1). A solid cylinder and a hollow cylinder of the same mass and radius, both ini

Question

1). A solid cylinder and a hollow cylinder of the same mass and radius, both initially at rest, roll down the same inclined plane without slipping. (a) Which reaches the bottom first? (b) How do their rotational kinetic energies about the center of mass compare at the bottom? (c) if the plane were frictionless which would reach the bottom first? 2). An ice skater starts to go into a spin having his arms and a leg somewhat away from his body. As the spin is begun he pulls his arms and leg closer to his body. What is the result of his angular speed as he does this and what physical property is being used?

Explanation / Answer

1)

a)
for solid cyllinder

use energy conservation

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

m*g*h = 0.5*m*v^2 + 0.5*(0.5*m*R^2)*w^2

m*g*h = 0.5*m*v^2 + 0.5*0.5*m*(R*w)^2

m*g*h = 0.5*m*v^2 + 0.25*m*v^2

m*g*h = 0.75*m*v^2

v = sqrt(g*h/0.75)

= sqrt(1.33*g*h)

for hollow cyllinder

use energy conservation

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

m*g*h = 0.5*m*v^2 + 0.5*m*R^2)*w^2

m*g*h = 0.5*m*v^2 + 0.5m*(R*w)^2

m*g*h = 0.5*m*v^2 + 0.5*m*v^2

m*g*h = m*v^2

v = sqrt(g*h)

= sqrt(g*h)

so, solid cyllinder reaches first.

b) KE(solid) = 0.5*I*w^2

= 0.5*0.5*m*R^2*(v/R)^2

= 0.25*m*v^2

= 0.25*m*(g*h/0.75)

= m*g*h/3

KE(hollow) = 0.5*I*w^2

= 0.5*m*R^2*(v/R)^2

= 0.5*m*v^2

= 0.5*m*(g*h)

= m*g*h/2

KE(solid)/KE(hollw) = 2/3

c) both reach at the same time.

2) here angular moemntum is conserved

Pi = Pf

I1*w1 = I2*w2

when the person pulls his hands, I2 < I1

so, w2>w1

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