Question

A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in figure. What will be the trajectory of the particle if the string is cut at

1. Point B?
2. Point C? 
3. Point X?

Answer 1

According to the situation shown above when a bob of mass m is whirled into a vertical circle, the required centripetal force is obtained from the net force towards the center at any point of time in the string. Tension in the string is variable and it is always towards center. But the gravitational force on the bob is always towards center. The speed of the body will be different at different points. So the equations of dynamical equilibrium (F_c = ma_c, F_t = ma_t) must be satisfied at all the points. Let when the string makes an angle of 9 with vertical, the speed of mass is v.

Apply Newton’s law perpendicular to the string:

Mg sin = ma, ⇒ a,= g sin

The above equation gives tangential acceleration as a function of the angle. At lowest point = 0° and at highest point = 180°. So at both points sin 9 = 0. Hence a, = 0 at both points L and H.

At point M, = 90°, then a₁ = g. It is the maximum value of a_t

Apply Newton’s law along the string: T – mg cos = ma_c

or  T = mgcos + mv²/r  ......(i)

As the body goes up, its velocity will go on decreasing and angle θ will go on increasing. The maximum speed of the body will be at the lowest point L and minimum at the highest point H. Then from the above relation we can find that tension will be maximum at the lowest point and minimum at the highest point.

Tension at the lowest point (θ – 0°, v = v_L): T_L = `mg + m v_L^2/r`  ......(ii)

Tension at the highest point (θ – 180°, v = v_H): T_H = `mg + m v_H^2/r`  ......(iii)

When a string is cut, the tension in string becomes zero and centripetal force is not provided. Hence, the bob tends to move in along the direction of its velocity.

1. If the string is cut at any point, then velocity of the body of mass m is along the tangent to the circle. A tangent at point B is vertically downward so the trajectory of the particle is a straight line.
2. The tangent at point C is horizontally towards the right.
   So the trajectory of the particle is the parabola.
3. The tangent at point X makes some angle with the horizontal. Again bob will follow a parabolic path with a vertex higher than C.