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 0.748 m, at what angle will a piece of cloth lose contact with the wall of the cylinder and fall down? (Take the +x axis to be pointing horizontally to the right.) degree counterclockwise from the +x axis

Answer 1

`\theta = 49.81^0`

Step-by-step explanation:

Given that:

`\omega = 41.5 \ rev/min\\\\\omega = 41.5 *(1)/(60)* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m`

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

`\\\\mg sin \theta = (mv^2)/(r)\\\\sin \theta = (2v^2)/(dg)\\\\\theta = sin^(-1) ((2v^2)/(dg))`

where;

`v = (\omega d)/(2)\\\\v = (4.45 *0.748)/(2)\\\\v = 1.6643\\\\v^2 = 2.77`

Again:

`\theta = sin^(-1)((2v^2)/(dg))\\\\\theta = sin^(-1)( (2*2.77)/(0.74*9.8))\\\\\theta = 49.81^0`