A turntable with a moment of inertia of 0.022 rotates feely at 2.7 rad/s. A circular disk of mass 370 g and diameter 30 cm, and initially not rotating, slips down a spindle and lands on the turntable.

Question

A turntable with a moment of inertia of   0.022 rotates feely at  2.7 rad/s. A circular disk of mass  370 g and diameter 30 cm, and initially not rotating, slips down a spindle and lands on the turntable.

anonymous · Answer

What's the question?

anonymous · Answer

This is probably an angular momentum related problem.

anonymous · Answer

sorry the question is 
a) Find the new angular speed. 
 rad/s 

b) what is the change in kinetic energy?

anonymous · Answer

That would probably be:$$(I_a \omega_a)/(I_b+I_a)=\omega$$ where I sub a is the moment of inertia of the turntable, and I sub is the moment of inertia of the circular disk. The moment of inertia of the circular disk is given by:
$$Mr^2=I_b$$ Where M is its total mass and r is its radius.

anonymous · Answer

I sub b is the moment of inertia of the disk*.

anonymous · Answer

I'm not sure about the change of kinetic energy of the system but that's probably:
$$\Delta KE =\frac{1}{2}(I_a+I_b)\omega -  \frac{1}{2} I_a \omega _a ^2$$
Where omega is the new angular speed

anonymous · Answer

Thanks that really helps

anonymous · Answer

oh the omega in change in KE equation is squared.