Physics Rolling

Green1114

2010-12-27 6:48 pm

A uniform disk is placed on an inclined plane at 30 degree above the horizontal. The axis of the disk is horizontal.What is the minimum value of the coefficient of friction if the disk is to roll down without slipping ? The moment of inertia of a uniform disk is 1/2MR^2.

ans 0.19

回答 (1)

天同

2010-12-27 7:22 pm

✔ 最佳答案

Let ff be the frictional force. Consider the rotational motion,
Ff x R = [MR^2/2] x (angular acceleration)where R is the disk radius, and M is the disk mass
But angular acceleration of the disk = a/R, where a is the linear acceleration if the disk rolls without slipping
hence, Ff x R = [MR^2/2]/(a/R)
u.Mgcos(30) x R = MR^2/2 x a/Rwhere u is the coefficient of friction and g is the acceleration due to gravitySimplifying, we have, ugcos(30) = a/2 ----------------- (1)Consider the linear motion,
Mg.sin(30) - Ff = Ma
i.e. Mg.sin(30) - uMgcos(30) = Ma
g.sin(30) - ugcos(30) = a
substituting value of a from (1)
g.sin(30) - ugcos(30) = 2ug.cos(30)
u = [tan(30)]/3 = 0.19

👍 0 👎 0