A playground merry-go-round has radius 2.00 m and moment of inertia 2300
kg . m^2
about a vertical axle through its center, and it turns with negligible friction.
1. A child applies an 21.0N force tangentially to the edge of the
merry-go-round for 17.0s . If the merry-go-round is initially at rest,
what is its angular speed after this 17.0s interval?
unit: rad/s
2. How much work did the child do on the merry-go-round?
unit: J
3. What is the average power supplied by the child?
unit: W
merry-go-round
2012-11-02 10:47 pm
回答 (1)
2012-11-02 11:29 pm
✔ 最佳答案
Torque applied by the child to the merry-go-round = 21 x 2 Nm = 42 N.mAngular acceleration of merry-go-round = 42/2300 rad/s^2 = 0.01826 rad/s^2
Use equation of motion: v = u + at
with u = 0 rad/s, a = 0.01826 rad/s^2, t = 17 s, v =?
v = 0.01826 x 17 rad/s = 0.3104 rad/s
Increase in kinetic energy of meryy-go-round
= (1/2).(2300).(0.3104)^2 J = 110.8 J
Hence, work done = 110.8 J
Average power = 110.8/17 w = 6.52 w
收錄日期: 2021-04-29 17:49:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121102000051KK00220