This article is a continuation of “Teaching Rolling Motion”. A description of the notation I use can be found in still another article “Teaching Rotational Dynamics”. In the rolling motion article, I analyzed the motion of a sphere rolling down an incline. I assumed no slipping 롤대리 and then determined what condition that assumption imposed on the coefficient of static friction. This article describes a situation for which that condition is not satisfied – so slipping occurs.
Problem. A sphere rolls down an incline steep enough that slipping occurs. What is the linear acceleration of the sphere’s center of mass? What is the sphere’s angular acceleration around the axis through its center of mass? The mass of the sphere is M, its radius is R, its moment of inertia around the center of mass is Icm = 2(MR**2)/5, the incline angle is th, and the coefficient of kinetic friction between the sphere and incline is Uk.
Analysis. We’ll use an inertial reference system with an x axis pointing down the incline and a y axis perpendicular to the incline. The incline exerts a normal force N and a frictional force f on the sphere. The weight of the sphere has components MGsin(th) along the incline and MGcos(th) perpendicular to the incline. With the help of a free-body diagram, we apply the equations of motion:
…………………………….Newton’s Second Law…………………………Rotational Equation of Motion
……………….SUM(Fx) = MAx…………………SUM(Fy) = May………………SUM(Text) = IcmA
…………MGsin(th) – f = MAx………………….N – MGcos(th) = 0…………………fR = IcmA
Since the hoop slips, f = UkN. From the y equation, N = MGcos(th). Combining these two, we find that f = UkMGcos(th). When this is substituted into the x equation, we have
………………………………………..MGsin(th) – UkMGcos(th) = Max,
so………………………………………..Ax = G(sin(th) – Ukcos(th)).
Notice that we don’t need the rotational equation to determine the acceleration. Also notice that this is the same acceleration a box sliding down the incline would have. For both of them, the net force down the incline is Mgsin(th) and the net force up the incline is UkMGcos(th). But we do need the rotational equation of motion to determine the angular acceleration; it is
……………………A = fR/Icm = (UkMGcos(th))R/(2MR**2)/5) = 5UkGcos(th)/2R.
We see that the slipping case results in an easier problem solution than that for no slipping. However, even for slipping the condition imposed on the coefficient of static friction must be determined with the no-slipping problem.