Question

A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular arc of radius $R$ and has a total length $L$ . Suppose the motorcycle starts from the highest point.
(a) What can its maximum velocity be for which the contact with the road is not broken at the highest point?
(b) If the motorcycle goes at speed $\frac{1}{\sqrt{2}}$ times the maximum found in part $(a)$ where will it lose the contact with the road?
(c) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?

Accepted Answer

A motorcycle is moving on the over bridge of length $L$ .

(a) When the motorcycle is at the highest point,

where, $F_{c} = \frac{m v^{2}}{R}$ (centrifugal force).

Now, the limiting condition, $N = 0,$
and if normal force $= 0$ ,
then, $F_{c} = m g = \frac{m v^{2}}{R}$ .

$∴ v = \sqrt{R g} \dots (1)$

Therefore, $v$ must be less than $\sqrt{R g}$ such that the motorcycle does not break its contact with the bridge at the highest point.

$v < \sqrt{R g}$ .

(b) To calculate the length of the path travelled, when the speed of the particle is $\frac{1}{\sqrt{2}}$ times.

Let the motorcycle be at some general point, let's say $θ$ angle form from the vertical and at that point it loses contact.

The forces acting on the block will be $m g$ , which is resolved into its two components $m g \cos θ$ and $m g \sin θ$ , the normal force and the centrifugal force $F_{c}$ , away from the centre of the bridge.

where, $F_{c} = \frac{m v^{2}}{R}$ .

Now, according to the question, the velocity at that point, $\frac{1}{\sqrt{2}}$ .

Substituting the value of maximum velocity from equation $(1)$ ,

$v^{'} = \frac{1}{\sqrt{2}} \sqrt{R g}$ .

Now, at this point, to break contact, $N = 0$ .

$∴ m g \cos θ = \frac{N + m {(v^{'})}^{2}}{R}$ by balancing forces.

Substituting $N = 0$ ,

$m g \cos θ = \frac{m {(v^{'})}^{2}}{R}$

$\Rightarrow \cos θ = \frac{{(v^{'})}^{2}}{g R} \dots (2)$

According to the question, speed at this point is $\frac{1}{\sqrt{2}}$ times of the maximum speed.
${(v^{'})}^{2} = \frac{R g}{2}$ , Substitute in equation $(2)$ ,

and $\cos (θ) = \frac{R g}{2 R g}$ ,

$\cos (θ) = \frac{1}{2}$ .

$∴ θ = 60^{°} = \frac{π}{3}$ .

Now, the length of bridge $= L$ and radius $= R$ .

Angle is equal to ratio of arc to radius,

$L = θ \times R$ ,

$L = \frac{π}{3} \times R = \frac{π R}{3}$ .

Therefore, at this length, the particle will lose contact when the speed of the particle is $\frac{1}{\sqrt{2}}$ times the maximum speed.

(c) In this part, it is mentioned that the particle does not break contact at any point of the bridge.

Now, for the last point, let the motorcycle make an angle with respect to vertical in the bridge.

Balancing the forces, $m g \cos (α) = N + \frac{m v^{2}}{R}$ .

Substituting $N = 0$ and obtaining an expression for velocity,

$v = \sqrt{R g \cos (α)}$

Also, length of the bridge, $L = R (2 α)$

and $α = \frac{L}{2 R}$ and putting $α$ in the above expression for velocity.

$v = \sqrt{R g \cos \frac{L}{2 R}}$ .

Important Questions on Circular Motion

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