Question

Two metal sphere's have a radius of 25 cm and the same mass, except one is solid and the other is hollow. Determine the ratio of the speed of the solid sphere to the speed of the hollow sphere after rolling down a 0.26 m high incline

Answer 1

Using energy conservation for hollow sphere

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 hollow sphere = 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

V_h = sqrt (6*g*h/5)

Using energy conservation for solid sphere

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 solid sphere = 2*m*r^2/5

m*g*h = 0.5*m*V^2 + 0.5*(2*m*r^2/5)*(v/r)^2

g*h = V^2/2 + V^2/5

V_s = sqrt (10*g*h/7)

V_s/V_h = sqrt (10*g*h*5/(7*6*g*h))

V_s/V_h = sqrt (50/42) = 1.09 m/sec

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