Question:

A 0.55-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75 N, what is the maximum speed the ball can have?

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer:

$13\textrm{ m/s}$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. We write down all the information we are given that the ball is 0.55 kilograms; it goes in a radius of 1.3 meters and the maximum centripetal force you can have is 75 newtons. Here's the centripetal force formula—*m v squared*over

*r*and we'll multiply both sides by

*r*and divide by

*m*and then switch the sides around and solve for

*v squared*; then take the square root of both sides to get

*v*. So

*v*is the square root of radius of curvature times centripetal force divided by mass. So square root of 1.3 times 75 newtons divided by 0.55 kilograms, which is, 13 meters per second, is the maximum speed it could have before the cord breaks.