Rotation Motion Problem: Finding Acceleration and Tensions in a Pulley System

[BrainMan]

Homework Statement

A mass m1 is connected by a light cord to a mass m2, which slides on a smooth surface. The pulley rotates about a friction less axle and has a moment of inertia I and radius R. Assuming the cord does not slip on the pulley, find (a) the acceleration of the two masses, (b) the tensions T1 and T2, and (c) numerical values for a, T1 and T2 if I = 0.5 kg * m^2, R = 0.3 m, m1 = 4 kg, and m2 = 3 kg. (d) What would your answers be if the inertia of the pulley was neglected?

Homework Equations

The Attempt at a Solution

I drew a free body diagram and tried to find the acceleration of the first block. So I did
-9.6m1 + T = m1a1
-9.8 + T/m = a1
unfortunately this is not even close to being the right answer. How should I approach this problem?

 

[BrainMan]

 

[Nathanael]

I drew a free body diagram and tried to find the acceleration of the first block. So I did
-9.6m1 + T = m1a1
-9.8 + T/m = a1
unfortunately this is not even close to being the right answer. How should I approach this problem?

No, that is indeed the correct answer. It's just that you haven't figured out what $T_1$ is yet.

To find $T_1$ you're going to have to draw the free body diagram of the pulley.

(Note: The two blocks have the same acceleration which is equal to R times the angular acceleration of the pulley)

 

[BrainMan]

No, that is indeed the correct answer. It's just that you haven't figured out what $T_1$ is yet.

To find $T_1$ you're going to have to draw the free body diagram of the pulley.

(Note: The two blocks have the same acceleration which is equal to R times the angular acceleration of the pulley)

OK I did that and did
Ta = r∂
Torque = I∂
Torque = 9.8m
9.8 m = I∂
Ta = r(9.8m/I)
What do I do from here to find the acceleration of both the blocks?

 

[Nathanael]

Torque = 9.8m

The torque on the pulley will not be 9.8m

The torque on the pulley comes from the two tensions, $T_1$ and $T_2$

The torque on the pulley will be the difference of the two tensions, since they act in opposite directions.
Note: Since you also don't know $T_2$, you will have to draw the free body diagram of the other block too (isn't this fun! hehe)

 

[BrainMan]

The torque on the pulley will not be 9.8m

The torque on the pulley comes from the two tensions, $T_1$ and $T_2$

The torque on the pulley will be the difference of the two tensions, since they act in opposite directions.



Note: Since you also don't know $T_2$, you will have to draw the free body diagram of the other block too (isn't this fun! hehe)

How do I find the acceleration of the second block. I feel like I'm going in circles. I must find the acceleration of the circle to find the acceleration of the second block. I must find the acceleration of the block to find the acceleration of the circle lol :)

 

[Nathanael]

How do I find the acceleration of the second block. I feel like I'm going in circles. I must find the acceleration of the circle to find the acceleration of the second block. I must find the acceleration of the block to find the acceleration of the circle lol :)

Yes it is circular in a way, but it will still be solvable because you know that the acceleration of the two blocks and the pulley are all related.

The acceleration of the two blocks is exactly the same, and that acceleration is proportional to the angular acceleration of the pulley.So you will end up with 3 equations, but you can still use the same "a" (for acceleration) in all the equations because they are all related.Once you write those 3 equations the problem will be almost solved.
(The 3 equations are the "net force equations" for both blocks and the pulley)

 

[BrainMan]

Yes it is circular in a way, but it will still be solvable because you know that the acceleration of the two blocks and the pulley are all related.

The acceleration of the two blocks is exactly the same, and that acceleration is proportional to the angular acceleration of the pulley.


So you will end up with 3 equations, but you can still use the same "a" (for acceleration) in all the equations because they are all related.


Once you write those 3 equations the problem will be almost solved.
(The 3 equations are the "net force equations" for both blocks and the pulley)

If the two accelerations of the blocks are against each other shouldn't they cancel out?

 

[Nathanael]

If the two accelerations of the blocks are against each other shouldn't they cancel out?

What do you mean against each other? They both accelerate together (one goes to the right, one goes down)

 

[BrainMan]

What do you mean against each other? They both accelerate together (one goes to the right, one goes down)

OK I am confused. I understand everything in parts but I am not getting the big picture of the forces

 

[Nathanael]

Ok I'll do my best to explain the big picture.

Block 1 accelerates downwards causing the pulley to rotate and the block 2 to accelerate. Since they are all attatched together by a rope, they all have the same (linear) acceleration.

The rope pulls up on the block 1 with some tension, T1, and so (by Newton's 3rd law) the block pulls down on the rope, which pulls down on the pulley (with the same force, T1).

