Question

A student sits at rest on a piano stool that can rotate withoutfriction. The moment of inertia of the student-stool system is 4.0kg*m^2. A second student tosses a 1.7 kg mass with a speed of 2.7m/s to the student on the stool, who catches it at a distance of0.35 m from the axis of rotation.

What is the resulting angular speed of thestudent and the stool?

= ? rad/s

Answer 1

`\omega_(f)=0.381 \ rad/s`

Step-by-step explanation:

given,

moment of inertia of the stool = 4 kg m²

mass tosses by the student = 1.7 Kg

speed = 2.7 m/s

distance = 0.35 m from axis of rotation

initial angular momentum of the system

`L_i = mvR`

final angular momentum of the system

`L_f = (I_1 +mR^2)\omega_(f)`

from conservation of angular momentum

`L_i = L_f`

`mvR= (I_1 +mR^2)\omega_(f)`

`\omega_(f)= (mvR)/(I_1 +mR^2)`

`\omega_(f)= (1.7 * 2.7 * 0.35)/(4 +1.7 * 0.35^2)`

`\omega_(f)=0.381 \ rad/s`