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 0.262m . An object of mass 4.40kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration.(Figure 1)

Part A

If the suspended object moves downward a distance of 2.65m in 2.06s , what is the mass of the wheel?

Express your answer with the appropriate units.

of 1

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 0.262m . An object of mass 4.40kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration.(Figure 1)

Part A

If the suspended object moves downward a distance of 2.65m in 2.06s , what is the mass of the wheel?

Express your answer with the appropriate units.

M =

of 1

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 0.262m . An object of mass 4.40kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration.(Figure 1) Part A If the suspended object moves downward a distance of 2.65m in 2.06s , what is the mass of the wheel? Express your answer with the appropriate units.

Explanation / Answer

x = x0 + v0t + 1/2at^2 solve for a:
2.65 = 0 + 0 + 1/2a(2.06)^2
a = 1.25

v = v0 + at solve for v:
v = 0 + (1.25)(2.06)
v = 2.575

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

(4.40)(9.8)(2.65) = 1/2(4.40)(2.575)^2 + 1/4M(.262)^2(2.575/.262)^2 solve for M:

M = 60.13 kg

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