Two boxes hang from a solid disk pulley that is free to rotate as the blocks ris
ID: 3898505 • Letter: T
Question
Two boxes hang from a solid disk pulley that is free to rotate as the blocks rise/fall. The left box has a mass m1 = 3.6 kg and the right box has a mass m2 = 2.7 kg. The pulley has mass m3 = 3.1 kg and radius R = 0.15 m.
What is the linear acceleration of the left box? (up is the positive direction)
What is the angular acceleration of the pulley? (let counter-clockwise be positive)
What is the tension in the string between the left mass and the pulley?
What is the tension in the string between the right mass and the pulley?
The boxes accelerate for a time t = 1.62 s.
What distance does each box move in the time 1.62 s?
What is the magnitude of the velocity of the boxes after the time 1.62 s?
What is the magnitude of the final angular speed of the pulley?
Explanation / Answer
I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)
Two boxes hang from a solid disk pulley that is free to rotate as the blocks rise/fall. The left box has a mass m1 = 4.4 kg and the right box has a mass m2 = 2.3 kg. The pulley has mass m3 = 2.8 kg and radius R = 0.15 m.
1)
What is the linear acceleration of the left box? (up is the positive direction)
2)
What is the angular acceleration of the pulley? (let counter-clockwise be positive)
3)
What is the tension in the string between the left mass and the pulley?
4)
What is the tension in the string between the right mass and the pulley?
5)
The boxes accelerate for a time t = 1.59 s.
What distance does each box move in the time 1.59 s?
6)
What is the magnitude of the velocity of the boxes after the time 1.59 s?
7)
What is the magnitude of the final angular speed of the pulley?
Answer
let the acceleration be a and the angular acceleration of the pulley be y.so,
writing the equations of motion,
a=b*r
4.4*9.8-x=4.4a
y-2.3*9.8=2.3a
r=0.15
writing the torque equation,
x-y=0.5*2.8a
so,
a=b*0.15
solving the four equations , we get
a=2.1667 m/s^2
b=14.4421 rad/s^2
x=33.59 N
y=27.52 N
1)linear acceleration=a=2.1667 m/s^2
2)angular acceleration=b=14.4421
3)tension in the left string=x=33.59 N
4)tension in the right string=y=27.52 N
5)distance moved be d.so,
d=ut+0.5at^2
=0.5*1.59^2*2.1667
=2.74 m
6)velocity=u+at
=1.59*2.167
=3.445 m/s
7)angular velocity=wo+b*t
=1.59*14.4421
=22.96 rad/s
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