Question

A 975-kg sports car (including driver) crosses the rounded top of a hill (radius = 88.0 m) at 18.0 m/s. Determine (a) the normal force exerted by the road on the car, (b) the normal force exerted by the car on the 62.0-kg driver, and (c) the car speed at which the normal force on the driver equals zero.

Short answer

(a) The normal force exerted by the road on the car is $5.97\times {10}^3\mathrm{N}$.

(b) The normal force exerted by the car on the 62.0 kg driver is $3.79\times {10}^2\mathrm{N}$.

(c) The car speed at which the normal force on the driver equals zero is 29.4 m/s.

Step-by-step solution

Step 1. Understanding the concept of centripetal force acting on the car

Whenever the car of mass m moves in a circular path of radius r with a constant speed v,then the car requires a force that keeps it moving in the circular motion. This force is described as the centripetal force. The direction of this force will be towards the centre of the circle.

The expression for centripetal force is given as:

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

Here,when car crosses the rounded top of the hill, it is acted upon by the centripetal force in the downward direction.

Step 2. Identification of the given information

• The mass of the car including driver is, M = 975 kg.
• The mass of the driver is, m = 62.0 kg.
• The radius of the rounded top of the hill is, r = 88.0 m.
• The speed of the car at the top of the hill is, v = 18.0 m/s.

Step 3. (a) Determination of the normal force exerted by the road on the car

When car reaches at the top of the hill, theforces acting on the car are. its weight (Mg) acting downwards and the normal force acting upwards (N).

Thus, the net force acting on the car in the downward direction is, $\left(Mg-N\right)$. This force must provide the centripetal force required by the car to cross the rounded top of the hill. Thus, the centripetal force acting on the car is:

$\frac{M{v}^2}{r}=Mg-N$

Thus, the normal force exerted by the road on the car is:

$N=M\left(g-\frac{v^2}{r}\right)$

Substituting the given values in above expression to get:

$\begin{array}{c}N=\left(975\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2-\frac{{\left(18.0\mathrm{m}/\mathrm{s}\right)}^2}{88.0\mathrm{m}}\right)\\ =\left(5965.2\mathrm{N}\right)\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ \approx 5.97\times {10}^3\mathrm{N}\end{array}$

Step 4. (b) Determination of the normal force exerted by the car on the 62.0-kg driver

Theforces acting on the driver at the top of the hill are. its weight (mg) acting downwards and the normal force acting upwards $\left(N\text{'}\right)$.

Thus, the net force acting on the driver in the downward direction is, $\left(mg-N\text{'}\right)$. This force must provide the required centripetal, i.e.,

$\frac{m{v}^2}{r}=mg-N\text{'}$

Thus, the normal force exerted by the car on the driver is:

$N\text{'}=m\left(g-\frac{v^2}{r}\right)$

Substituting the given values in above expression to get:

$\begin{array}{c}N\text{'}=\left(62.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2-\frac{{\left(18.0\mathrm{m}/\mathrm{s}\right)}^2}{88.0\mathrm{m}}\right)\\ =\left(379.33\mathrm{kgm}/{\mathrm{s}}^2\right)\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ \approx 3.79\times {10}^2\mathrm{N}\end{array}$

Step 5. (c) Determination of the speed of the car at which the normal force on the driver equals zero

When normal force on the driver becomes zero, i.e., $N\text{'}=0$, then:

$\begin{array}{c}N\text{'}=m\left(g-\frac{v^2}{r}\right)\\ 0=m\left(g-\frac{v^2}{r}\right)\\ g=\frac{v^2}{r}\\ v=\sqrt{gr}\end{array}$

On substituting the given values in the above expression, you will get the speed of the car as:

$\begin{array}{c}v=\sqrt{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \left(88.0\mathrm{m}\right)}\\ =\sqrt{862.4{\mathrm{m}}^2/{\mathrm{s}}^2}\\ =29.4\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed of the car at which the normal force on the driver is zero is 29.4 m/s.