UNIVERSITY PHYSICS （20 MARKS）

DAMON

2013-10-25 5:46 am

A solid wood door 1.00 m {\rm m} wide and 2.00 m {\rm m} high is hinged along one side and has a total mass of 45.0kg kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500kg kg , traveling perpendicular to the door at 14.0m/s m/s just before impact.

Q1 Find the final angular speed of the door.
Q2 Does the mud make a significant contribution to the moment of inertia?

回答 (1)

天同

2013-10-25 7:21 am

✔ 最佳答案

1. Moment of inertia of door about the hinge
= (45 x 0.5^2)/3 kg.m^2 = 3.75 kg.m^2

Moment of inertia of mud about the door hinge
= 0.5 x 0.5^2 kg.m^2 = 0.125 kg.m^2

Initial angular momentum of mud
= (0.125 x 14/0.5) kg.m^2/s = 3.5 kg.m^2/s

Angular momentum of door and mud after imapct
= (3.75 + 0.125)w where w is the angular velocity of door after impact

By conservation of angular momentum,
3.5 = (3.75+0.125)w
w = 0.903 rad/s

2. No. The momentum of inertia of the mud contributes to only about 3% of that of the door.

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