Question

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad>s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.
a. Is angular momentum conserved?
b. Find the change in kinetic energy of the block, in J.
c. How much work was done in pulling the cord? in J.

Answer 1

Answer:

W= $K_2-K_1==9.12\times10^{-3} J$

Explanation:

a) Yes, In the absence of external torques acting on the system, the angular momentum is conserved.

b) By the law of conservation of energy angular momentum

$L_1=L_2$

$I_1\omega_1=I_2\omega_2$

$mr_1^2\omega_1=mr_2^2\omega_2\\\omega_2=(\frac{r_1}{r_2} )^2\omega_1$

$\omega_2=(\frac{0.3}{0.15})^2\times2.85$

$\omega_2=5.7\text{ rad/sec}$

c) work done in pulling the chord W= Final kinetic energy(K_2)-Initial Kinetic energy(K_1)

$K_1=\frac{1}{2} mr_1^2\omega_1^2$

$K_1=\frac{1}{2} \times0.025\times0.3^2\times2.85^2$

$=9.12\times10^{-3} J$

Now,

$K_2=\frac{1}{2} mr_2^2\omega_2^2$

$K_2=\frac{1}{2} \times0.025\times0.15^2\times5.7^2$

$K_2=18.24\times10^{-3}$ J

Therefore, Work done W= $K_2-K_1==9.12\times10^{-3} J$