A model train with a mass of #4 kg# is moving along a track at #9 (cm)/s#. If the curvature of the track changes from a radius of #180 cm# to #24 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 #9 (cm)/s#.  If the curvature of the track changes from a radius of #180 cm# to #24 cm#, by how much must the centripetal force applied by the tracks change?

Nathan Ashcroft · Accepted Answer

The change in centripetal force is #=0.177N#

The centripetal force is

#F=(mv^2)/r#

mass, #m=(4)kg#

speed, #v=(0.09)ms^-1#

radius, #=(r) m#

The variation in centripetal force is

#DeltaF=F_2-F_1#

#F_1=mv^2/r_1=4*0.09^2/1.8=0.018N#

#F_2=mv^2/r_2=4*0.09^2/0.24=0.135N#

#DeltaF=0.135-0.018=0.177N#

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:

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

where:
$ m = 4 \, \text{kg} $ (mass of the train),
$ v = 9 \, \text{cm/s} $ (velocity of the train),
$ r_{initial} = 180 \, \text{cm} $ (initial radius of curvature),
$ r_{final} = 24 \, \text{cm} $ (final radius of curvature).

Initial centripetal force:
$$ F_{c\_initial} = \dfrac{4 \, \text{kg} \times (9 \, \text{cm/s})^2}{180 \, \text{cm}} $$

Final centripetal force:
$$ F_{c\_final} = \dfrac{4 \, \text{kg} \times (9 \, \text{cm/s})^2}{24 \, \text{cm}} $$

Change in centripetal force:
$$ \Delta F_c = F_{c\_final} - F_{c\_initial} $$

$$ \Delta F_c = \dfrac{4 \, \text{kg} \times (9 \, \text{cm/s})^2}{24 \, \text{cm}} - \dfrac{4 \, \text{kg} \times (9 \, \text{cm/s})^2}{180 \, \text{cm}} $$

$$ \Delta F_c = \dfrac{4 \times 81}{24} - \dfrac{4 \times 81}{180} $$

$$ \Delta F_c = 13.5 \, \text{N} - 1.8 \, \text{N} $$

$$ \Delta F_c = 11.7 \, \text{N} $$

So, the centripetal force applied by the tracks must change by $ 11.7 \, \text{N} $.