Question

A small block on a frictionless horizontal surface has a mass of 0.0280 kg . It is attached to a massless cord passing through a hole in the surface. (See the figure below (Figure 1) .) The block is originally revolving at a distance of 0.310 m from the hole with an angular speed of 1.80 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.115 m . You may treat the block as a particle.

QUESTIONS:

1. Is angular momentum conserved? Yes or No

2. Why or why not?

The tension in the cord is directed _(toward/away from)_ the axis of rotation and produces no net torque. there is _(no net/ a net)_ external torque

3. What is the new angular speed?

4. Find the change in kinetic energy of the block.

5. How much work was done in pulling the cord?

Answer 1

a)yes

b)A block of mass0.028 is rotating. It is attached to a string. There are no external torques is acting on the body so the so angular momentum is conserved.

The tension in the cord is directed toward the axis of rotation and produces no net torque. there is no net external torque

c)mr²$\omega$₀ = mR²$\omega$₁ -------------m is common in both sides , so cancels

Let us substitute the values

(0.31m)² * 1.8rad/s = (0.115m)² * $\omega$₁

$\omega$₁ =13.07 rad/s

D) KE = 0.5I$\omega$²

I=moment of inertia=mr²

$\omega$=angular speed

initial KE = 0.5 * 0.028kg * (0.31m)² * (1.8rad/s)² = 0.00435 J

final KE = 0.5* 0.028kg * (0.115m)² * (13.07 rad/s)² = 0.316 J

dKE = 0.027 J

E) That would be work = 0.027 J