Question

A spring has a force constant of 133 N/m and an unstretched length of 0.06 m. One end is attached to a post that is free to rotate in the center of a smooth table, as shown in the top view above. The other end is attached to a 2 kg disc moving in uniform circular motion on the table, which stretches the spring by 0.06 m. Friction is negligible. What is the centripetal force on the disc?

Answer 1

Answer:

7.98 N

Explanation:

In this problem, the centripetal force that keeps the mass in circular motion is provided by the restoring force acting in the spring.

This means that the centripetal force is actually equal to the restoring force in the spring, which is given by (in magnitude):

$F=k(x-x_0)$

where

k is the spring constant

x is the total length of the spring

x0 is the natural length of the spring

In this problem:

k = 133 N/m is the spring constant of the spring

$x_0 = 0.06 m$ is the natural length of the spring

$x=0.06+0.06 = 0.12 m$ is the total length of the spring with the mass attached

So, the centripetal force is

$F=(133)(0.12-0.06)=7.98 N$