Consider an analog clock that is working perfectly (keeping perfect time). It ha
ID: 2303315 • Letter: C
Question
Consider an analog clock that is working perfectly (keeping perfect time). It has an hour hand, a minute hand and a second hand. The Hour hand is 8.0 cm long with a mass of 8.00 g, the Minute hand is 11.0 cm long with a mass of 14.00 g and the second hand is 11.0 cm long with a mass of 6.00 g. All of the hands of the clock can be modeled as uniform rods rotating around one end so that the moment of inertia is given by (ML2)/3. What is the TOTAL angular momentum of the hands on the clock? Note: before you ask about the angular velocity of the hands, think about how they would move in a real clock.
Explanation / Answer
Angular velocity of Hour hand = (pie/6) radians/ hour = (pie/6) radians / 60 minutes
= (pie/6) radians/ 3600 seconds = (pie/6)/3600 rad/sec
Angular momentum of Hour hand = I x w = (ML2/3) x ((pie/6)/3600))
= (0.008 * (0.08)2/3 )x(1.45 x 10-4) = 2.48 x 10-9 Kg/m2
Angular velocity of minute hand = 2 pie radians/ hour = 2 pie radians/ 3600 sec = 0.0017 rad/sec
Angular momentum of minute hand = (ML2/3) x (0.0017) = 9.59 x 10-8 Kg/m2
Angular velocity of second hand = 2 pie rad/ 1 minute = 2 pie rad/ 60 sec = 0.104 rad/sec
Angular momentum of second hand = (ML2/3) x (0.104) = 2.51 x 10-6 Kg/m2
Total angular momentum = 2.48 x 10-9 Kg/m2 + 9.59 x 10-8 Kg/m2 + 2.51 x 10-6 Kg/m2
= 2.1 x 10-6 Kg/m2