Q. The bob of a pendulum at rest is given a horizontal velocity of $\sqrt {7gl}$ , where is $l$ length of string of pendulum. The tension when bob will reach the highest point of its circular motion is (m is mass of bob)
Solution:
From energy conservation
$\frac{1}{2}mv^2_1+mg.2l=\frac{1}{2}mv^2$
$\frac{1}{2}mv^2_1=\frac{7}{2}mgl-2mgl$
$mv^2_1=3mgl...(1)$
Now,
T = $\frac{mv^2_1}{l}-mg=\frac{3mgl}{l}-mg$
T = 2mg
