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

Answer:

$\omega = \dfrac{I_o\omega_o}{(I_o+mR^2)}$

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

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

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

$\omega = \dfrac{I_o\omega_o}{(I_o+mR^2)}$