Physics-rotation and inertia

2007-04-02 6:36 am

A uniform rod of 3m is suspended at one end so that it can move about an axis perpendicular to its length. The moment of inertia about the end is 6kgm^2 and ths mass of the rod is 2kg. If the rod is initially horizontal and then released, find the angular velocity of the rod when
a) it is inclined at 30degrees to the horizontal,
b) it reaches the vertical.

回答 (1)

魏王將張遼

2007-04-02 8:24 pm

✔ 最佳答案

(a) The scenario can be illustrated by the diagram below:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/April07/MI1.jpg

Therefore, the rotational k.e. of the rod should be equal to its loss in gravitational p.e. which is given by:
mgh = 2 × 10 × 0.75 = 15J
Also by the formula of rotational k.e.:
(1/2)Iω2 = 15 where I and ω are moment of inertial and angular velocity respectively.
(1/2) × 6 × ω2 = 15
ω2 = 5
ω = 2.24 rad/s
(b) When the rod is vertical, its loss in gravitational p.e. is:
mgh = 2 × 10 × 1.5 = 30 J
Also by the formula of rotational k.e.:
(1/2)Iω2 = 30 where I and ω are moment of inertial and angular velocity respectively.
(1/2) × 6 × ω2 = 30
ω2 = 10
ω = 3.16 rad/s

參考: My physics knowledge

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