Question

The bob of a simple pendulum is given minimum velocity in horizontal direction when it is at lowest position, such that the bob describes vertical circle of radius equal to the length of pendulum. If the velocity of bob at an angular displacement $60°$ from the lowest point is $8m/s$, then it's velocity at an angular displacement $60°$ from highest point is:

• A. $2\sqrt{2}m/s$
• B. $4m/s$
• C. $4\sqrt{2}m/s$
• D. $6m/s$

Answer 1

Answer: (C)

$4\sqrt{2}m/s$

Apply WET between pt. A and B,

$mg2R(1-cos60)=\frac{-m}{2}{v}^2+\frac{m}{2}{8}^2$
$\frac{v^2}{2}=32-gR$ _____ (I)
Now, WET between A and C,
$mg\frac{3R}{2}=\frac{1}{2}m64-\frac{m}{2}gR$

$2gR=\frac{64}{2}$
$R=1.6m$
Put R in (I), we get $v=\sqrt{32}m/s$
$=4\sqrt{2}m/s$