Question

A model train, with a mass of `2 kg`, is moving on a circular track with a radius of `3 m`. If the train's kinetic energy changes from `3 j` to `5 j`, by how much will the centripetal force applied by the tracks change by?

Answer 1

`=1.dot3N`

Explanation:

For Circular Motion, Centripetal Force `F_c= - mr omega^2`

Where `m` is the mass of the body in circular motion. `r` is the radius of the circle and `omega` is angular velocity.
`-` sign means that the force is opposite to the radius vector and is directed towards the center.
Now velocity `v=romega`
Therefore, magnitude of the force `|F_c|=(mv^2)/r`
Kinetic energy is given by
`KE=1/2mv^2`
Let the velocity of the train change from `v_i` to `v_f`
Given is change in kinetic energy `Delta KE=1/2mv_f^2-1/2mv_i^2`
`1/2mv_f^2-1/2mv_i^2=5-3=2`
or `m/2(v_f^2-v_i^2)=2`
or `m(v_f^2-v_i^2)=4` ........(1)

Change in the centripetal force applied by the tracks `=(mv_f^2)/r-(mv_i^2)/r`
`=m/r(v_f^2-v_i^2)` ........(2)
Inserting value of `m(v_f^2-v_i^2)` from (1) and given value of `r`

Change in the centripetal force applied by the tracks `=1/3*4`
`=1.dot3N`