Question

In Fig. 6-57, a stuntman drives a car (without negative lift) over the top of a hill, the cross section of which can be approximated by a circle of radius R = 250 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill?

Short answer

The greatest speed at which he can drive without the car leaving the road at the top of the hill 178 km/h .

Step-by-step solution

Identification of the given data

The given data is shown as below,

The radius of the path is,R = 250 m

Definition of centripetal force and Newton’s second law

Newton’s second law of motion states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Also, it involvescentripetal force.

A centripetal force is a force that makes a body follow a curved path.

Determination of the greatest speed at which the stuntman can drive without the car leaving the road at the top of the hill

At the top of the circular hill the vertical forces on the car are the upward normal force exerted by the ground and the downward pull of gravity. Write the expression for the force.

$mg-F_N=\frac{m{v}^2}{R}$

Here, mis the mass of the car, g is the acceleration due to gravity, $F_N$is the upward normal force exerted by the ground, v is the greatest speed at the given conditions, Rand is the radius of the path.

It is known that $F_N=0$. So, substitute it in the above expression and rearrange it.

$$\begin{gathered} mg-0=\frac{m{v}^2}{R} \\ mg=\frac{m{v}^2}{R} \\ g=\frac{v^2}{R} \\ {v}^2=gR \\ v=\sqrt{gR} \end{gathered}$$

Substitute 9.8 m/s² for g, and 250 m forR in the above expression.

$$\begin{gathered} v=\sqrt{9.8\mathrm{m}/{\mathrm{s}}^2\times 250\mathrm{m}} \\ =49.5\mathrm{m}/\mathrm{s} \\ =49.5\times \frac{3600}{1000}\mathrm{km}/\mathrm{h} \\ =178\mathrm{km}/\mathrm{h} \end{gathered}$$

Thus, the greatest speed at which he can drive without the car leaving the road at the top of the hill 178 km/h .