Rolling Down an Incline 1. Once upon a time, we showed that an object rolling down an incline of angle θ (from the horizontal) would have acceleration a = g sin θ. Strictly speaking, this isn’t quite true. Now that we know a bit about rotational inertia, let’s find the actual acceleration a of a radially symmetric object (could be a disc, cylinder or sphere, and could be hollow or solid) of mass m and radius R with rotational inertia I around the center of mass.

Putting it all together…and solving for a:

Aside: Show that the acceleration you found above gives the expected result (a = g sin θ) when we ignore rotational inertia: that is, let I → 0.

Find vf for the object rolling down an incline of height h starting from rest, using an your expression for a boxed above (which includes rotational effects) and an equation of motion. (Get vf in terms of m, h, I and R.)

Aside: What do you expect the expression for vf to simplify to, in the absence of rotational effects? Does it?

Draw all the forces acting on the object.Translational Dynamics: Start by applying Newt’s 2nd Law:

Notice that you have 2 unknowns, fs and a. Use rotational dynamics to get another equation involving fs and a. Apply the torque equation around the center of mass…Solve for fs in terms of I and a.

a =

vf =

fs =

m, IR

θ

1

2. OK, now let’s find vf (including rotational effects) the easy way. Use the energy approach to find vf (again starting from rest) in terms of m, h, I and R. Don’t forget to label the diagram, including the reference level. Include rotational kinetic energy.

Which way do you like better, the forces or the energy approach?

m, I

h

R

θ

2