Question

A physics student of mass 59.0 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⋅m2 for rotation about the axle, how long does it take her to reach the sidewalk? Ignore friction.

t = ?

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

v = ?

Answer 1

T = tension in the rope

$\alpha$ = angular acceleration of wheel

a = linear acceleration

r = radius = 0.3

I = moment of inertia = 9.60

T = tension in rope

force equation for the student is given as

mg - T = ma                                     eq-1

Torque equation for the wheel is given as

T r = I $\alpha$

T = Ia/r²                            eq-2

using eq-1 and eq-2

mg - Ia/r²  = ma     

a = mg / (I/r² + m)

a = 59 x 9.8 / (9.60/0.3² + 59)

a = 3.5 m/s²

using conservation of energy

Potential energy at top = KE at bottom + rotational KE of wheel when student is at bottom

mgh = (0.5) m v² + (0.5) I w²

mgh = (0.5) m v² + (0.5) I (v/r)²

59 x 9.8 x 12 = (0.5) (59) v² + (0.5) (9.60) (v/0.3)²

V = 9.15 m/s

Vi = initial velocity of student = 0

a = acceleration

t = time

d = distance moved = 12 m

using the equation

d = V_i t + (0.5) a t²

12 = 0 x t + (0.5) (3.5) t²

t = 2.62 sec