Part A
Consider a ball placed on a ramp inclined at 30° above the horizontal. What is the acceleration of the ball's center of mass as it rolls down the ramp? Assume that the ball rolls without slipping.
Solution: We consider two methods.
Method 1
System: The ball is treated as a rigid body subject to external influences from the earth (gravity) and the ramp (friction and normal force).
Model: Point Particle Dynamics and [Angular Momentum and Torque].
Approach: The ball will translate and rotate as it rolls down the slope. The relevant free body diagram is shown at right. From this free body diagram we can construct Newton's 2nd Law for the ball's center of mass and also the angular version of Newton's 2nd Law for rotations of the ball about the center of mass.
The x-component of Newton's 2nd Law is:
\begin
[\sum F_
= mg - F_
= ma_
] \end
and the sum of torques about the center of mass is:
\begin
[ \sum \tau = F_
= I\alpha ] \end
We can eliminate the friction force from the two equations to find:
\begin
[ mg = ma_
+ I\alpha] \end
Now, because the ball is rolling without slipping, we can relate the angular acceleration to the linear acceleration of the center of mass:
\begin
[ \alpha R = a_
] \end
Substituting this expression allows us to express the acceleration as:
\begin
[ a = \frac
{1 + \frac
{\displaystyle mR^
}} ]\end
Using the result that the moment of inertia for a sphere is 2/5 m R2, we have:
\begin
[ a =\frac
{1+\frac
{\displaystyle 5}} = \frac
= \mbox
]\end
Note that we have been given no information whatever about the mass or radius of the ball! The acceleration is independent of the mass and the radius.
Method 2
System: The ball as a rigid body subject to external influences from the earth (gravity) and the ramp (normal force and friction).
Model: [Angular Momentum and Torque].
Approach: Instead of breaking the ball's motion into translation of the center of mass and rotation about the center of mass, we could instead choose to simply consider the movement as angular momentum about a fixed axis.