Part A A thin, light wire is wrapped around the rim of a wheel, as shown in (Fig

Question

Part A A thin, light wire is wrapped around the rim of a wheel, as shown in (Figure 1). The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.190 m and moment of inertia for rotation about the axle of 0.490 What is the magnitude of the angular velocity of the wheel after the block has descended 3.70 m kg m2. A small block with mass 0.350 kg is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does -7.50 J of work as the block descends 3.70 m rad/s Submit My Answers Give Up de Feedback Continue Figure 1 of 1

Explanation / Answer

Apply conservation of energy

Means, energy lost = energy gained
energy is lost in the form of
-->potential energy of block
-->due to frictional force in bearings
energy is gained in the form of
--> kinetic energy of block
-->rotational energy of the wheel

Given that, mass of the block,m = 0.350 kg
velocity of the block,v
moment of inertia of wheel,I = 0.490 kg*m2
radius of wheel,r = 0.190 m
angular velocity of the wheel,
earth gravity,g =9.81 m/s2
energy lost due to friction,F = 7.50 J
height descended by block,h = 3.70 m

therefore, by conservation law,
PEblock + Fbearing = KEwheel + KEblock
mgh + F = mv2/2 + I2/2
at any moment the the point of ocntact of string anf wheel is stationary i.e v=r
mgh + F = m(rw)2/2 + I2/2
mgh + F = [mr2/2 + I/2]2
2 = [mgh + F]/ [mr2/2 + I/2]
=> = ([mgh + F]/ [mr2/2 + I/2])

put the values -
= [(0.350*9.81*3.70 + 7.50)/(0.35*0.190/2 + 0.49/2)] = sqrt[(20.20 / 0.278)] = 8.52 rad/s

Hence the requisite angular velocity = 8.52 rad/s.

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