A model train with a mass of #2 kg# is moving along a track at #6 (cm)/s#. If the curvature of the track changes from a radius of #12 cm# to #8 cm#, by how much must the centripetal force applied by the tracks change?

Question

A model train with a mass of #2  kg# is moving along a track at #6 (cm)/s#.  If the curvature of the track changes from a radius of #12 cm# to #8 cm#, by how much must the centripetal force applied by the tracks change?

Nora Arzate · Accepted Answer

The change in centripetal force is #=0.03N#

The centripetal force is

#vecF_C=(mv^2)/r*vecr#

The mass is of the train #m=2kg#

The velocity of the train is #v=0.06ms^-1#

The radii of the tracks are

#r_1=0.12m#

and

#r_2=0.08m#

The variation in the centripetal force is

#DeltaF=F_2-F_1#

The centripetal forces are

#||F_1||=2*0.06^2/0.12=0.06N#

#||F_2||=2*0.06^2/0.08=0.09N#

#DeltaF=F_2-F_1=0.09-0.06=0.03N#

Sebastian Acosta · Answer

undefined The centripetal force required to keep an object moving in a circular path is given by the formula:

$$ F = \frac{mv^2}{r} $$

Where:
- $ F $ is the centripetal force,
- $ m $ is the mass of the object,
- $ v $ is the velocity of the object, and
- $ r $ is the radius of the circular path.

Using this formula, we can calculate the initial and final centripetal forces with the given information. Then, we find the difference between these two values to determine the change in centripetal force.

Initial radius $ r_1 = 12 $ cm and final radius $ r_2 = 8 $ cm.

Initial centripetal force $ F_1 = \frac{2 \times (6)^2}{12} $.

Final centripetal force $ F_2 = \frac{2 \times (6)^2}{8} $.

Change in centripetal force = $ F_2 - F_1 $.