A simple pendulum of length 1.8 m having a bob of mass 2.2 kg is pulled aside a horizontal distance of 20 cm and released. Calculate the
i. velocity of bob at the lowest point of the swing.
ii. maximum kinetic energy.

Solution

Length of pendulum (l) = 1.8 m

Mass (m) = 2.2 kg

Amplitude (r) = 20 cm = 0.2 m

i. velocity at lowest point (v) =? (Mean position)

ii. Maximum K.E. =?

We know that,

[latex]T = 2\pi\sqrt{\dfrac{l}{g}}[/latex]

[latex]= 2\pi\sqrt{\dfrac{1.8}{10}}[/latex]

[latex]= 2.666\ s[/latex]

Again, [latex]\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{2.666} = 2.357\ rad/s[/latex]

Now, at mean position, [latex]v = \omega r = 2.357 \times 0.2 = 0.471\ m/s[/latex]

Now, maximum K.E. [latex]= \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\omega^2r^2[/latex]