Chapter 5, Problem 82GP

(a)

To determine

To find: The normal force acting on different points A, B and C.

Explanation of Solution

Introduction:

The resultant of normal reaction and weight of cool provides the centripetal acceleration.

At point A the expression for normal is

  $N=mg-\frac{m{v}^2}{r}$

So normal is minimum at A

At point B the normal is

  $N=mg$

So normal at B is more than A

At point C the normal is

  $N=mg+\frac{m{v}^2}{r}$

So normal at C is maximum

i.e., $N_A<N_B<N_C$

Conclusion:

Therefore normal at A is least and at C it is maximum.

(b)

To determine

To find: That where the driver will feel heaviest.

Explanation of Solution

Introduction:

The resultant force of normal and weight of driver will provide the centripetal acceleration.

The driver will feel heaviest at point C as the normal reaction at C is given as

  $N=mg+\frac{m{v}^2}{r}$

The driver feels heavy when the normal force is larger than his weight.

The driver feels light when the normal force is smaller than his weight.

Then at point A he feels light.

At point B he feels his weight.

At point C he feels heavy.

Conclusion:

Therefore driver will feel heaviest at ‘C’.

(c)

To determine

To find: The maximum safe speed at A.

Answer to Problem 82GP

  $v_{\max }=\sqrt{Rg}$

Explanation of Solution

Given Data:

Radius $=R$

Normal $\left(N\right)=0$

Formula Used:

  $N=mg-\frac{m{v}^2}{R}$

Calculation:

Substituting the value

  $0=mg-\frac{m{v}^2}{R}$

i.e., $mg=\frac{m{v}^2}{R}$

  $v=\sqrt{Rg}$

Conclusion:

Therefore the maximum safe speed is $v=\sqrt{Rg}$ .