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

2 answers

Let's assume the situation is that a weight sits on a rotating horizontal turntable and there is frictional contact. Then the centripetal acceleration is w^2*r, where w is the angular rate in rad/sec. At the angular rate ws at which the weight slips, the accel. is ws^2*r and the centripetal force is m*ws^2*r. The normal (downward) force is m*g. The coefficient of friction is thus m*ws^2*r/m*g = ws^2*r/g.

2006-10-22 06:23:45 · answer #1 · answered by kirchwey 7 · 0 0

The question is overly simplistic. I CAN solve whatever problem you're working on, but I need to see the whole problem.

Messenger me on yahoo, my ID is
fortitudinousskeptic

We'll discuss it.

2006-10-22 06:06:49 · answer #2 · answered by Anonymous · 0 0

fedest.com, questions and answers