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s2008m [1.1K]
2 years ago
5

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
Physics
1 answer:
frozen [14]2 years ago
6 0

Answer:

\theta = 49.81^0

Explanation:

Given that:

\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{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 = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})

where;

v = \frac{\omega d}{2}\\\\v =  \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77

Again:

\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0

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Answer:

\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}

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