Determine the tangential and centripetal components of the net force exerted on the car (by the ground) in Example 5–8 when its speed is 15 m/s. The car’s mass is 950 kg.

When the net force exerted on the car of mass m moving in a circular path of radius r having a constant speed v does not act towards the center of the circle but acts at an angle, then the motion of car is termed as non-uniform circular motion.

This force has two components, tangential component which acts along the tangent and centripetal component which acts along the centre of the circle.

The magnitude of the centripetal acceleration is given as:

$a_{\mathrm{R}}=\frac{v^2}{r}$

The magnitude of the tangential acceleration is given as:

$a_{\mathrm{t}}=\frac{\Delta v}{\Delta t}$

• The speed of the car is, v = 15 m/s.
• The mass of the car is, m = 950 kg.
• The radius of the circular track is, r = 500 m.
• The tangential acceleration is, $a_{\mathrm{t}}=3.2\mathrm{m}/{\mathrm{s}}^2$.

The tangential component of the net force is given as:

$F_{\mathrm{t}}=m{a}_{\mathrm{t}}$

Substitute all the values in the above equation.

$\begin{array}{c}{F}_{\mathrm{t}}=\left(950\mathrm{kg}\right)\left(3.2\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =3040\mathrm{N}\\ \approx 3.0\times {10}^3\mathrm{N}\end{array}$

Thus, the tangential component of the net force exerted on the car is $3.0\times {10}^3\mathrm{N}$.

The centripetal component of the net force is given as:

$F_{\mathrm{R}}=\frac{m{v}^2}{r}$

Substitute all the values in the above equation.

$\begin{array}{c}{F}_{\mathrm{R}}=\frac{\left(950\mathrm{kg}\right){\left(15\mathrm{m}/\mathrm{s}\right)}^2}{\left(500\mathrm{m}\right)}\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =427.5\mathrm{N}\\ \approx 4.3\times {10}^2\mathrm{N}\end{array}$

Thus, the centripetal component of the net force exerted on the car is $4.3\times {10}^2\mathrm{N}$.