Question

A wet socks clinging to the inside of a washing machine drum which is spinning at a speed of

4.7 m/s. The radius of the drum is 30 cm.

(a) (Marks: 2) What is the minimum coefficient of static friction

between sock and the drum so that the sock will not slip down?

(b) (Marks: 1) At the same rotational speed and coefficient of

static friction between sock and the drum, would a sock of twice

the mass slide down the wall? Explain your answer.

(c) (Marks: 2) As the drum slows down after the washing is done

at an angular retardation of 3.25 rad/sec2

. Find the time taken and

number of rotations before it stops.

Answer 1

a) speed of the drum, v=4.7m/s

radius of the drum, r=30cm=0.3m

angular speed $\omega=v/r=4.7/0.3=15.667rad/s$

From Newton's second law equation,

$F=ma_c$

$a_c$ is centripetal acceleration $=\omega^2r$

$\mu_s F=mg=\mu_sm\omega ^2r$

$\mu_s=\frac{g}{\omega ^2r}=\frac{9.8}{(15.667^2\times 0.3)}=0.133$(minimum coefficient of static friction)

b) Coefficient of static friction $\mu_s = \frac{g}{\omega ^2r}$

From the above equation coefficient of static friction depends on acceleration due to gravity, angular speed, radius of the drum only. It does not depend on the mass of the sock. So a sock of twice the mass of the first sock does not slide down the drum wall.

c) Angular retardation is 3.25rad/$s^2$

Angular acceleration $\alpha =-3.25rad/s^2$

final angular speed $\omega_f=0rad/s^2$

$w_f=w_i+\alpha t$

$t=(w_f-w_i)/\alpha=\frac{0-15.667}{-3.25}=4.82s$ to come to rest.

Now $\omega^2_f-\omega^2_i=2\alpha \theta$

$\theta=\frac{\omega^2_f-\omega^2_i}{2\alpha}=\frac{0^2-15.667^2}{2\times (-3.25)}=37.76rad=\frac{37.76}{2\pi}=6rev$

Therefore it makes 6 revolutions before coming to rest.