An object with a mass of #200 kg# is hanging from an axle with a radius of #6 cm#. If the wheel attached to the axle has a radius of #32 cm#, how much work would it take to turn the wheel a length equal to the circumference of the axle?

2 Answers
Sep 9, 2017

Answer:

The work is #=138.5J#

Explanation:

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The load is #L=200gN#

The radius of the axle is #r=0.06m#

The radius of the wheel is #R=0.32m#

The effort is #=FN#

Taking moments about the center of the axle

#F*0.32=200g*0.06#

#F=200g*0.06/0.32=367.5N#

The force is #F=367.5N#

The distance is #d=2pir=2*pi*0.06=(0.12pi)m#

The work is #W=Fd=367.5*0.12pi=138.5J#

Sep 9, 2017

Answer:

The work done is 138.5 J.

Explanation:

The circumference of the axle is #2*pi*0.06 m#. The circumference of the wheel is #2*pi*0.32 m#. Therefore since the wheel is turned a length equal to the circumference of the axle, the wheel, and also the axle, is rotated a fraction of 1 revolution. The fraction is the ratio, k, of the 2 circumferences
#k = (cancel(2)*cancel(pi)*0.06 m)/(cancel(2)*cancel(pi)*0.32 m) = 0.1875#

Therefore the object is lifted a distance, #Deltah#, that is #0.1875*2*pi*0.06 m#.

By the principle of conservation of energy, the work that was done is equal to the increase in the object's gravitational potential energy, GPE.
#DeltaGPE = m*g*Deltah#
#DeltaGPE = 200 kg * (9.8 m)/s^2 * 0.1875*2*pi*0.06 m = 138.5 J#

This could also be calculated by multiplying the torque by the angular translation. The torque would be equal and opposite the torque that the object's weight applies to the axle. The angular translation would be k2pi radians.
#W = Tau*Theta = m*g*r*k*2*pi#
#W = 200 kg*9.8 m/s^2*0.06 m*0.1875*2*pi = 138.5 J#

I hope this helps,
Steve