A playground merry-go-round has radius 2.60m and moment of inertia 2700kg⋅m 2 about a vertical axle through its center, and it turns with negligible friction.
Q1 ： A child applies an 20.5N {\rm N} force tangentially to the edge of the merry-go-round for 25.0s {\rm s} . If the merry-go-round is initially at rest, what is its angular speed after this 25.0s {\rm s} interval?
Q2： How much work did the child do on the merry-go-round?
Q3： What is the average power supplied by the child?
UNIVERSITY PHYSICS
2013-10-25 9:07 pm
回答 (1)
2013-10-25 10:23 pm
✔ 最佳答案
1. Angular acceleration of merry-go-round= (20.5 x 2.6)/2700 rad/s^2 = 0.01974 rad/s^2
Angular speed after 25 s
= 0.01974 x 25 rad/s = 0.4935 rad/s
2. Increase of kinetic energy of merry-go-round
= (1/2).(2700).(0.4935^2) J
= 328.8 J
Hence, work done by child = 328.8 J
3. Average power
= 328.8/25 J/s
= 13.15 watts
收錄日期: 2021-04-21 22:39:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131025000051KK00076