Solve Rotational Dynamics Homework: Min Putty Velocity to Rotate Rod

[Delzac]

Homework Statement

We have a rod length l with mass m. This rod is nail at its end to a wall. A putty with mass m also then collide with the rod right at the other end. The collision is perpendicular to the rod.
Question is what is the minimum velocity the putty need to travel to rotate the whole rod (with itself attached) up vertically above the pivot.

The Attempt at a Solution

I thought of 2 ways to do this (both are wrong):

First, after collision, we know that centre of mass is 3/4 l away from pivot. To bring it up vertically above pivot,

[tex] E = mgh [/tex]

I then conveniently equate kinetic energy with gravitational energy:

[tex] V_r = \sqrt(2gh) = \sqrt(2g*1.5l) [/tex]

Since collision is inelastic:

[tex] mv_p = 2mv_r [/tex]
[tex] v_p = 2\sqrt(3gl) [/tex]

This is wrong according to the answer given, correct ans is sqrt(8gl)

Then i used any method:

I kept linear velocity required to bring rod up to vertical position as what i have calculated above.

Then by conservation of angular momentum:

Iw (initial) = Iw (rod) + Iw (putty)

[tex] ml^2(\frac{v}{l}) = (\frac{ml^2}{3})(\frac{v_r}{0.5l}) + ml^2\frac{v_r}{l} [/tex]

This gives me V putty = 5/3 sqrt(3gl) which is also not sqrt(8gl).

1) So the question here is why are both methods incorrect?
2) Why does 2 methods give me different answer? ( Now that i am typing this, i think i didn't account for inelastic collision in my second method did i?)

Any help will be greatly appreciated.

 

[Delzac]

Up for the day. Any help will be great.

Also, i have another question.

I remember in the MIT open courseware, the lecturer mention that for object undergoing pure roll, rolling without slipping, V centre of mass = V circumference. However inside my physics textbook, they seem to write V top of cylinder = 2 V centre of mass.

Why is there such a difference?

 

[Delzac]

Up once more. Any help would be great.

 

[Delzac]

Any help will be great.

 

[rl.bhat]

First of' all let us apply conservation of angular momentum.
Before collision,
angular momentum of putty = mv*l
angular momentum of the rod = 0
total initial angular momentum Li = mv*l.
After collision,
angular momentum of putty = m(ωl)*l
angular momentum of the rod = 1/3*m*l^2*ω
total final angular momentum Lf = mω(l^2 + 1/3*l^2) = mω(4l^2/3)
Since Li = Lf,
mvl = mω(4l^2/3) or ω = 3v/4l
Now let us apply the conservation of energy.
Initial KE of putty and rod = 1/2*1/3*ml^2*ω^2 + 1/2*ml^2*ω^2 = 2/3*ml^2*ω^2
Initial PE of putty and rod = 0
final KE of putty and rod = 0
Final PE of putty and rod = mgl + mg*2l = 3mgl
hence 2/3*ml^2*ω^2 = 3mgl
substitute the value of ω and solve for v.

 

[Delzac]

I have a question: Why is it that i can't use inelastic collision to work out the answer?

 

[rl.bhat]

Since collision is inelastic:
[tex]mv_p = 2mv_r[tex]

you cannot write this. Because after collisional the rod and putty are in rotational motion.

In the first part we have considered inelastic collision. Hence we have not considered the KE of the putty. But in the second part the system of putty and rod is in pure rotation. So we have applied the conservation of energy.

 

[Delzac]

In the last part of the equation you wrote this:

[tex] \frac{2}{3}ml^2\omega^2 = 3mgl [/tex]

And this works out to be:

[tex] \frac{2}{3} m v^2 = 3mgl [/tex]

Where v is the linear velocity at that instant. This i agree.

But, why can't i obtain linear velocity via this method:

[tex] \frac{1}{2} (2m)v^2 = 3mgl [/tex]

This is also from conservation of energy, except that i assume that this object is traveling upwards, not rotating. Why is there such a difference in the result, in the energy requirement?

 

[rl.bhat]

In the conservation of energy I am considering rotational KE and the PE of the center of mass and the putty. In that the linear velocity does not come into the picture.

Here KE = 1/2*I*ω^2

KE = 1/2*(2m)v^2 is not possible because center of mass and the putty are not moving with the same linear velocity.

 

[Delzac]

Can't we consider the centre of mass for the whole system, so the rod and the putty becomes a single object with centre of mass 3/4l away from the pivot? And we take the centre of mass as traveling vertically upwards instead of rotating. The total kinetic energy (be it translational or rotational) require whether the object rotates upwards or move vertically directly to it should be the same isn't it?

