A ball rolls from rest down a roof without slipping. the roof has pitch of 30 de

Question

A ball rolls from rest down a roof without slipping. the roof has pitch of 30 degree until there is a flat, horizontal section near the edge of the roof. the ball lands 2 meters away from the horizontal position of the edge of the roof. The edge of the roof is located 5m above the ground.
a. how far did the ball roll down the sloped portion of the roof?
b. what was its angular momentum when it left the roof?
c. what was its total kinetic energy just before it hit the ground?

thanks bunch! please answer

Explanation / Answer

At the time when the ball leaves contact of the roof(its horizontal part),it has only horizontal velocity vh. Now the ball lands 2m away from the edge.The vertical hieght that it travelled was 5m and the vertical velocity was 0. So,the ball travelled 5m with 0 initial velocity under an acceleration of g=9.8m/s2 in time t.

s=ut+0.5at2

5=0+0.5x9.8xt2

=>t=1.01s

now,horizontally,the ball had no accleration and travelled 2m in time t=1.01

2=vhxt

=>vh=2/1.01=1.98 or 2m/s

veloctiy of the ball when it left the slope was 2m/s

as the ball was released from rest,total energy gained by the ball should be equal to the potential energy stored in the ball,

0.5mvh2=mx9.8xh

h=4/19.6m

since the angle of the slope is 300,

a). the ball rolled hcot30=4/19.6 x 1.7332

=0.3535m

b). angulur momentum=I

where =vr and I=mr2

angulur momentum=mvr3=2mr3

c). The total energy of the ball is =mx9.8x(5+h)=51m Joules

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