A Classical Approach of Newtonian Mechanics

8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion




centre of mass

v

b

vt

vt

cylinder

surface

Figure 83: A cylinder rolling over a rough surface.

kinetic energy: Kt = (1/2) M v^2 , where v is the cylinder’s translational velocity;

and, secondly, rotational kinetic energy: Kr = (1/2) I ω^2 , where ω is the cylin-

der’s angular velocity, and I is its moment of inertia. Hence, energy conservation

yields

M g h =

1

M v^2 +

1

I ω^2. (8.90)

2 2

Now, when the cylinder rolls without slipping, its translational and rotational

velocities are related via Eq. (8.89). It follows from Eq. (8.90) that

v^2 =

2 g h

1 + I/M b^2

. (8.91)

Making use of the fact that the moment of inertia of a uniform cylinder about

its axis of symmetry is I = (1/2) M b^2 , we can write the above equation more

explicitly as

v^2 =

4

g h. (8.92)

3

Now, if the same cylinder were to slide down a frictionless slope, such that it fell

from rest through a vertical distance h, then its final translational velocity would

satisfy

v^2 = 2 g h. (8.93)

A comparison of Eqs. (8.92) and (8.93) reveals that when a uniform cylinder rolls

down an incline without slipping, its final translational velocity is less than that

obtained when the cylinder slides down the same incline without friction. The

reason for this is that, in the former case, some of the potential energy released