A wheel and axle is a type of compound pulley which a smaller A wheel and axle i
ID: 2254758 • Letter: A
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A wheel and axle is a type of compound pulley which a smaller
A wheel and axle is a type of compound pulley which a smaller "axle" is attached to a larger "wheel". It is a type of simple machine in which a smaller mass can be used to lift a much larger mass. In the figure shown the axle has a diameter of 20 cm and a mass of 5 kg. The larger wheel has a mass of 50 kg and a diameter of 1.0 m. If this system is in equilibrium, how large is the mass attached to the wheel? What principle is required to solve this problem? Suppose the smaller mass is 24 kg and the system is released from rest. By how much would the gravitational potential energy of both masses change if the small mass drops a distance of 0.75 m?Explanation / Answer
In the figure shown the axle has a diameter of 20 cm and a mass of 5 kg. The larger wheel has a mass of 50 kg and a diameter of 1.0 m.
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If this system is in equilibrium, how large is the mass attached to the wheel?
The net torque must be zero(because the system is in equilibrium):
m1 g r1 = m2 g r2
==> 90 * 9.8 * 0.20 = m2 * 9.8 * 1
==> m2 = 18 Kg
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What principle is required to solve this problem?
The system is in equilibrium therefore the net torque must be zero.
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Suppose the smaller mass is 24 kg and the system is released from rest. By how much would the gravitational potential energy of both masses change if the small mass drops a distance of 0.75 m?
When the 24Kg mass drops d2=0.75m, the wheel and the axle rotate N turns:
N = 0.75/(2 pi r2) = 0.75/(2*3.1416*1) = 0.11937 turns
therefore the 90Kg mass displacement is:
d1 = N (2 pi r1) = 0.11937 * (2*3.1416*0.20) = 0.15001 m
gravitational potential energy:
dU = m2 g d2 - m1 g d1 = 24 * 9.8 * 0.75 - 90 * 9.8 * 0.15001 = 44.09118 J = 44.1 J
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Use conservation of energy to find the rotational speed of the wheel and axle after the small mass has dropped 0.75 m. Make certain you account for all the kinetic energy this system of objects has as this event takes place!
dK = dU
0.5 m1 vf1^2 + 0.5 m2 vf2^2 + 0.5 I_wheel w^2 + 0.5 I_axle w^2 = dU
0.5 m1 vf1^2 + 0.5 m2 vf2^2 + 0.5 (0.5 (Mw) (Rw)^2) w^2 + 0.5 (0.5 (Ma) (Ra)^2) w^2 = dU
Mw = mass of the wheel
Ma = masss of the axle
0.5 m1 vf1^2 + 0.5 m2 vf2^2 + 0.5 (0.5 (Mw) (Rw)^2) w^2 + 0.5 (0.5 (Ma) (Ra)^2) w^2 = dU
0.5 m1 (R1 w)^2 + 0.5 m2 (R2 w)^2 + 0.5 (0.5 (Mw) (Rw)^2) w^2 + 0.5 (0.5 (Ma) (Ra)^2) w^2 = dU
0.5*90*(0.20*w)^2 + 0.5*24*(1*w)^2 + 0.5*(0.5*(50)*(1)^2) * w^2 + 0.5*(0.5*5*(0.20)^2) * w^2 = 44.09118
0.5*90*(0.20)y2 + 0.5*24*(1)y2 + 0.5*(0.5*(50)*(1)y2) + 0.5*(0.5*5*(0.20)y2) = 44.09118
==> w = 1.29 rad/s
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