In an automatic clothes drier, a hollow cylinder moves the clothes on a vertical circle (radius r = 0.364 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 è 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 è = 66.0 °?
mg down
m v^2/r radial = m r w^2 where w = 2 pi f
vertical component of radial =m rw^2 sin e
falls when total vertical force = 0
m g = m r w^2 sin e
or
w^2 = g/(r sin e)
w^2 = 9.8 /(.364 sin 66)
find w
find f = w/2pi
To determine the number of revolutions per second the cylinder should make for the clothes to lose contact with the wall, we can use the concept of centripetal acceleration.
1. First, we need to find the radius in meters, denoted as "r." In this case, the given radius is 0.364 m.
2. The angle at which the clothes lose contact with the wall is given as 66.0°, denoted as "θ."
3. We can find the value of the centripetal acceleration using the formula a = rω², where "a" is the centripetal acceleration and "ω" is angular velocity.
4. The centripetal acceleration can also be expressed as a = r(ω^2) = r((2πf)^2), where "f" is the frequency (revolutions per second).
5. Rearrange the formula: (2πf)^2 = (a / r).
6. Substitute the given values: a = g * tan θ, where "g" is the acceleration due to gravity (9.8 m/s²).
7. Therefore, (2πf)^2 = (g * tan θ / r).
8. Solve for "f": f = sqrt((g * tan θ) / (4π² * r)).
9. Substitute the given values: f = sqrt((9.8 * tan 66.0°) / (4π² * 0.364)).
10. Calculate the result using a calculator: f ≈ 0.183 revolutions per second.
Therefore, the cylinder should make approximately 0.183 revolutions per second for the clothes to lose contact with the wall when the angle is 66.0°.