p08_035.pdf

N and, according to Newton’s second law, this must be

equal to mv 2 /R ,where v is the speed of the boy. At the point where the boy leaves the ice N =0, so

g cos θ = v 2 /R . We wish to ﬁnd his speed. If the gravitational potential energy is taken to be zero when

he is at the top of the ice mound, then his potential energy at the time shown is U =

−

mgR (1

−

cos θ ).

He starts from rest and his kinetic energy at

thetimeshownis 2 mv 2 . Thus conservation

of energy gives 0 = 2 mv 2

. . . .

. . .

−

mgR (1

−

cos θ ), or

•

v 2 =2 gR (1

cos θ ). We substitutethis expres-

sion into the equation developed from the sec-

ond law to obtain g cos θ =2 g (1

−

. .

cos θ ). This

gives cos θ =2 / 3. The height of the boy above

the bottom of the mound is R cos θ =2 R/ 3.

−

35. The free-body diagram for the boy is shown below. N is the normal force of the ice on him and m is

his mass. The net inward force is mg cos θ