Question

A group of students perform the same "Conservation of Mechanical Energy" experiment that you performed in lab by allowing a solid sphere and then a solid cylinder to roll down the ramp. Both objects were released from the same position at the top of the ramp. If the speed vsphere of the solid sphere at the bottom of the ramp was 1.45 m/s, what would be the speed vcylinder at the bottom of the ramp?

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 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

g*h = 0.7*V^2

h = (0.7*V^2/g)

Using given values:

h = (0.7*1.45^2/9.81) = 0.15 m

Part B

for cylinder

PEi = KEtrans. + KErot

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

w = V/r

I = moment of inertia of cylinder = 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*0.15/3)

V = 1.40 m/sec

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