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?

1 Answer
Jul 22, 2017

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