Question

A 200 g, 40 cm-diameter turntable rotates on frictionless bearings at 60 rpm (rounds per minute). A 10 g block sits at the center of the turntable. A compressed spring shoots the block radially outward along a frictionless groove in the surface of the turntable. You may treat the block as a point mass. Calculate the rotational energy before and after the spring release. Is energy conserved?

Answer 1

Moment of inertia of turntable

I_t =(1/2)M_tR² =(1/2)*0.2*(0.4/2)²=4*10^-3 Kg-m²

Moment of inertia of block

I_B=M_BR²=(0.01)(0.4/2)²=4*10^-4 Kg-m²

By Conservation of angular Momentum

I₁W₁=I₂W₂

(4*10^-3)*60 =(4*10^-4+4*10^-3)W₂

W₂=54.55 rpm

in rad/s

W₁=60*2pi/60 =6.283 rad/s

W₂=54.55*2pi/60 =5.712 rad/s

Rotational Kinetic energy before and and after

K_before=(1/2)I₁W₁² =(1/2)*(4*10^-3)*6.283² =0.078957 J =0.079 J

K_after= (1/2)I₂W₂²=(1/2)*(4.4*10^-3)*5.712²=0.071779 J =0.0718 J

NO energy is not conserved