- #1

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When a circular section object (cylinder, sphere of radius r) rolls down an inclined plane (at angle theta) it experiences both linear and angular acceleration due to gravity. I have read multiple times that:

mg sin(theta) - friction = m a

where friction = Torque/r

From there on, everything is clear:

moment of inertia * angular acceleration = friction * radius of ball = Torque

a= angular acceleration * r

and finally:

a = (mg sin(theta)) / (m + moment of inertia * r^-2);

friction = moment of inertia * a * r^-2

But, going back to the first equation there is a question that has always profoundly bothered me:

mg sin(theta) - friction = m a

Why does friction appears here? Isn't it a force applied at distance r in the contact point with the surface and directed down, parallel to it?

I understand that friction is needed to avoid slipping so it should exist. Moreover, that the weight must be reduced by some force to represent the final rotational energy. I mean, if we had just

mg sin(theta) = m a

and presume rotation then we would have the same final kinetic energy as a pure traslation with no friction (slipping) but with an impossible extra rotational energy.

Summarizing, I recognize that:

At centre of mass:

mg sin(theta) - Unknown force = m a

At contact point:

Torque = moment of inertia * angular acceleration = friction * radius of ball

But how do we get Unknown force = friction?

Thanks in advance!