A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v₀ as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v₀ is:

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(2) ((sin θ)/(2+3 sin θ))^1/2. Given: the bob starts at the lowest point with speed v₀, and the string goes slack at P, an angle θ above the horizontal through the centre. Idea: slack string means zero tension, so at P the whole centripetal force comes from the component of the weight along the string. At P: mg sin θ=(mv²)/l, so gl=(v²)/(sin θ). P stands a height l+l sin θ above the lowest point. Energy: 1/2v₀²=1/2v²+gl(1+sin θ). Substituting gl: (v₀²)/2=(v²)/2+(v²)/(sin θ)+v², so v₀²=v²(3+2/(sin θ)). v/(v₀)=((sin θ)/(2+3 sin θ))^1/2.

A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v₀ as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v₀ is:

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