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

Answer:

#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#