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Solved: CP A thin, light wire is wrapped around the rim of
Chapter 9, Problem 49E(choose chapter or problem)
A thin, light wire is wrapped around the rim of a wheel, as shown in Fig. E9.49. 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 \(R=0.280 \mathrm{~m}\) An object of Mass \(\mathrm{m}=4.20\) 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 \(3.00 \mathrm{~m}\) in \(2.00 \mathrm{~s}\), what is the mass of the wheel?
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QUESTION:
A thin, light wire is wrapped around the rim of a wheel, as shown in Fig. E9.49. 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 \(R=0.280 \mathrm{~m}\) An object of Mass \(\mathrm{m}=4.20\) 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 \(3.00 \mathrm{~m}\) in \(2.00 \mathrm{~s}\), what is the mass of the wheel?
Step 1 of 3
We shall have to calculate the acceleration of the suspended object in order to proceed with the other calculations.
Distance moved by the object \(s=3.00 \mathrm{~m}\)
Time \(t=2.00 s\)
From the equation \(s=\frac{1}{2}\) at 2 , we can calculate the acceleration as follows. \(a=2 s / t^{2}\)
\(a=2 s / t^{2}\)
\(a=2 \times 3.00 /(2.00)^{2} \mathrm{~m} / \mathrm{s}^{2}\)
\(a=1.5 \mathrm{~m} / \mathrm{s}^{2}\)