Even-numbered problems for this week

Numerical answers for the even-numbered problems this week:

Due tonight:
15.24 x < 0.23m
15.56 T~2.0s; 18.4 N-m/rad

Due Thursday:
16.34 10W; 20W; 40W; 26W; 0
16.60 0.846 g/m

Random hints:

15.111 First, for the linear acceleration, you can always just apply the center of mass equations:

m a_com = F_net

Even though rotation is involved, the center of mass stuff still works.

For the second part, you want to find the net torque to get angular acceleration. You have a force F applied some distance x from the point at which the bat is held (point O), and that's it. The distance x is 2/3 of the length of the bat, given. This torque must equal the angular acceleration (alpha) times the moment of inertia about the end of the bat (so (1/3)ml^2).

Finally, if the bat is rotating after the hit, the angular acceleration implies that the tips of the bat have an additional acceleration in the tangential direction. If no one were holding the bat (or if it is held lightly, as the problem states), the rotation will just be about the center of mass. The tangential acceleration at the ends of the bat has to be (alpha)(L/2) in that case. In the end, this acceleration due to rotation acts in the opposite direction compared to the acceleration of the impact of the ball itself. At the "sweet spot" these two cancel exactly, and you shouldn't feel anything when you hit the ball - if you hit anywhere else, it 'stings' because of the net acceleration at your end of the bat.

In other words, there is (say) a forward acceleration due to the force of the ball hitting the bat, and a backward acceleration at the hands due to the resulting rotation. You'd like the two to exactly cancel for your force to be used most efficiently, which is just what happens at the sweet spot.