Blocks falling while attatched to pulley

In summary, the two blocks, m1 = 3.3 kg and m2 = 4.8, connected by a massless rope and a pulley with a 12 cm diameter and 2.0 kg mass, are released from rest at a height of 1.0 m. Friction at the pulley's axle exerts a torque of 0.64 N · m. To determine how long it takes the 4.8 kg block to reach the floor, the torque on the pulley must be calculated and converted to linear acceleration. By representing the masses as part of the pulley, the correct torque equation can be used to find the linear acceleration and, subsequently, the time it takes for the
  • #1
Jamie_Pi
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Homework Statement


The two blocks, m1 = 3.3 kg and m2 = 4.8, in the figure below are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.64 N · m. If the blocks are released from rest, how long does it take the 4.8 kg block to reach the floor from a height of h = 1.0 m? (Note: If your random numbers do not create movement between the masses enter 0 for your answer.)
Screen Shot 2017-11-04 at 4.56.19 PM.png


Homework Equations


Torque=Force*Radius
Torque=moment of inertia*angular acceleration
angular acceleration = linear acceleration/radius
s=1/2*linear acceleration*time^2

The Attempt at a Solution


I tried to calculate the torque on the pulley and convert that to linear acceleration, and find the time from that.
Torque total= m*g*r-friction
Torque total=4.8*9.8*0.06-4.8*9.8*0.06-0.64
Torque total=0.4004
moment of inertia=1/2*mass*radius^2
moment of inertia=0.0036
0.4004=0.0036*angular acceleration
angular acceleration=111.22
111.22=linear acceleration/0.06
linear acceleration=6.6733
s=3.337*t^2
1=3.337*t^2
t=sqrt(1/3.337)
t=0.5441

I did something wrong, and I'm not sure what. I'm not very good with rotation and torque, though, so I don't know where my mistake could be. The real answer is 2.12 seconds.
 

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  • #2
Jamie_Pi said:
Torque total=4.8*9.8*0.06-4.8*9.8*0.06-0.64
I assume you meant 4.8*9.8*0.06-3.3*9.8*0.06-0.64
Jamie_Pi said:
Torque total=0.4004
Seems a bit too much. Please post your working.
Jamie_Pi said:
0.4004=0.0036*angular acceleration
You cannot take the entire torque and apply it all to the pulley. Some of it will go into accelerating the masses.
Either:
- Assign distinct variables to the two tensions and write the three acceleration equations for the two masses and the pulley,
or:
- represent the masses in the torque equation (as mr2)
 
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  • #3
haruspex said:
I assume you meant 4.8*9.8*0.06-3.3*9.8*0.06-0.64

Seems a bit too much. Please post your working.

You cannot take the entire torque and apply it all to the pulley. Some of it will go into accelerating the masses.
Either:
- Assign distinct variables to the two tensions and write the three acceleration equations for the two masses and the pulley,
or:
- represent the masses in the torque equation (as mr2)
Ok, I re did it and I must have done some math wrong before.
Torque=0.242

If I understand your suggestions correctly, I think representing the masses as part of the pulley is the best idea, so I'll do:
moment of inertia= 1/2(2+4.8+3.3)*0.06^2

So from there,
torque=moment of inertia * angular acceleration
0.242=1/2*(2+4.8+3.3)*0.06^2*a/0.06 ('a' being linear acceleration)
a=(2*0.242)/(2+4.8+3.3+0.06)
a=0.476
s=1/2a*t^2 ('s' is 1)
t=sqrt(2*1/a)

this gives me t=2.049*, which is a bit off, but better. I must have misunderstood what you meant about representing the mass in the torque equation, could you explain that a little bit more? I feel like I'm really making progress on this one.

*this is an edit, earlier I messed up my math and had the incorrect number here.
 
Last edited:
  • #4
Jamie_Pi said:
representing the masses as part of the pulley is the best idea, so I'll do:
moment of inertia= 1/2(2+4.8+3.3)*0.06^2
The masses are not disks rotating about their centres. All of the mass acts at the same radius as far as the pulley is concerned.
 
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  • #5
haruspex said:
The masses are not disks rotating about their centres. All of the mass acts at the same radius as far as the pulley is concerned.
Ok, that makes sense. I'm not sure what to represent them as, though, because they aren't actually spinning at all. I tried representing them as points rotating around the central axis, and that got me somewhat close:

torque=(1/2*2*0.06^2+(4.8+3.3)*0.06^2)*a/0.06
which I simplified into
torque=(4.8+3.3)*0.06*a

and plugging this back into the same thing:
0.242=0.06*8.1*a
a=0.4979
time=sqrt(2/0.4979)

time=2.004 seconds, which is pretty close.
 
  • #6
Jamie_Pi said:
torque=(4.8+3.3)*0.06*a
You left out the pulley.
 
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  • #7
haruspex said:
You left out the pulley.
Well, since the pulley is 2kg and it is multiplied by 1/2, it became 1 and wasn't really doing anything.

Ahh, but of course! It's being added not multiplied.
torque=0.06*(8.1+1)*a
0.242/(0.06*9.1)=a
a=0.4432 (looks good so far)
time=sqrt(2/0.4432)

time= 2.12
Yay! Thanks a bunch! Again.
 

Related to Blocks falling while attatched to pulley

1. How does a pulley affect the motion of a falling block?

A pulley can change the direction of the force acting on a falling block, making it move in a horizontal direction rather than just falling straight down. This is because the pulley changes the direction of the force applied by the rope or string attached to the block.

2. What factors affect the speed at which a block falls while attached to a pulley?

The speed at which a block falls while attached to a pulley is affected by the mass of the block, the length and tension of the rope or string, and the angle of the pulley. The more mass the block has, the faster it will fall. The longer and tighter the rope or string, the faster the block will fall. The angle of the pulley can also affect the speed at which the block falls, with steeper angles resulting in faster speeds.

3. Can the use of multiple pulleys affect the motion of a falling block?

Yes, the use of multiple pulleys can affect the motion of a falling block. By adding more pulleys, the force required to lift the block is spread out over a larger distance, making it easier to lift the block. However, this does not affect the speed at which the block falls.

4. How does gravity play a role in the motion of a falling block attached to a pulley?

Gravity is the force that causes the block to fall in the first place. The pulley does not change the force of gravity, but it does change the direction of the force, making the block move horizontally rather than just falling straight down. Gravity also causes the block to accelerate as it falls, increasing its speed until it reaches the ground or another surface.

5. Are there any real-life applications of blocks falling while attached to a pulley?

Yes, there are many real-life applications of blocks falling while attached to a pulley. One common example is in elevators, where pulleys are used to lift and lower the elevator car. Pulleys are also used in cranes to lift heavy objects, in window blinds to raise and lower them, and in flagpoles to raise and lower the flag. Pulleys are also used in many sports and recreational activities, such as rock climbing and zip lining.

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