Question

A spin dryer tub can be approximated as a hollow cylinder, of radius 0.3 m, and mass 10.0 kg. It is originally spinning at 3000 rpm. At time t=0, a small piece of material, with μK = 0.2 is pressed against the side of the tub, with a force of 200N.

a) How long does it take to come to a stop?

b) how many times does it go around (in revs) during the braking process ?

c) How many Joules of heat are produced in the brake by friction?

Answer 1

given
R = 0.3 m
M = 10 kg
wo = 3000 rpm = 3000*2*pi/60 = 314 rad/s
mue_k = 0.2
F = 200 N

a) frictional force acting on the cyllinder, fk = mue_k*F

= 0.2*200

= 40 N

Moment of inertia of the cyllinder, I = 0.5*M*R^2

= 0.5*10*0.3^2

= 0.450 kg.m^2

angular acceleration of the cyllinder, alfa = Torque/I

= -fk*R/I

= -40*0.3/0.450

= -26.7 rad/s^2

time taken to stop, t = (wf -wo)/alfa

= (0 - 314)/(-26.7)

= 11.8 s <<<<<<<<<<<------------------Answer

b) angular displacement, theta = wo*t + (1/2)*alfa*t^2

= 314*11.8 + (1/2)*(-26.7)*11.8^2

= 1846 rad

= 1846/(2*pi)

= 294 revolutions <<<<<<<<<<<------------------Answer

c) Heat produced = initial kinetic energy of the cyllinder

= (1/2)*I*wo^2

= (1/2)*0.45*314^2

= 2.22*10^4 J <<<<<<<<<<<------------------Answer