A model train with a mass of #2 kg# is moving along a track at #2 (cm)/s#. If the curvature of the track changes from a radius of #3 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 #2 (cm)/s#.  If the curvature of the track changes from a radius of #3 cm# to #8 cm#, by how much must the centripetal force applied by the tracks change?

Benjamin Asiedu · Accepted Answer

The centripetal force changes by #=0.017N#

The centripetal force is

#F=(mv^2)/r#

mass, #m=2kg#

speed, #v=0.02ms^-1#

radius, #=r#

The variation in centripetal force is

#DeltaF=F_2-F_1#

#F_1=2*0.02^2/0.03=0.0267N#

#F_2=2*0.02^2/0.08=0.01N#

#DeltaF=0.0267-0.01=0.0167N#

Carson Alexander · Answer

undefined To calculate the change in centripetal force applied by the tracks, we first need to find the initial and final centripetal forces acting on the train. Centripetal force is given by the equation:

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

Where:
- $m$ = mass of the train (2 kg)
- $v$ = velocity of the train (2 cm/s)
- $r$ = radius of the curvature of the track

Initial centripetal force ($F_{initial}$) when the radius is 3 cm:
$$F_{initial} = \frac{2 \times (2)^2}{3}$$

Final centripetal force ($F_{final}$) when the radius changes to 8 cm:
$$F_{final} = \frac{2 \times (2)^2}{8}$$

To find the change in centripetal force, subtract the initial force from the final force:
$$Change\ in\ F = F_{final} - F_{initial}$$