A model train with a mass of #4 kg# is moving along a track at #21 (cm)/s#. If the curvature of the track changes from a radius of #84 cm# to #70 cm#, by how much must the centripetal force applied by the tracks change?

Question

A model train with a mass of #4  kg# is moving along a track at #21 (cm)/s#.  If the curvature of the track changes from a radius of #84 cm# to #70 cm#, by how much must the centripetal force applied by the tracks change?

Nora Arzate · Accepted Answer

The change in centripetal force is #=0.042N#

The centripetal force is

#F=(mv^2)/r#

The mass, #m=(4)kg#

The speed, #v=(0.21)ms^-1#

The radius, #=(r) m#

The variation in centripetal force is

#DeltaF=F_2-F_1#

#F_1=mv^2/r_1=4*0.21^2/0.84=0.21N#

#F_2=mv^2/r_2=4*0.21^2/0.70=0.252N#

#DeltaF=0.252-0.21=0.042N#

Benjamin Asiedu · Answer

undefined To find the change in centripetal force, we first need to calculate the initial and final centripetal forces using the formula:

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

Where:
- $ m $ = mass of the train (4 kg)
- $ v $ = velocity of the train (21 cm/s)
- $ r_1 $ = initial radius of curvature (84 cm)
- $ r_2 $ = final radius of curvature (70 cm)

Initial centripetal force:
$$ F_{c1} = \frac{(4 kg) \times (21 cm/s)^2}{84 cm} $$

Final centripetal force:
$$ F_{c2} = \frac{(4 kg) \times (21 cm/s)^2}{70 cm} $$

Change in centripetal force:
$$ \Delta F_c = F_{c2} - F_{c1} $$

$$ \Delta F_c = \frac{(4 kg) \times (21 cm/s)^2}{70 cm} - \frac{(4 kg) \times (21 cm/s)^2}{84 cm} $$

$$ \Delta F_c = (4 kg) \times (21 cm/s)^2 \left( \frac{1}{70 cm} - \frac{1}{84 cm} \right) $$

$$ \Delta F_c \approx (4 kg) \times (21 cm/s)^2 \left( \frac{1}{70 cm} - \frac{1}{84 cm} \right) $$

$$ \Delta F_c \approx 15.9 N $$

So, the centripetal force applied by the tracks must change by approximately 15.9 Newtons.