Question

A small blob of putty of mass m falls from the ceiling and lands on the outer rim of a turntable of radius R and moment of inertia I0 that is rotating freely with angular speed ω0 about its vertical fixed-symmetry axis. (Use any variable or symbol stated above as necessary.) What is the postcollision angular speed of the turntable-putty system?

Answer 1

`\omega = (I_o\omega_o)/((I_o+mR^2))`

Step-by-step explanation:

Mass = m

Radius of turn table = R

Initial moment off inertia of table = Io

Initial angular velocity is =ωo

As we know that if there is no any external torque on the system then its angular momentum will remain conserve .

So initial angular momentum of table = Io x ωo

Now lets take speed of the mass and turn table system will become ω

Final inertia of system is I

`I+I_o+mR^2`

Final linear momentum

`L_f=(I_o+mR^2)\omega`

`I_O\omega_0=(I_o+mR^2)\omega`

`\omega = (I_o\omega_o)/((I_o+mR^2))`