English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

I get an answer of 11.2 rev/min and this could be the problem to me obtaining the proper answer or it could be my conversions. please help!

A playground merry-go-round of radius 2.00 m has a moment of inertia I = 230 kgm2 and is rotating about a frictionless vertical axle. As a child of mass 25.0 kg stands at a distance of 1.00 m from the axle, the system (merry-go-round and child) rotates at the rate of 14.5 rev/min. The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge?

2006-11-01 15:00:24 · 4 answers · asked by Anonymous in Science & Mathematics Physics

4 answers

One revolution = 2pi radians. One minute = 60 seconds. So 1 rev/min = 2pi/60 rad/s. Multiply 11.2 rev/min by 2pi/60 to get 1.17286 rad/s.

2006-11-01 15:04:14 · answer #1 · answered by Amy F 5 · 0 0

That seems about right: the rotational speed has decreased by less than half as the child moves twice the distance from the centre.

Does the moment of inertia of the merrygoround include the child?

(2 pi radians) / 60 is 1 rpm

2006-11-01 23:07:37 · answer #2 · answered by ? 3 · 0 0

1 rev is 2 pi radians. 1 min is 60 seconds

2006-11-01 23:01:46 · answer #3 · answered by arbiter007 6 · 0 0

use dimensional analysis

14.5 rev 1 min 2 pi rad
------- * ----------- * -----------------
1 min 60 seconds 1 rev

2006-11-02 00:32:30 · answer #4 · answered by ioedvhsd 1 · 0 0

fedest.com, questions and answers