEngineer

Answer:

10.93 rad/s

Explanation:

If we treat the student as a point mass, her moment of inertia at the rim is

[tex]I_r = mr^2 = 67.8*3.7^2 = 928.182 kgm^2[/tex]

So the system moment of inertia when she's at the rim is:

[tex]I_1 = I_d + I_r = 274 + 928.182 = 1202.182 kgm^2[/tex]

Similarly, we can calculate the system moment of inertia when she's at 0.456 m from the center

[tex]I_2 = I_d + 67.8*0.456^2 = 274 + 14.1 = 288.1 kgm^2[/tex]

We can apply the law of angular momentum conservation to calculate the post angular speed when she's 0.456m from the center:

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

[tex]\omega_2 = \omega_1\frac{I_1}{I_2} = 2.62*\frac{1202.182}{288.1} = 10.93 rad/s[/tex]

Otras preguntas