Question

A pendulum bob swings along a circular path on a smooth inclined plane as shown in Fig. 8.280, where m = 3 kg, l = 0.75 m, $\theta ={37}^0$. At the lowest point of the circle the tension in the string is T = 274 N. Take g = 10 $m{s}^{-1}$.
The tension in the string at the highest position is

• A. 166 N
• B. 67.56 N
• C. 68 N
• D. 152 N

Answer 1

The correct option is A 166 N

All formula will be same as in circular motion in vertical plane. Only replace, acceleration due to gravity (g) by $g=g\sin \theta$ Due to wedge.

$T_{Bottom}=T_{Top}+6mg$ (For vertical plane)

$T_{Bottom}=T_{Top}+6mgsin37$ (For our case)

$\therefore T_{Top}=274-6\times 3\times 10\times \sin 37$

$T_{Top}=165.67\ne 166N$