Question

A model train with a mass of `3 kg` is moving along a track at `8 (cm)/s`. If the curvature of the track changes from a radius of `12 cm` to `18 cm`, by how much must the centripetal force applied by the tracks change?

Answer 1

`Δa=-0.053m/s^2`

Explanation:

The centripetal acceleration is give by the foluma :

`a_c=(mv^2)/r`

Let's calculate first the acceleration with radius of `12cm=0.12m`

`a_{c_1}=(mv^2)/r_1=(3*0.08^2)/0.12=0.16m/s^2`

Now for the radius of `18cm=0.18m`

`a_{c_2}=(mv^2)/r_2=(3*0.08^2)/0.18=0.1067m/s^2`

So the change is :

`Δa=a_{c_2}-a_{c_1}=0.1067-0.16=-0.053m/s^2`