Chapter 6, Problem 58A

a.

To determine

Acceleration of the car.

Answer to Problem 58A

Acceleration is equal to $9.64{\text{ m/sec}}^2$ .

Explanation of Solution

Given :

Mass of the car $m=615\text{ kg}$

Radius of the track $r=50\text{ m}$

Time taken to complete one revolution $t=14.3\text{ sec}$

Formula used:

Total distance traveled to complete one revolution is,

  $d=2\pi r$

Velocity is given as

  $v=\frac{d}{t}$

Centripetal acceleration is given by,

  $a_c=\frac{v^2}{r}$

Calculation:

Total distance traveled by car to complete one revolution is;

  $\begin{array}{c}d=2\times 3.14\times 50\\ =314\text{ m}\end{array}$

Now, velocity of the car is,

  $\begin{array}{c}v=\frac{d}{t}\\ =\frac{314}{14.3}\\ =21.95\text{ m/sec}\end{array}$

Centripetal acceleration will be,

  $\begin{array}{c}{a}_c=\frac{{21.95}^2}{50}\\ =9.64{\text{ m/sec}}^2\end{array}$

Conclusion:

Therefore, acceleration is equal to $9.64{\text{ m/sec}}^2$ .

b.

To determine

Force exerted by track on the tire.

Answer to Problem 58A

Force exerted by track on the tire is equal to 5926.17 N .

Explanation of Solution

Given :

Mass of the car $m=615\text{ kg}$

Radius of the track $r=50\text{ m}$

Time taken to complete one revolution $t=14.3\text{ sec}$

Formula used :

Rotational force is,

  $F_c=\frac{m{v}^2}{r}$

Here, m is mass, v is velocity and r is radius.

Calculation:

Force exerted by track on the tire will be equal to centripetal force.

Centripetal force is given by,

  $F_c=\frac{m{v}^2}{r}$

Therefore force is equal to,

  $\begin{array}{c}{F}_c=\frac{615\times {21.95}^2}{50}\\ =5926.17\text{ N}\end{array}$

Conclusion:

Therefore, the force exerted by track on the tire is equal to 5926.17 N.