A model train, with a mass of 33 kgkg, is moving on a circular track with a radius of 33 mm. If the train's kinetic energy changes from 1818 JJ to 00 JJ, by how much will the centripetal force applied by the tracks change by?

1 Answer
Feb 4, 2016

The centripetal force is given by F=(mv^2)/rF=mv2r. We are given the mass and the radius and can calculate the velocity from the kinetic energy. The change is -1212 NN.

Explanation:

The centripetal force on an object of mass mm kgkg in circular motion at velocity vv ms^-1ms1 in a circle of radius rr mm is given by:

F=(mv^2)/rF=mv2r

In this case we know the mass is 33 kgkg and the radius is 33 mm, but the velocity is not immediately obvious. We are given the kinetic energy at two moments in time, though, and we know that kinetic energy is:

E_k=1/2mv^2Ek=12mv2

We can rearrange this to make vv the subject:

v=sqrt((2E_k)/m)v=2Ekm

Since the kinetic energy at the second time is 00 JJ, the velocity is also 00 ms^-1ms1, and therefore the centripetal force is also 00 NN.

At the first time, the kinetic energy is 1818 JJ, so:

v=sqrt((2*18)/3)=sqrt(36/3)=sqrt12v=2183=363=12 ms^-1ms1

This means the centripetal force is:

F=(mv^2)/r = (3*12)/3 = 12F=mv2r=3123=12 NN (since (sqrt12)^2=12(12)2=12)

So the centripetal force was 1212 NN at the beginning and 00 NN at the end. Its change was therefore -1212 NN.