The rope also pulls the block 2 forward with some tension T2, and so the block 2 pulls on the pulley (via the rope) with a tension T2

You don't know what either force of tension is, but you do know one crucial fact: They all accelerate at the same rate!

So you must write an equation for each object (block 1, block 2, and the pulley) that relates the tension with the acceleration. Once you write all 3 equations you'll be able to solve the problem for the acceleration by using algebra to get rid of ("cancel out") the forces of tension.
(Once you know the acceleration you can go back and find the forces of tension, since the problem asked for them)


Did this help at all?

 

[BrainMan]

Ok I'll do my best to explain the big picture.

Block 1 accelerates downwards causing the pulley to rotate and the block 2 to accelerate. Since they are all attatched together by a rope, they all have the same (linear) acceleration.

The rope pulls up on the block 1 with some tension, T1, and so (by Newton's 3rd law) the block pulls down on the rope, which pulls down on the pulley (with the same force, T1).

The rope also pulls the block 2 forward with some tension T2, and so the block 2 pulls on the pulley (via the rope) with a tension T2

You don't know what either force of tension is, but you do know one crucial fact: They all accelerate at the same rate!

So you must write an equation for each object (block 1, block 2, and the pulley) that relates the tension with the acceleration. Once you write all 3 equations you'll be able to solve the problem for the acceleration by using algebra to get rid of ("cancel out") the forces of tension.
(Once you know the acceleration you can go back and find the forces of tension, since the problem asked for them)


Did this help at all?

It helped me visualize the big picture but I am still struggling with the mathematics so the tension in block 1 should be m1a and the tangential force on the block should be m1a so the tension in the second block should be m2a. How do I relate all of this?

 

[Nathanael]

It helped me visualize the big picture but I am still struggling with the mathematics so the tension in block 1 should be m1a and the tangential force on the block should be m1a so the tension in the second block should be m2a. How do I relate all of this?

No those are not the tensions. You don't know what the tensions are. Don't even try to figure out the tension because at this point it could really be anything (it's not something simple like m1a)

For the first block you've already wrote the equation correctly in your very first post:
"-9.8 + T/m = a"
That is equation number one.

Now do the same thing for the other block, then do the same thing for the pulley (with torque)

 

[ehild]

The tension is the force exerted by the string. The string pulls the objects connected at its ends by the same force. You have two pieces, the horizontal and vertical one, with two different tensions. These tensions act also at the rim of the pulley, and exert opposite torques RT1 and RT2.

Show the equation you get for each object.

The hanging ball moves vertically with acceleration a. The forces applied are gravity and the tension T1. So m₁a = ?
The ball pulls the string. The vertical piece gets longer, the horizontal piece gets shorter with the same amount. The string pulls the box on the horizontal surface, with the same speed and acceleration as the ball moves. But the tension is different in the horizontal piece of the string, it is T₂. So m₂a₂=T₂

The tensions act also at the rim of the pulley and rotate it. The string does not slide on the pulley, sot the linear acceleration of the rim is the same as the acceleration of both the box and the ball. As the ball moves downward, the pulley rotates clockwise. Show the equation you get for the relation between the torque and the angular acceleration of the pulley. You know that the angular acceleration is equal to a/R.

You have three equations for the unknowns T₁, T₂, a. Cancel the tensions. Show your work.


ehld

 

[dean barry]

You could work with pulley equivalent mass, use an example velocity for m1 (say 1.0 m/s), then calculate the KE values for m1 and the pulley.
The equivalent mass of the pulley = ( KE pulley / KE (m1) ) * m1
Now you can treat the pulley as a linear mass to solve some of this problem.

 

[BrainMan]

You could work with pulley equivalent mass, use an example velocity for m1 (say 1.0 m/s), then calculate the KE values for m1 and the pulley.
The equivalent mass of the pulley = ( KE pulley / KE (m1) ) * m1
Now you can treat the pulley as a linear mass to solve some of this problem.

I'm sorry but I don't understand.

 

[ehild]

I'm sorry but I don't understand.

