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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 $(a)$ point $B$ ? $(b)$ point $C$ ? $(c) $ point $X$ ?
Solution
When bob is whirled into a vertical circle, the required centripetal force is equal to the tension in the string. When string is cut, tension becomes zero and centripetal force also hence, bob start to move in a straight line path along the direction of its velocity.
$(a)$ At point $B$, the velocity of $B$ is vertically downward, therefore, when string is cut at $B$, bob moves vertically downward.
$(b)$ At point $C$, the velocity is along the horizontal towards right, therefore, when string is cut at $C$, bob moves horizontally towards right.
Also, the bob moves under gravity simultaneously with horizontal uniform speed. So it traversed on a parabolic path with vertex at $\mathrm{C}$.
$(c)$ At point $\mathrm{X}$, the velocity of the bob is along the tangent drawn at point $\mathrm{X}$, therefore when string is cut at point $\mathrm{C}$, bob moves along the tangent at that point $\mathrm{X}$.
Also, the bob move under gravity simultaneously with horizontal uniform speed. So, it taversed on a parabolic path with vertex higher than $\mathrm{C}$.
