Question

PLEASE HELP .........

A cylinder has a 21.02 in. diameter. The linear speed of a point on the​ cylinder's surface is 18.12 ​ft/sec. What is the angular speed of the​ cylinder, in revolutions per​ hour?

Answer 1

11860.22 rev/hour

Explanation:

The diameter if a cylinder, d = 21.02 inches

Radius, r = 10.51 inches

or

r = 0.875 feet

The linear speed of a point on the cylinder's surface is 18.12 ft/sec.

We need to find the angular speed of the cylinder.

The relation between the linear speed and the angular speed is given by :

`v=r\omega`

Where

`\omega` is angular speed

`\omega=(v)/(r)\\\\\omega=(18.12)/(0.875)\\\\\omega=20.7\ rad/s`

or

`\omega=11860.22\ rev/hour`

So, the required angular speed of the cylinder is 11860.22 rev/hour.