You do not need any equivalent mass. The moment of inertia I and the radius of the pulley R are given. How is the angular acceleration of the pulley related to the tensions, R, and I?


ehild

 

[dean barry]

Sorry, i misled you there

 

[dean barry]

equivalent mass

I found this anomaly when helping solve a problem involving a non - slipping sphere rolling down an incline. )

For the wheel eqivalent mass calculation , use either :
= ( KE wheel / Ke m1 ) * m1
or :
= ( KE wheel / Ke m2 ) * m2

( The result is the same )

Because the wheel rotates without slipping its KE is always proportional to the KE of either m1 or m2 (use either, the result is the same), if you could define the wheels impact on the system as a linear mass ( as shown previously, let's call it m3 ), then the acceleration (a) of the system could be (more easily) defined as :
( a = f / m )
a = ( m * g ) / ( m1 + m2 + m3 )

As far as string tension is concerned, the tension T2 would then be :
= m2 * a
And the tension T1 :
= m3 * a

Ignoring the wheel :
a = ( m1 * g ) / ( m1 + m2 )
T1 ( & T2 ) = m2 * a


You can check part of this theory by dropping m1 an arbitrary 1.0 metre (h), calculating the final velocity with the pre calculated acceleration, then figure the KE's of all : m1, m2 and the wheel, and added together this should equal :
m1 * g * h
( final total system KE = original potential energy (PE) of m1 )

 

[ehild]

I do not understand what you suggest. There is no wheel in the problem. If you mean pulley, its moment of inertia is given. You can define an equivalent mass as I/R² for the pulley, but the problem is easy to solve, why do you complicate it with something nobody understands what it is? And what is the fourth mass m?
You ignore the wheel (pulley), why do you calculate its equivalent mass then? And the problem requires that the moment of inertia of the pulley should be taken into account.
The acceleration of any of the masses is determined by the sum of all forces acting on it and the tension is only one of the forces.


ehild

 

[dean barry]

Sorry, clearly i don't know how to tune into "real" physics , and contribute any ideas, i thought this was an ideas forum, silly me, ill slide downhill to a lower station if that's ok with you.

 

[BrainMan]

The tension is the force exerted by the string. The string pulls the objects connected at its ends by the same force. You have two pieces, the horizontal and vertical one, with two different tensions. These tensions act also at the rim of the pulley, and exert opposite torques RT1 and RT2.

Show the equation you get for each object.

The hanging ball moves vertically with acceleration a. The forces applied are gravity and the tension T1. So m₁a = ?
The ball pulls the string. The vertical piece gets longer, the horizontal piece gets shorter with the same amount. The string pulls the box on the horizontal surface, with the same speed and acceleration as the ball moves. But the tension is different in the horizontal piece of the string, it is T₂. So m₂a₂=T₂

The tensions act also at the rim of the pulley and rotate it. The string does not slide on the pulley, sot the linear acceleration of the rim is the same as the acceleration of both the box and the ball. As the ball moves downward, the pulley rotates clockwise. Show the equation you get for the relation between the torque and the angular acceleration of the pulley. You know that the angular acceleration is equal to a/R.

You have three equations for the unknowns T₁, T₂, a. Cancel the tensions. Show your work.


ehld

OK so I got four equations
(1) a1 = 9.8 - T1/m1
(2) torque = T1-T2
(3) a2 = (T1-T2/I)R
(4) a3 = T2/m2
What do I do now?

 

[Nathanael]

OK so I got four equations
(1) a1 = 9.8 - T1/m1
(2) torque = T1-T2
(3) a2 = (T1-T2/I)R
(4) a3 = T2/m2
What do I do now?

Good, but your equation (2) is wrong. (The units are different.) It should be "torque = (T1-T2)R"

Only equation 1, 3 and 4 are important (you only needed equation 2 to get to equation 3)

Using the correct torque for equation 3, you get "a2 = ((T1-T2)/I)R^2"

There is actually one more equation, and then the problem will be solvable.

The final equation is
a1=a2=a3
(So you can simply call it "a" in all three equations)Now to solve it you have to use algebra. The normal way to solve a system of equations (like this) is to plug one variable into the other equation (for example you could plug "a" from one equation into the other)
(Since there are 3 equations you will have to do this more than once.)The algebra is just the dirty work, the problem is essentially solved

 

[BrainMan]

Good, but your equation (2) is wrong. (The units are different.) It should be "torque = (T1-T2)R"

Only equation 1, 3 and 4 are important (you only needed equation 2 to get to equation 3)

Using the correct torque for equation 3, you get "a2 = ((T1-T2)/I)R^2"

There is actually one more equation, and then the problem will be solvable.

The final equation is
a1=a2=a3
(So you can simply call it "a" in all three equations)


Now to solve it you have to use algebra. The normal way to solve a system of equations (like this) is to plug one variable into the other equation (for example you could plug "a" from one equation into the other)
(Since there are 3 equations you will have to do this more than once.)


The algebra is just the dirty work, the problem is essentially solved

What should happen when I plug one variable from one equation into another?

 

[Nathanael]

What should happen when I plug one variable from one equation into another?

It will "eliminate" the variables.

