A pulley with a rotational inertia of 1.0 x 10-3 kg m^2 about its axle and a radius of 10 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.50t + 0.30t^2, with F in newtons and t in
seconds. The pulley is initially at rest. At t = 3.0 s what are (a) its angular acceleration and (b) its angular speed?
ans: a)4.2 x 10^2 rad/s, b) 5.0 x 10^2 rad/s
thanks
newton's 2nd law (rotation)
2011-10-19 5:45 am
回答 (1)
2011-10-19 7:10 am
✔ 最佳答案
(a) Force at t = 3s is= (0.5 x 3 + 0.3 x 3^2) N = 4.2 N
Torque = 4.2 x (10/100) N.m = 0.42 N.m
Hence, angular acceleration = 0.42/10^-3 rad/s^2 = 420 rad/s^2
(b) Angular velocity = integral[ (angular acceleration).dt
the limit of integration is from t = 0s to t = 3s
Hence, angular velocity = integral [(0.5t + 0.3t^2].dt with t = 0s to 3 s
= [0.5t^2/2 + 0.3t^3/3] with t = 0s to t = 3s
= (0.5.(3)^2/2 + 0.3.(3)^3/3) rad/s = 4.95 rad/s
2011-10-19 09:00:12 補充：
Sorry that in (b), I forgot to divide the integral by the rotational inertia of the pulley.
Hence, 4.95 (in unit of N.s) is in fact the impulse. Dividing the impulse by 10^-3 kg-m^2 gives the angular speed. Thus, the answer is 495 rad/s.
2011-10-19 15:18:41 補充：
I made a mistake in (b) again. The correct solution is as follows:
use: impulse = torque x time = change of angular momentum
hence, impulse = integral[(0.5t + 0.3t^2](0.1).dt with t = 0s to 3 s
(cont'd)...
2011-10-19 15:19:26 補充：
since the value of the integral = 0.495 N.m.s
i.e. 0.495 = change of angular momentum
but angular momentum = rotational inertia x angular velocity
hence, 0.495 = 10^-3 x (angular velocity)
i.e. angular velocity = 0.495/10^-3 rad/s = 495 rad/s
收錄日期: 2021-04-29 17:44:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111018000051KK00859