A model train with a mass of #1 kg# is moving along a track at #6 (cm)/s#. If the curvature of the track changes from a radius of #8 cm# to #9 cm#, by how much must the centripetal force applied by the tracks change?

Curve trajectories require centripetal force to be applied in order to maintain their curvature. This force is defined as follows and is directed toward the curve's center:

Using an 8-cm radius, the centripetal force is:

where SI units have been used.

The centripetal force will now drop with a radius of 9 cm, assuming constant mass and speed:

The explanation is simple: the train is easier to maintain on its trajectory and requires less force when it is in a larger circle, as depicted by the train.

To calculate the change in centripetal force, we can use the formula:

[ F = \dfrac{mv^2}{r} ]

Where:

• ( F ) is the centripetal force,
• ( m ) is the mass of the train (1 kg),
• ( v ) is the velocity of the train (6 cm/s),
• ( r ) is the radius of curvature of the track.

First, we calculate the initial centripetal force with ( r = 8 ) cm:

[ F_{\text{initial}} = \dfrac{(1 , \text{kg})(6 , \text{cm/s})^2}{8 , \text{cm}} ]

[ F_{\text{initial}} = \dfrac{1 , \text{kg} \times 36 , \text{cm}^2/\text{s}^2}{8 , \text{cm}} ]

[ F_{\text{initial}} = 4.5 , \text{N} ]

Then, we calculate the final centripetal force with ( r = 9 ) cm:

[ F_{\text{final}} = \dfrac{(1 , \text{kg})(6 , \text{cm/s})^2}{9 , \text{cm}} ]

[ F_{\text{final}} = \dfrac{1 , \text{kg} \times 36 , \text{cm}^2/\text{s}^2}{9 , \text{cm}} ]

[ F_{\text{final}} = 4 , \text{N} ]

The change in centripetal force is:

[ \Delta F = F_{\text{final}} - F_{\text{initial}} ]

[ \Delta F = 4 , \text{N} - 4.5 , \text{N} ]

[ \Delta F = -0.5 , \text{N} ]

Therefore, the centripetal force applied by the tracks must decrease by 0.5 N.