A circular pulley with a radius of 10 cm is turning at 12 revolutions per minute. How

fast is a point on the edge of the pulley rising when it is 5 cm higher than the center of
the pulley?

Draw a diagram. Consider point P on the circle, at height h above the center of the pulley. Let the radius from the center to P make angle θ with the horizontal.

h/r = sinθ

since r = 10,

h = 10 sinθ
dh/dt = 10 cosθ dθ/dt

Since the wheel is turning at 12 rpm, dθ/dt = 24π/min

dh/dt = 10 * √3/2cm * 24π/min = 653 cm/min

Thanks Steve

To find the speed at which a point on the edge of the pulley is rising, we can use the concept of angular velocity.

First, let's define some variables:
- r = radius of the pulley (10 cm)
- ω = angular velocity (in radians per minute)
- h = height of the point above the center of the pulley (5 cm)
- v = speed at which the point is rising (what we need to find)

The angular velocity, ω, is the rate at which an object rotates around a fixed axis. In this case, it is given that the pulley is turning at 12 revolutions per minute. To convert this to radians per minute, we multiply it by 2π since there are 2π radians in one revolution.
So, ω = 12 revolutions/minute * 2π radians/revolution = 24π radians/minute.

Now, let's consider a right triangle formed by the radius, the height of the point, and the line connecting the point on the edge to the center of the pulley. The length of this line is the radius (r) plus the height (h), which is 10 cm + 5 cm = 15 cm.

Using the Pythagorean theorem, we have:
(r + h)^2 = r^2 + v^2

Substituting the known values, we have:
(10 cm + 5 cm)^2 = 10 cm^2 + v^2
(15 cm)^2 = 100 cm^2 + v^2
225 cm^2 = 100 cm^2 + v^2
v^2 = 225 cm^2 - 100 cm^2
v^2 = 125 cm^2
v = √(125) cm
v ≈ 11.18 cm/min

Therefore, the point on the edge of the pulley is rising at a speed of approximately 11.18 cm/min.