Q110P - University Physics with Modern Physics

Question

A road heading due east passes over a small hill. You drive a car of mass   at constant speed   over the top of the hill, where the shape of the roadway is well approximated as an arc of a circle with radius  . Sensors have been placed on the road surface there to measure the downward force that cars exert on the surface at various speeds. The table gives values of this force versus speed for your car:

Treat the car as a particle. (a) Plot the values in such a way that they are well fitted by a straight line. You might need to raise the speed, the force, or both to some power. (b) Use your graph from part (a) to calculate   and  . (c) What maximum speed can the car have at the top of the hill and still not lose contact with the road?

Step-by-Step Solution

Answer

(a) The required graph is as follows.

(b) The values of m and R are 897 kg , and 49.5 m , respectively.

(c) The maximum speed of the car is  22 m/s.

Step 1: Identification of given data:

The given data is as follows:

• The mass of the car is m.
• The radius of the circular path of the road is R.

Step 2: Concept/Significance centripetal force:

If a particle moves in a circular arc of radius R at constant speed V, the particle is said to be in uniform circular motion. It then experience a net centripetal force $\overrightarrow{\mathbf{F}}$ and a centripetal acceleration $\overrightarrow{a_c}$. The magnitude of the force is given by,

$$\begin{gathered} \mathbf{F}\mathbf{=}{\mathbf{ma}}_{\mathbf{c}} \\ \mathbf{ }\mathbf{ }\mathbf{=}\mathbf{m}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}} \end{gathered}$$ 

Step 3: (a) Plot the values in such a way that they are well fitted by a straight line:

The free-body diagram for the given situation as follows.

The free-body diagram shows the forces acting on the car. Here, n is the normal force exerted by the surface on the car acting upward, and mg is the gravitational weight of the car acting downwards.

 

The equation of net force in  -direction is given by,

$$\begin{gathered} \sum {\mathrm{F}}_{\mathrm{y}}=-{\mathrm{ma}}_{\mathrm{c}} \\ \mathrm{n}-\mathrm{mg}=-\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{R}} \\ \mathrm{n}=\mathrm{mg}-\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{R}} \end{gathered}$$ 

 

According to the Newton’s third law, the force exerted by the car on the road is equal to in magnitude to the normal force exerted by the road on the car.

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

 

Therefore, the graph F  versus $v^2$  is a straight line with a slope $-\frac{m}{R}$ and y -intercepts mg.

 Calculate the square value of speed as shown in the following table.

The graph between square of speed and the force exerted on the body by car is as follows.

Therefore, the acceleration of each block is $2.212\mathrm{m}/{\mathrm{s}}^2$.

Step 4: (b) Find m and R from graph:

Since the car moves along the circular road, the net forces measured are the difference between the weight of the car and the centripetal force.

$F=mg-\left(\frac{m{v}^2}{R}\right)$ 

At zero velocity,

F=m g 

 

The best curve fir line from the graph is,

$\mathrm{F}=-18.124{\mathrm{v}}^2+8794.5$ 

At zero velocity,

F =8794.5 N 

 

Equate the value of F from the above two equations.

$$\begin{gathered} m=\frac{8794.5}{g} \\ =\frac{8794.5}{9.8} \\ =897 kg \end{gathered}$$ 

 

 

The slope of the car from the graph and the best fit curve is given by,

$$\begin{gathered} \mathrm{R}=\frac{\mathrm{m}}{18.124} \\ =\frac{897}{18.124} \\ =49.5 \mathrm{m} \end{gathered}$$ 

 

 

Therefore, the values of m and R are 897 kg , and  49.5 m, respectively.

Step 5: (c) Find the maximum speed of the car:

The car will lose contact with the road when the normal force n exerted by the road is zero. In order to find the maximum speed of the car put n=0 in equation $n=mg-m\frac{v^2}{R}$.

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

 

Substitute $9.8\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$ for g , and 49.5 m  for R in equation  ${\mathrm{v}}_{\max }=\sqrt{\mathrm{gR}}$.

$$\begin{gathered} {\mathrm{v}}_{\max }=\sqrt{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(49.5 \mathrm{m}\right)} \\ =22 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Therefore, the maximum speed of the car is 22 /s.