A Classical Approach of Newtonian Mechanics

8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion

i.e.,

b =

l

. (8.84)
2
Moreover, the moment of inertia of the bat about a perpendicular axis passing

through one of its ends is

I =

1
M l^2 (8.85)
3
(this is a standard result). Combining the previous three equations, we obtain

JJ = −

1 −

3 h

!

J = −

1 −

h

!

J, (8.86)

where

2 l

h 0 =

h 0

2

l. (8.87)

3

Clearly, if h = h 0 then no matter how hard the ball is hit (i.e., no matter how

large we make J) zero impulse is applied to the hitter’s hands. We conclude that
the “sweet spot”—or, in scientific terms, the centre of percussion—of a uniform

baseball bat lies two-thirds of the way down the bat from the hitter’s end. If

we adopt a more realistic model of a baseball bat, in which the bat is tapered

such that the majority of its weight is located at its hitting end, we can easily

demonstrate that the centre of percussion is shifted further away from the hitter

(i.e., it is more that two-thirds of the way along the bat).

8.11 Combined translational and rotational motion

In Sect. 4.7, we analyzed the motion of a block sliding down a frictionless incline.

We found that the block accelerates down the slope with uniform acceleration

g sin θ, where θ is the angle subtended by the incline with the horizontal. In this

case, all of the potential energy lost by the block, as it slides down the slope, is

converted into translational kinetic energy (see Sect. 5 ). In particular, no energy

is dissipated.

There is, of course, no way in which a block can slide over a frictional surface
without dissipating energy. However, we know from experience that a round