A physics student of mass 47.0 kg is standing at the edge of the flat roof of a

Question

A physics student of mass 47.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriendly dog is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student grabs the end of the rope and steps off the roof.

A. If the wheel has radius 0.300 m and a moment of inertia of 9.60 kgm2 for rotation about the axle, how long does it take her to reach the sidewalk? Ignore friction.

B. How fast will she be moving just before she lands?

Explanation / Answer

given that

moment of inertia   I = 9.60 kg*m^2

r = 0.300 m

m = 47 kg

h = 12 m

part(A)

using conservation of energy

energy before = m*g*h

energy after = 1/2*m*v^2 + 1/2*I*w^2

we know

s = t* (v+u) / 2

as u = 0 for both v=2s/t so

v=2*h / t and w = 2* h /(r*t)

energy before = energy after

So

m*g*h = 1/2*m*(2*h/t) ^2 + 1/2*I*(2*h/r*t)^2

put the values in above expression

47*9.8*12 = 0.5*47*(2*12/t)^2 + 0.5*9.60*(2*12/0.300*t)^2

11054.4 = 47*576/t^2 + 9.60 *6400/t^2

11054.4 = 27072/t^2 + 61440/t^2

11054.4 = 88512/t^2

t^2 = 88512 / 11054.4

t = sqrt(8.00694)

t = 2.82 s

part(B)

w = w0+g*t

w = 2*h/r*t + g*t

w = 2*12/0.3*2.82 + 9.8*2.82

w = 28.36 + 27.63

w = 56 rad/s

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