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 0.278m . An object of mass 4.40kg is suspended from the free end of the wire. The system is released from rest and the suppended object descends with constant acceleration.

If the suspended object moves downward a distance of 2.6m in 1.93s, what is the mass of the wheel?

Explanation / Answer

x = x0 + v0t + 1/2at^2 solve for a:
2.6 = 0 + 0 + 1/2a(2.14)^2
a = 1.135

v = v0 + at solve for v:
v = 0 + (1.135)(2.14)
v = 2.42

mgh = 1/2mv^2 + 1/4MR^2(v/R)^2
where m = mass of object
M = mass of wheel
R = radius

(4.4)(9.8)(2.6) = 1/2(4.4)(2.42)^2 + 1/4M(.278)^2(2.42/.278)^2 solve for M:
112.101 = 12.88 + 1.46M
M = 67.8 kg

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