In an automatic clothes dryer, a hollow cylinder moves the clothes on a vertical

Question

In an automatic clothes dryer, a hollow cylinder moves the clothes on a vertical circle (radius r = 0.39 m), as the drawing shows. The appliance is designed so that the clothes tumble gently as they dry. This means that when a piece of clothing reaches an angle of theta above the horizontal, it loses contact with the wall of the cylinder and falls onto the clothes below. How many revolutions per second should the cylinder make in order that the clothes lose contact with the wall when theta = 68.0 degree?

Explanation / Answer

Consider the dryer is rotateing with an angular speed of w

Here the normal reaction is zero

If the dryer is rorating with angle theta baove the horizontal then

N+mgsintheta =mrw2

N =0 then

mgsinthea =mrw2

gsinthea=rw2

Then w =Sqrt(gsinthea/r) =Sqrt(9.81*sin68/0.39)=4.82rad/sec

Then we know that

0ne revolution =2pi radian

one minute =60s

1rev/min *(2piradian/1revolution)*(1min/60s) =1rad/s

Then 4.82rad/sec =4.82*2pi/60 =0.504rev/min

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