The two masses (m 1 = 4.81 kg and m 2 = 2.86 kg) in the Atwood\'s machine shown
ID: 2144123 • Letter: T
Question
The two masses (m1 = 4.81 kg and m2 = 2.86 kg) in the Atwood's machine shown in the figure below are released from rest, with m1 at a height of 0.786 m above the floor.
When m1 hits the ground its speed is 0.204 m/s. Assuming that the pulley is a uniform disk with a radius of 11.6 cm, calculate the pulley's mass.
The two masses (m1 = 4.81 kg and m2 = 2.86 kg) in the Atwood's machine shown in the figure below are released from rest, with m1 at a height of 0.786 m above the floor. When m1 hits the ground its speed is 0.204 m/s. Assuming that the pulley is a uniform disk with a radius of 11.6 cm, calculate the pulley's mass.Explanation / Answer
Finally, angular velocity of the pulley = v/r = 0.204/0.116 = 1.76 rad/s
Angular momentum of pulley = 1/2 m*r^2 = 6.73*10^-3 m, where m is mass of pulley
Energy lost = m1g*h = 4.81*9.8*0.786 = 37.05 J
Energy gained = PE of m1 + KE of m1 + KE of m2 + rotational K.E of pulley
= 2.86*9.8*0.786 + 1/2 * (4.81+2.86)*0.204^2 + 1/2 * I *w^2
= 22.03 + 0.16 + 0.0104m
Energy lost = Energy gained
37.05 = 22.19+0.0104m
m = 1428kg
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