0.29 Phy1260: circular motion -- the mathematics of circular motion  (Page 7/7)

Radial acceleration

A student ties a 10 kg mass onto a fishing line with a breaking strength of 5 newtons, and then starts swinging the mass around over his head. The studenttries very hard to cause the path to be circular. The distance from the center of the circle to the mass is 3 meters.

As time goes on, the student swings the mass faster and faster until the fishing line breaks. What is the tangential velocity of the mass when thefishing line breaks.

Solution:

Centripetal force = mass * (centripetal acceleration), or

f = m * Ar, or

f = m * v^2/r, or

v^2 = f*r/m, or

v = (f*r/m)^(1/2), or

v = (5 newtons * 3 meters/10 kg)^(1/2), or

v = 1.22 m/s

The magnitude of the tangential velocity of the mass when the fishing line breaks is 1.22 meters/second

Check the solution using angular velocity, w

w = v / r, or

w = (1.22 m/s) / 3m = 0.407 radians/second

The angular velocity for a tangential velocity of 1.22 m/s is 0.407 radians/second

Ar = (w^2)*r, or

Ar = (( 0.407 /second)^2)*3m, or

Ar = 0.5 m/s^2

The magnitude of the radial acceleration is 0.5 m/s^2

f = m * Ar, or

f = 10kg*(0.5 m/s^2), or

f = 5 newtons, or

the force equals 5 newtons, which matches the breaking strength of the fishing line.

Work through the examples

I encourage you to work through the examples that I have presented in this module toconfirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Makecertain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

-end-