In an automatic clothes dryer, a hollow cylinder moves the clothes on a vertical circle (radius r = 0.31 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 θ 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 θ = 70.0°?

To find the required number of revolutions per second for the clothes to lose contact with the wall, we can use the concept of centripetal acceleration.

The clothes lose contact with the wall when the centripetal acceleration acting on them is equal to the gravitational acceleration.

The formula for centripetal acceleration is given by:
ac = (v^2) / r

Where ac is the centripetal acceleration, v is the linear velocity of the clothes, and r is the radius of the vertical circle.

In this case, when the clothes lose contact with the wall, the only force acting on them is the force due to gravity. So, the centripetal acceleration should equal the gravitational acceleration (9.8 m/s^2).

We can express the linear velocity, v, in terms of the angular velocity, ω, and the radius, r, by using the formula:
v = rω

Substituting this value of v into the formula for centripetal acceleration, we get:
ac = (rω^2) / r

Simplifying, we get:
ac = rω^2

Setting the centripetal acceleration equal to the gravitational acceleration, we have:
9.8 = (0.31ω^2)

Now we can solve for ω:
ω^2 = 9.8 / 0.31
ω^2 ≈ 31.61
ω ≈ √31.61
ω ≈ 5.62 radians/s

Since 1 revolution is equal to 2π radians, we can convert the angular velocity to revolutions per second:
ω = 5.62 radians/s * (1 revolution / 2π radians) ≈ 0.897 revolutions per second

Therefore, the cylinder should make approximately 0.897 revolutions per second in order for the clothes to lose contact with the wall when θ = 70.0°.