# 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?

Jul 22, 2017

Δa=-0.053m/s^2

#### Explanation:

The centripetal acceleration is give by the foluma :

${a}_{c} = \frac{m {v}^{2}}{r}$

Let's calculate first the acceleration with radius of $12 c m = 0.12 m$

${a}_{{c}_{1}} = \frac{m {v}^{2}}{r} _ 1 = \frac{3 \cdot {0.08}^{2}}{0.12} = 0.16 \frac{m}{s} ^ 2$

Now for the radius of $18 c m = 0.18 m$

${a}_{{c}_{2}} = \frac{m {v}^{2}}{r} _ 2 = \frac{3 \cdot {0.08}^{2}}{0.18} = 0.1067 \frac{m}{s} ^ 2$

So the change is :

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