p08_011.pdf

11. We use Eq. 8-18, representing the conservation of mechanical energy (which neglects friction and other

dissipative eﬀects).

(a) In the solution to exercise 5 (to which this problem refers), we found ∆ U = mgL as it goes to the

highest point. Thus, we have

∆ K +∆ U =0

K top −

K 0 + mgL =0

which, upon requiring K top =0,gives K 0 = mgL and thus leads to

v 0 = 2 K 0

= 2 gL .

(b) We also found in the solution to exercise 5 that the potential energy change is ∆ U =

−

mgL in

going from the initial point to the lowest point (the bottom). Thus,

∆ K +∆ U =0

K bottom −

K 0 −

mgL =0

which, with K 0 = mgL ,leadsto K bottom =2 mgL . Therefore,

v bottom = 2 K bottom

= 4 gL

which implies ∆ K = 0. Consequently, the speed is the same as what it was initially ( √ 2 gL ).

(d) It is evident from the above manipulations that the results do not depend on mass. Thus, a diﬀerent

mass for the ball must lead to the same results.

which simpliﬁes to 2 √ gL .