Question #269588

A motor exerts torque on a carousel for it to attain a speed of 2.5 rev/s starting from rest in 3.5 s. Find the work done by the motor if the carousel has a radius of 27 m and mass of 1.75 x10^5kg. Consider


the carousel to be a thin-walled cylinder rotating about its center.

1
Expert's answer
2021-11-21T17:32:09-0500

for thin-walled cylinder:\text{for thin-walled cylinder:}

I=mr2I = mr^2

I=1.751052721.28108 kgm2I = 1.75*10^5*27^2\approx1.28*10^8\ kg*m^2

W=Iϕ,where ϕthe rotation angle of the carousel in radiansW = I\phi,\text{where }\phi -\text{the rotation angle of the carousel in radians}

w0=0w_0 = 0

w1=2.5revs=5πradsw_1= 2.5\frac{rev}{s}=5\pi\frac{rad}{s}

t=3.5st = 3.5s

ϵ=w1w0t=4.49rads2\epsilon= \frac{w_1-w_0}{t}=4.49\frac{rad}{s^2}

ϕ=w0t+ϵt22=4.493.522=27.50 rad\phi = w_0t+\frac{\epsilon t^2}{2}= \frac{4.49*3.5^2}{2}=27.50\ rad

W=Iϕ=1.2810827.50=3.52109JW = I\phi= 1.28*10^8*27.50 = 3.52*10^9 J


Answer: 3.52109J\text{Answer: }3.52*10^9 J


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