物理問題!angular speed

用戶91254

2013-06-21 7:56 am

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.600kg , traveling perpendicular to the door at 14.0m/s just before impact.

Find the final angular speed of the door.

我試過用p=mv搵mud既momentum,之後用v=p/m搵door既speed,再用v=rw搵angular speed. 但答案0.37 rad/s並唔正確。

我亦試過用L_initial=L_door + L_mud既approach去搵答案,計到0.99 rad/s都唔正確。

請高手指教。

回答 (1)

天同

2013-06-21 8:48 am

✔ 最佳答案

Moment of inertia of door about the hinged side = (45 x 1^2)/3 kg.m^2/s
= 15 kg.m^2/s
Moment of inertia of mud = 0.6 x 0.5^2 kg.m^2/s = 0.15 kg.m^2/s
Hence, total moment of inertia of door and mud = (15 + 0.15) kg.m^2/s = 15.15 kg.m^2/s

Initial angular momentum of mud = 0.6 x 14 x 0.5 kg.m^2/s^2 = 4.2 kg.m^2/s^2
By conservation of angular momentum,
4.2 = 15.15w
where w is the final angular speed of the mud and door
w = 4.2/15.15 rad/s = 0.277 rad/s

2013-06-21 17:17:20 補充：
Sorry, I have used the wrong unit for Moment of Inertia. It should be "kg.m^2" instead of "kg.m^2/s".

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