Question

6. A small block on a frictionless, horizontal surface has a mass "m". It is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance "r" with an angular speed "o". The cord is pulled below, shortening the radius to half it value. Given [r, m, ω.], Determine: a. The final angular speed. b. The amount of work done in pulling the cord.

Answer 1

6] The moment of Inertia initial is:

I = mr²

once the cord is pulled, r' = r/2

so, new moment of inertia is: I' = m(r/2)² = mr²/4

a] The angular momentum before and after the change in length will be conserved.

so,

The moment of inertia initial is: I = mr^2 once the cord is pulled, r' = r/2 so, new moment of inertia is: I' = m (r/2)^2 = mr^2/4 a) The angular momentum before and after the change in length will be conserved so, mr^2 omega_1 = mr^2/4 omega_f omega_f = 4 omega_1 this is the final angular speed b) From work energy theorem, the work done in pulling the cord = change in its rotational kinetic energy W = KE_f - KE_i = 1/2 mr^2/4 (4 omega_1)^2 - 1/2 mr^2 omega_1^2 W = 2mr^2 omega_1^2 - 1/2 mr^2 omega_1^2 = 3/2 mr^2 omega_1^2

=> $\omega_f = 4\omega_1$

this is the final angular speed

b] From Work Energy theorem, the work done in pulling the cord = change in its rotational kinetic energy

$W = KE_f-KE_i = \frac{1}{2}\frac{mr^2}{4}(4\omega_1)^2 - \frac{1}{2}mr^2\omega_1^2$

=> $W = 2mr^2\omega_1^2 - \frac{1}{2}mr^2\omega_1^2 = \frac{3}{2}mr^2\omega_1^2$ .