A Classical Approach of Newtonian Mechanics

8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion

parallel to the slope, yields

M ̇v = M g sin θ − f, (8.95)

where ̇v is the cylinder’s translational acceleration down the slope.

Let us, now, examine the cylinder’s rotational equation of motion. First, we

must evaluate the torques associated with the three forces acting on the cylin-
der. Recall, that the torque associated with a given force is the product of the

magnitude of that force and the length of the level arm—i.e., the perpendicular

distance between the line of action of the force and the axis of rotation. Now, by

definition, the weight of an extended object acts at its centre of mass. However,

in this case, the axis of rotation passes through the centre of mass. Hence, the

length of the lever arm associated with the weight M g is zero. It follows that

the associated torque is also zero. It is clear, from Fig. 84 , that the line of action

of the reaction force, R, passes through the centre of mass of the cylinder, which

coincides with the axis of rotation. Thus, the length of the lever arm associated
with R is zero, and so is the associated torque. Finally, according to Fig. 84 , the

perpendicular distance between the line of action of the friction force, f, and the

axis of rotation is just the radius of the cylinder, b—so the associated torque is

f b. We conclude that the net torque acting on the cylinder is simply

τ = f b. (8.96)

It follows that the rotational equation of motion of the cylinder takes the form,

I ω ̇ = τ = f b, (8.97)

where I is its moment of inertia, and ω ̇ is its rotational acceleration.

Now, if the cylinder rolls, without slipping, such that the constraint (8.89) is

satisfied at all times, then the time derivative of this constraint implies the follow-

ing relationship between the cylinder’s translational and rotational accelerations:

̇v = b ω ̇. (8.98)

It follows from Eqs. (8.95) and (8.97) that

̇v =

g sin θ

1 + I/M b^2

, (8.99)