Question

Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of 4.00 m/s. Ignore frictional losses. (a) What is the height of the hill? m (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom? m/s

Answer 1

Using energy conservation

KEi + PEi = KEf + PEf

PEf = 0, at ground

KEi = 0, intially speed is zero

PEi = KEf

PEi = KEtrans. + KErot

m*g*h = 0.5*m*V^2 + 0.5*I*w^2

w = V/r

I = moment of inertia of basketball = 2*m*r^2/3

m*g*h = 0.5*m*V^2 + 0.5*(2*m*r^2/3)*(v/r)^2

g*h = V^2/2 + V^2/3

h = (V^2/g)*(5/6)

Using given values:

h = (4^2/9.81)*(5/6) = 1.36 m

Part B

for can of juice

PEi = KEtrans. + KErot

m*g*h = 0.5*m*V^2 + 0.5*I*w^2

w = V/r

I = moment of inertia of can = m*r^2/2

m*g*h = 0.5*m*V^2 + 0.5*(m*r^2/2)*(v/r)^2

g*h = V^2/2 + V^2/4

V = sqrt (4*g*h/3)

Using above value pf h

V = sqrt (4*9.81*1.36/3)

V = 4.22 m/sec

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