For example, let's say you just wanted the acceleration, and didn't care about the tension. You would solve for the tension in one equation and then plug it into the other equations (thus the tension will "dissapear" from the equations)

For example, from equation (4) we get:
$T_2=m_2a$

And now we can plug that into equation 3, and we will get:
$a=(T_1-m_2a)\frac{R^2}{I}$

Now, instead of having two unknowns in equation 3, there is only one unknown.

To get zero unknowns, solve for T1 (in equation 1) and then plug it in (to equation 3) and you will have your answer for the acceleration. (You'll have to solve for it using normal algebra.)

(There are many different ways to do this elimination, but they will all lead to the same answer)

Make sense?

 

[BrainMan]

It will "eliminate" the variables.

For example, let's say you just wanted the acceleration, and didn't care about the tension. You would solve for the tension in one equation and then plug it into the other equations (thus the tension will "dissapear" from the equations)

For example, from equation (4) we get:
$T_2=m_2a$

And now we can plug that into equation 3, and we will get:
$a=(T_1-m_2a)\frac{R^2}{I}$

Now, instead of having two unknowns in equation 3, there is only one unknown.

To get zero unknowns, solve for T1 (in equation 1) and then plug it in (to equation 3) and you will have your answer for the acceleration. (You'll have to solve for it using normal algebra.)

(There are many different ways to do this elimination, but they will all lead to the same answer)

Make sense?

Alright I got a = (m1(9.8) - m1a-m2a) R^2/I
How do I get rid of the a's?

 

[Nathanael]

Alright I got a = (m1(9.8) - m1a-m2a) R^2/I
How do I get rid of the a's?

Whaaat why do you want to get rid of the a's? The problem asked you to solve for a Rearrange the equation so that all a's are on one side and then you can factor the a out and then you will only have 1 a (instead of 3) and then isolate it and there's your answer

 

[BrainMan]

Whaaat why do you want to get rid of the a's? The problem asked you to solve for a


Rearrange the equation so that all a's are on one side and then you can factor the a out and then you will only have 1 a (instead of 3) and then isolate it and there's your answer

OK I was able to correctly solve for a and T2. Now I am having trouble with the equation for T1.
T1 = m1g-m1a
So I substituted a into this equation and got
T1 = m1g- m1(m1g((I/R^2) + m1 + m2)^-1)
It says the correct answer is T1 = m1g(I + m2R^2)[I + (m1 + m2)R^2]^-1
Is this just a different representation of my answer?

 

[Nathanael]

I'm sorry but it's very difficult to discern your equation.

The second equation (the "correct answer") looks like:

$\frac{m_1g(I+m_2R^2)}{I+(m_1+m_2)R^2}$

I think that is how you meant it to be.

But your first equation looks like:
$m_1g-(m_1)(m_1g)[\frac{I}{R^2}+m_1+m_2]^{-1}$

But I'm pretty sure that's not how you meant it to be.
If your other answers were correct then I am very confident that your last answer (for T1) will be correct.
(I think it's impossible for it not to be correct, but I can't prove it )

Try multiplying your answer by (R/R) that way yours and the other answer have the same denominator.

If that doesn't work maybe double check your algebra?

 

[ehild]

OK I was able to correctly solve for a and T2. Now I am having trouble with the equation for T1.
T1 = m1g-m1a
So I substituted a into this equation and got
T1 = m1g- m1(m1g((I/R^2) + m1 + m2)^-1)
It says the correct answer is T1 = m1g(I + m2R^2)[I + (m1 + m2)R^2]^-1
Is this just a different representation of my answer?

Your equation is equivalent to the given result, but you can simplify it a bit.

So you got the acceleration correctly

$$a=\frac{m_1 g}{I/R^2+m_1+m_2}$$

With that, $$T=m_1g-m_1a=m_1g-m_1\frac{m_1g}{I/R^2+m_1+m_2}$$

It is the same formula you wrote.

Factor out m1g, and use common denominator to do the subtraction.

$$T=m_1g\left(1-\frac{m_1}{I/R^2+m_1+m_2}\right)=m_1g \left(\frac{I/R^2+m_1+m_2-m_1}{I/R^2+m_1+m_2}\right)=m_1g \frac{I/R^2+m_2}{I/R^2+m_1+m_2}$$

Multiply both the numerator and the denominator by R² and you get the desired form.

ehild

 

[ehild]

But your first equation looks like:
$m_1g-(m_1)(m_1g)[\frac{I}{R^2}+m_1+m_2]^{-1}$

But I'm pretty sure that's not how you meant it to be.

That formula is correct for T1, Nathanael.

ehild