Question

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the clothes. If the diameter of the cylinder is 0.692 m, at what angle will a piece of cloth lose contact with the wall of the cylinder and fall down

Answer 1

Answer:

The angle that the piece of cloth will lose contact with the wall would be  $37.31^{o}$

Explanation:

for a body moving in a circular path The net force is equal  to the centripetal force and can be calculated thus;

mgsin θ = m$w^{2}$r

isolating sine  θ

sin θ =  $w^{2}$r/ g

where w is the angular velocity = 41.5 rev/min = 41.5 rev/min x 2π/60 secs = 4.34587 rad/s

r is the radius = = d/2 = 0.692/2 = 0.3145 m

g is the acceleration due to gravity = 9.8 m/$s^{2}$

sin θ = ($4.34587^{2}$ x 0.3145)/ 9.8

sin θ  = 0.6061

θ = $sin^{-1}$0.6061

θ = $37.31^{o}$

Therefore the angle that the piece of cloth will lose contact with the wall would be  $37.31^{o}$