The radius of a wheel is 0.450 m. A rope is wound around the outer rim of the wh

Question

The radius of a wheel is 0.450 m. A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude 5.05 N, unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope.

a) how much rope unwinds while the wheel makes 1.00 revolution?


(b) How much work is done by the rope on the wheel during this time?
J

(c) What is the torque on the wheel due to the rope?
N · m

(d) What is the angular displacement , in radians, of the wheel during 1.00 revolution?
rad

(e) Show that the numerical value of the work done is equal to the product .
=  J

Explanation / Answer

a) The amount of rope unwound is given by the circumference of the wheel times the number of revolutions made, i.e.   xrope = 2R · (number of revolutions)

=  2*0.45*1 = 2.83 m

b) The work done by the rope on the wheel is given by the torque applied to the wheel times the angle through which it turns.

W =

The torque is given by the radius of the wheel times the magnitude of the force the rope is pulled with, since they are perpendicular. is the angle it turned in radians. One revolution is equal to 2 radians, so the work done is

W = rF = 0.45*5.05*2 = 14.28 J

c) As we said before, the torque is given by multiplying the radius of the wheel by the magnitude of the force.

= rF = 0.45*5.05 = 2.2725 Nm

d) As the wheel rotates 1 time, the wheel rotates 360 = 2 * radians
This is the angular displacement delta theta, in radians, of the wheel during 1.00 revolution.

e) Shown in part b)

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