What is the connection between linear and angular momentum?

[mech-eng]

Homework Statement

problem in in the picture

Homework Equations

I can't understand some parts of the equation.

The Attempt at a Solution

 

[mech-eng]

here on the right hand side of the equation why we summed the moment involing inertia with moment of net force
on the block.This seems to me as lefthand is Fnet x r=I x a + Fnet x r , in which r is half of diameter and
I is moment of inertia of the disk and a is angular acceleration, respectively. I think I x a is arises due to Fnet x
r so they are the same and equation is Fnet x r = Fnet x r + Fnet x r and so it is wrong.

 

[PhanthomJay]

Equation 1 of the solution manual is correct, but I am not in favor generally of using inertial pseudo forces. I also don't understand your attempt. Personally, I find it simpler to take a free body diagram diagram of the mass and apply Newton's 2nd Law for translation to solve for the tension in the portion of the cord attached to the weight . Then use the net torque equation taking a FBD of the pulley.

 

[mech-eng]

Equation 1 of the solution manual is correct, but I am not in favor generally of using inertial pseudo forces. I also don't understand your attempt. Personally, I find it simpler to take a free body diagram diagram of the mass and apply Newton's 2nd Law for translation to solve for the tension in the portion of the cord attached to the weight . Then use the net torque equation taking a FBD of the pulley.

thanks for the answer.my teacher used your method as well.But what is inertial pseudo forces.
I think inertial forces are imaginary and we add them to the system while implying D'lambert
principle for taking the system to equilibrium.meanwhile my attempt was wrong and ı thought
tension in that portion is equal to the weight of the block.my another question about this equation
is can we use moment of inertia of the block in spite of its not turning like we a mass turning
free end of a pendulum.have a nice day.

 

[PhanthomJay]

thanks for the answer.my teacher used your method as well.

Yes, that seems like the easier way.

But what is inertial pseudo forces.
I think inertial forces are imaginary and we add them to the system while implying D'lambert
principle for taking the system to equilibrium.

that is correct, but i am not sure why I said inertial forces were used since I can't open up your attachment anymore. It may have been an incorrect conclusion on my part.

meanwhile my attempt was wrong and ı thought
tension in that portion is equal to the weight of the block.

yes, the tension is not equal to the block weight when acceleration is involved

my another question about this equation
is can we use moment of inertia of the block in spite of its not turning like we a mass turning
free end of a pendulum.have a nice day.

well I guess you could use that approach where the inertia of the hanging mass, treated as a particle, is mr^2, where r is the pulley radius and m is the mass of the block; then you look at the entire system without including the internal tension force, where the sum of torques , mgr -F_s(r), is equal to the Iα of the pulley plus the Iα of the hanging mass, where the Iα of the hanging mass is mr^2(α), which maybe is what the book solution was doing, to get the same result, but this is not a good way of doing it, in my opinion.

 

[mech-eng]

"that is correct, but i am not sure why I said inertial forces were used since I can't open up your attachment anymore. It may have been an incorrect conclusion on my part. "

I can open the attachment.I think when you tried you were not logged on and you can look up it
again in the book.the book belongs to hibbeler, last chapter of engineering dynamics.

 

[mech-eng]

well I guess you could use that approach where the inertia of the hanging mass, treated as a particle, is mr^2, where r is the pulley radius and m is the mass of the block; then you look at the entire system without including the internal tension force, where the sum of torques , mgr -F_s(r), is equal to the Iα of the pulley plus the Iα of the hanging mass, where the Iα of the hanging mass is mr^2(α), which maybe is what the book solution was doing, to get the same result, but this is not a good way of doing it, in my opinion.[/QUOTE]

But block is doing a translational motion and moment of inertia is only for rotational motions but
there is a relation between them in that problem.Can you discuss with other mentor friends.

 

[PhanthomJay]

well I guess you could use that approach where the inertia of the hanging mass, treated as a particle, is mr^2, where r is the pulley radius and m is the mass of the block; then you look at the entire system without including the internal tension force, where the sum of torques , mgr -F_s(r), is equal to the Iα of the pulley plus the Iα of the hanging mass, where the Iα of the hanging mass is mr^2(α), which maybe is what the book solution was doing, to get the same result, but this is not a good way of doing it, in my opinion.

But block is doing a translational motion and moment of inertia is only for rotational motions but
there is a relation between them in that problem.Can you discuss with other mentor friends.

I would like for you to consider the case of a vertical massless frictionless ideal pulley of radius r = 0.5 m hung from the ceiling, and with a rope wrapped around it hanging down from both sides of the pulley. At one end of the rope is a 2 kg mass m, and at the free end of the rope on the other side you pull down on it with a force of P = 30 N.

What is the acceleration of the mass??

Using the 'conventional' method (translational Newton's 2nd law), the tension, T, on the rope is 30 N, and the acceleration of the mass per F_net = ma is
T - mg = ma
30 -20 = 2a
a = 5 m/s^2.

Now using Newton's 2nd law for rotation,

Net Torque = Iα
T(r) - mg(r) = mr^2(α)
30(.5) -20(.5) = 2(.5)^2)(α)
5 = .5α
α = 10 rad/sec^2
a = αr = (10)(.5) = 5 m/sec^2, same result.

Or using letters for better understanding

Method 1: T - mg = ma

Method 2: T(r) - mg(r) = mr^2(α), divide both sides by r,
T - mg = mr(α)
T- mg = mr(a/r)
T - mg = ma

Same result. Is moment of inertia only for rotational motions??

 

[mech-eng]

But the example you gave is that of one degree of freedom i.e there is only one independent
motion.because rotatinal motion of the pulley and translational motion of the block are dependent to each other (can we say they are actually the same? ) your approach is true but it shouldn't be always true.

 

[PhanthomJay]

But the example you gave is that of one degree of freedom i.e there is only one independent
motion.because rotatinal motion of the pulley and translational motion of the block are dependent to each other (can we say they are actually the same? ) your approach is true but it shouldn't be always true.

I see you are not convinced. It is not an easy concept to grasp. Try to avoid using it when practicable. It stems from the definition of moment of inertia, and the calculus behind it.

Let me give you one more simple example. No pulleys, no friction, no air resistance, not even rotation. It is the example of dropping an object of mass m from a building, an object in 'free fall' if you will.

What is the aceleration of the object?

Now it was over 400 years ago when Galileo told us that all objects accelerate at the same rate.

Let's do the Physics using Newton 2:

F_{net} =ma

The only force acting is the objects weight, mg, so the net force is mg.

mg = ma
a = g

That was sure easy.

Now let's try it using angular motion about some imaginary axis an arbitrary distance r away, passing perpendicularly through the plane of your screen. We write

T = I\alpha
mgr = mr^2\alpha
mgr = mr^2(a/r)
mgr = mra
a = g !

Convinced?

 

[mech-eng]

But an always falling object can't make an angular motion about an axis.I am not sure but this concept might related with linear and angular momentum.Let's postpone this after I worked momentum.

 

[PhanthomJay]

But an always falling object can't make an angular motion about an axis.I am not sure but this concept might related with linear and angular momentum.Let's postpone this after I worked momentum.

Fair enough, that is good insight. You should find that an object with linear momentum also possesses angular momentum about a chosen axis, in the same way that an object with linear acceleration also possesses angular acceleration about a chosen axis. Even when there is not rotation per se. Perhaps it may be helpful to replace the word "linear" with "tangential".