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

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Answer 1 · 2017-07-22T15:42:12

$Δa=-0.053m/s^2$

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 :

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

A model train with a mass of 3 kg3kg3 kg is moving along a track at 8 (cm)/s8cms8 (cm)/s. If the curvature of the track changes from a radius of 12 cm12cm12 cm to 18 cm18cm18 cm, by how much must the centripetal force applied by the tracks change?