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 0.290 m . An object of mass 4.00 kg 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 2.80 m in 2.10 s , what is the mass of the wheel?

Second time asking this question. the answer is not 63.1kg. Please be deatiled and sure before you post. Thanks so much!

Explanation / Answer

Here, use the equation -

x = x0 + v0t + 1/2at^2

Now, solve for a:
2.8 = 0 + 0 + 1/2a(2.10)^2
=> 2.8 = 2.205*a

=> a = 1.27

Again,

v = v0 + at

Solve for v:
v = 0 + (1.27)(2.10)
v = 2.67

Further, use the equation of energy conservation -

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

(4.0)(9.8)(2.8) = 1/2(4.0)(2.67)^2 + 1/4M(0.29)^2(2.67/0.29)^2

And solve for M:
=> 109.76 = 14.26 + 1.78*M

=> M = 53.65 Kg.

I hope you understand the solution..!

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