 

[Delzac]

Up for the day.

 

[rl.bhat]

By using center of mass, you can find the increase in PE of the system.
PE = 2m*g*2*3/4l = 3mgl.
But you cannot find the initial rotational KE, because it depends on the point of pivot and the mass distribution.

 

[Delzac]

So base on the increase in PE required we can calculate the amount of kinetic energy that the system must have to reach a certain height.

So, is there any proofs that you can provide that shows that using linear velocity of CM of this system doesn't work?

My thinking is that those nearer to the pivot will have less linear velocity, those further away will have more, but eventually, the centre of mass will be the average.

 

[Delzac]

Or can i think of it as such:

The putty and rod system, at an instant after the collision, the centre of mass of system is undergoing translational motion. But an instant later, the pivot sets into convert all the translational kinetic energy to rotational kinetic energy, and the pivot will start to rotate. So base of this analogy, it sounds reasonable to calculate via linear velocity of centre of mass.

 

[rl.bhat]

When you want to calculate the total energy, it is better to calculate in terms of the angular velocity ω which is constant for the rod and the putty. And using the conservation of the angular momentum, you can find the relation between the initial angular velocity and the linear velocity of the putty which is required. I have used this straight forward approach to find the required result.

 

[Delzac]

Yes. I understand the method that goes into calculating the required results.

But i am interested to know why can't i use linear velocity of the instant after the collision as the total kinetic energy. As long as the system has that energy, if should be able to rise up to the required position, regardless of which path the system take to rise to that position.

 

[rl.bhat]

The total energy of a rigid body cannot be calculated by knowing the linear velocity of a point on that body.
The linear velocity of the instant after the collision is not the velocity of the putty. To calculate that velocity, you cannot use the conservation of linear momentum. You cannot calculate the total energy using that linear velocity after collision.

 

[Delzac]

Okay got it. Just another question, when you equate angular momentum initial to final, does inelastic collision automatically works out (since there is energy lost in inelastic)? Or in fact, it works for both elastic and inelastic collisions.

 

[rl.bhat]

Okay got it. Just another question, when you equate angular momentum initial to final, does inelastic collision automatically works out (since there is energy lost in inelastic)? Or in fact, it works for both elastic and inelastic collisions.

Conservation of linear and angular moment works for both elastic and inelastic cases.

 

[Delzac]

Yeah got it. Thanks

 

[Delzac]

Well actually i have one more question that is still left unanswered in my second post.

I remember in the MIT open courseware, the lecturer mention that for object undergoing pure roll, rolling without slipping, V centre of mass = V circumference. However inside my physics textbook, they seem to write V top of cylinder = 2 V centre of mass.

Why is there such a difference?

 

[rl.bhat]

V centre of mass = V circumference. However inside my physics textbook, they seem to write V top of cylinder = 2 V centre of mass.

The second part is correct. In the first part v circumference is not same at every point. Which reference point the lecturer mentioned?

 

[Delzac]

This is the page.
http://ocw.mit.edu/courses/physics/8...es/lecture-24/

Following is the transcript:
If here is an object, the cylinder is here with radius R, and I'm going to rotate it like this and roll it in this direction, the center is called point Q.

Once it has made a complete rotation, if then the point Q has moved over a distance 2pi R, then we call that pure roll.

When we have pure roll, the velocity of this point Q, and the velocity of the circumference, if you can read that-- I'll just put a c there-- are the same.

In other words, vQ is then exactly the same as v circumference, and v circumference is always omega R.

This part always holds, but for pure roll, this holds.

 

[rl.bhat]

The velocity of the point on the circumference with respect to the center is constant and is equal to the velocity of the center with respect to the ground. The second statement is true when you find the velocity of the point on the circumference with respect to the point of contact of the cylinder with the ground

 

[Delzac]

I don't get you. What is true and not true? Can you be more exact?

 

[rl.bhat]

If time taken to complete one rotation is t, linear velocity of the center is 2πR/t.
The velocity of the point c on the circumference is also 2πR/t. = ωR.
While pure rolling, displacement of point of contact with respect to the ground is zero. Hence the velocity of c at the top of the cylinder with respect to the ground is ω(2R) = 2v.

 

[Delzac]

The velocity of the point c on the circumference is also 2πR/t. = ωR.
While pure rolling, displacement of point of contact with respect to the ground is zero. Hence the velocity of c at the top of the cylinder with respect to the ground is ω(2R) = 2v.

I don't get this part. You sound like you are contradicting yourself. Is there a difference between c on circumference and on top of cyliner?