p08_009.pdf

9. (a) If K i is the kinetic energy of the ﬂake at the edge of the bowl, K f is its kinetic energy at the

bottom, U i is the gravitational potential energy of the ﬂake-Earth system with the ﬂake at the top,

and U f is the gravitational potential energy with it at the bottom, then K f + U f = K i + U i .Taking

the potential energy to be zero at the bottom of the bowl, then the potential energy at the top is

U i = mgr where r =0 . 220 m is the radius of the bowl and m is the mass of the ﬂake. K i =0 since

the ﬂake starts from rest. Since the problem asks for the speed at the bottom, we write 2 mv 2 for

K f . Energy conservation leads to

2 mv 2 =

⇒

v = 2 gr = 2(9 . 8)(0 . 220) =2 . 08 m / s .

(b) We note that the expression for the speed ( v = √ 2 gr ) does not contain the mass of the ﬂake. The

speed would be the same, 2 . 08m / s, regardless of the mass of the ﬂake.

mgr = 1

U f .Since K i is greater than before, K f is

greater. This means the ﬁnal speed of the ﬂake is greater.