A physics student of mass 51.0 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 kg ⋅m^2 for rotation about the axle, how long does it take her to reach the sidewalk?
b. How fast will she be moving just before she lands?

Explanation:

The wheel and falling student will have common acceleration .

For rotational motion of wheel

Tx r = I α , T is tension in the crank , α is angular acceleration of wheel  , I is moment of inertia , r is radius of the wheel.

= I a / r

T = I a / r²

For motion of student

Mg - T = Ma , M is mass of the wheel.

Mg - I a / r²  = Ma

Mg = Ma +I a / r²

Mg = (M +I  / r²)a

a = Mg / (M +I  / r²)

= 51 x 9.8 / ( 51 + 9.6 / .3² )

499.8 / (51+ 106.67 )

= 499.8 / 157.67

= 3.17 m / s².

If time t is taken to  fall by 12 m

12 = 1/2 a t²

24 / a = t²

24 / 3.17 =t²

t²= 7.57

t = 2.75 s

velocity to reach sidewalk

v = u + at

= 3.17  x 2.75

= 8.72 m / s

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