Question: 7. A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v_{0} as shown in figure. If the string gets slack at some point P making an angle \theta from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v_{0} is:
(1) (\sin \theta)^{1 / 2}
(2) \left(\frac{1}{2+3 \sin \theta}\right)^{1 / 2}
(3) \left(\frac{\cos \theta}{2+3 \sin \theta}\right)^{1 / 2}
(4) \left(\frac{\sin \theta}{2+3 \sin \theta}\right)^{1 / 2}
Answer: Option (4)
Explanation:
Let the angle between the string and the downward vertical at point P be \phi. From the geometry of the figure, when the string makes angle \theta with the horizontal,
\phi=90^\circ+\theta.1. Height gained by the bob when it moves from the lowest point to P is
h=l\bigl(1-\cos\phi\bigr)=l\bigl(1-\cos(90^\circ+\theta)\bigr)=l\bigl(1+ \sin\theta\bigr).2. Use energy conservation between bottom (speed v_0) and point P (speed v):
\tfrac{1}{2}m v_0^2=\tfrac{1}{2}m v^2 + m g h \implies \tfrac{1}{2}m v_0^2=\tfrac{1}{2}m v^2 + m g l(1+\sin\theta).3. The string goes slack when the tension becomes zero. At that instant the required centripetal force m v^2 / l is provided entirely by the appropriate component of gravity. This gives
\dfrac{m v^2}{l}=m g \sin\theta \quad\Rightarrow\quad v^2=g l\sin\theta.4. Substitute v^2 from step 3 into the energy equation from step 2:
\tfrac{1}{2}m v_0^2=\tfrac{1}{2}m (g l\sin\theta) + m g l(1+\sin\theta) \implies v_0^2=g l\bigl(2+3\sin\theta\bigr)- Therefore the required ratio is
So
\dfrac{v}{v_0}=\left(\dfrac{\sin\theta}{2+3\sin\theta}\right)^{1/2}.This matches Option (4).