A thin, light wire is wrapped around the rim of a wheel, as shown in the followi

Question

A thin, light wire is wrapped around the rim of a wheel, as shown in the following figure. The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius 0.270 . An object of mass 4.35 is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration.


If the suspended object moves downward a distance of 3.15 in 2.07 , what is the mass of the wheel?

Explanation / Answer

First we can find the acceleration of the mass that moves downward

Using d = vot + .5at2 will solve for that knowing that the systems started from rest

(3.15) = (0) + (.5)(a)(2.07)2

a = 1.47 m/s2

Then knowing the equations for torque...

= FL = I    we can solve for the mass of the wheel

The force to use in the equation comes from the tension in the srting caused by the mass hanging from it. Froms Newton's second law, F = ma, so

mg - T = ma

T = mg - ma

T = (.00435)(9.8) - (.00435)(1.47)

T = .0362 N

The moment arm, L, is the radius of the wheel

The moment of inertia of a disk = .5MR2

The angular acceleration is found from = a/r = (1.47)/(.270) = 5.44 rad/s2

So we can plug into the torque equation

(.0362)(.270) = (.5)(M)(.270)2(5.44)

M = .0493 kg    which is 49.3 grams

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