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Question
A uniform rod of mass 300 g and length 50 cm rotates at a uniform angular speed of 2 rad/s about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of rotation, (b) the speed of the centre of the rod and (c) its kinetic energy.
Solution
Given
Mass of the rod = m = 300 gm
Length of the rod = l = 50 cm
Angular velocity of the rod = \[\omega =2\text{ rad/s}\]
(a) Moment of inertia about one end of the rod,
\[I = \frac{m L^2}{3} = \frac{\left\{ 0 . 3 \times \left( 0 . 50 \right)^2 \right\}}{3}\]
\[ \Rightarrow I = \frac{0 . 25}{10} = 0 . 025 kg - m^2\]
Angular momentum about that point,
\[L = I\omega\]
L = 0.025 × 2 = 0.05 kg-m2/s
(b) Speed of the centre of the rod,
\[v = \omega r = 2 \times \frac{0 . 50}{2} = 0 . 5 m/s\]
(c) Kinetic energy,
\[K = \frac{1}{2}I \omega^2\]
\[\Rightarrow K = \frac{1}{2} \times 0 . 025 \times 2^2 = \frac{1}{2} \times 0 . 025 \times 4 = 0 . 05\text{ joule}\]
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