Item 5 Part A A thin, light wire is wrapped around the rim of a wheel, as shown

Question

Item 5 Part A 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 0.298 m . An object of mass 4.40 kg 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) If the suspended object moves downward a distance of 3.10 m in 2.11 s, what is the mass of the wheel? Express your answer with the appropriate units. M=1 Value kg Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

Translated energy (PE to KE) = m * g * h
= 4.4 * 9.81 * 3.10
= 133.67 Joules

The acceleration (a) rate of the falling mass :
a = ( s - ( u * t ) ) / ( ½ * t ² )
a = 3.1/ (2.11^2)/2

a = 1.392 m/s2

The final velocity (v) of the falling mass :
v = u + ( a * t )
v = 1.392*2.11 m/s
v = 2.938 m/s
The final linear KE of the falling mass = ½ * 4.4 * 2.938 ²
= 18.99 Joules

The final rotating KE of the wheel = 133.67 - 18.99
= 114.67 Joules

The final rotation rate of the wheel :
= v / r
= 2.938 / 0.149
= 19.71 rad / sec

So, if :
114.67 = ½ * mass moment of inertia (i) * ²
Then :
i = 114.671 / ( ½ * ² )
i =0.589kg-m²

So, assuming the wheel is a solid disc ( for which i = ½ * mass * r ² )
Then :
0.589= ½ * mass * 0.149 ²

mass = 53.06 Kg

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