Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end.
Solution 39E Mathematically, the equation for angular momentum is given by L = I , where I is the moment of inertia of the second hand and is the angular velocity of the hand. Now, the hand is a rod rotating with constant angular velocity about one end. Therefore, 2 its moment of inertia is I = ML3 , where M is the mass of the hand and L is its length. Given, M = 6.00 g = 6.00 × 10 3kg = 0.006 kg L = 15.0 cm = 0.15 m 2 2 The moment of inertia of the hand is I = 0.006 kg×(0.15) m= 4.5 × 10 5kg.m 2 3 The second hand takes 60 s to make one complete revolution, therefore its angular speed is, = 2/T = 2 × 3.14/60 rad/s = 0.104 rad/s Now, substituting the values if I and in L = I , we get Angular momentum L = 4.5 × 10 5kg.m × 0.104 rad/s = 0.468 × 10 5 kg.m /s 6 2 L = 4.68 × 10 kg.m/s This is the required value of angular momentum of the second hand.