Advanced Mechanics Advanced Mechanics Physics Help Forum 
Sep 12th 2012, 11:27 PM

#1  Member
Join Date: Jul 2012
Posts: 67
 Another Angular Momentum Prob
A uniform rod with a length of L=0.6 m and a mass of M=2 kg is connected at its upper end to an axis, and is held at an angle of theta=20 deg. The rod is released from rest, and collides a fully elastic collision with a point mass of m=1 kg which is at rest just at the edge of a table that is at height of H=0.8 m above the ground.
1. What is the maximal angle the rod achieves after the collision?
2. What is the horizontal distance m achieves when it strikes the ground?
3. Assuming that after the collision the rod performs Smallangle approximation vibration around its equillibrium point, what is the small angle frequency?
first of all I have no idea how to solve 3 . About 1: I think ther's no conservation of either linear or angular momentum because of the force and the torque the axis applies on the rod, so I think you need conservation of energy but why does the force applied by the axis on the rod not doing any work?
Last edited by assaftolko; Sep 12th 2012 at 11:29 PM.

 
Sep 13th 2012, 06:45 AM

#2  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,289

There is indeed conservation of both energy (since it's an elastic collision) and momentum. The axle applies zero torque to the bar. It does apply a force to counteract the weight and centripedal force of the rotating bar, but the work it performs is 0 (remember that work = force times distance, and the axle doesn't move so distance = 0).
As for the frequency of vibration  does your text book cover how to set up a differential equation for a vibrating bar? You will need to know some characteristics of the bar not mentioned in your post, such as the material's density and young's modulus (E) and its crosssection moment of inertia (I).
Last edited by ChipB; Sep 13th 2012 at 06:52 AM.

 
Sep 13th 2012, 07:14 AM

#3  Member
Join Date: Jul 2012
Posts: 67

Originally Posted by ChipB There is indeed conservation of both energy (since it's an elastic collision) and momentum. The axle applies zero torque to the bar. It does apply a force to counteract the weight and centripedal force of the rotating bar, but the work it performs is 0 (remember that work = force times distance, and the axle doesn't move so distance = 0).
As for the frequency of vibration  does your text book cover how to set up a differential equation for a vibrating bar? You will need to know some characteristics of the bar not mentioned in your post, such as the material's density and young's modulus (E) and its crosssection moment of inertia (I). 
I meant there's no conservation of momentum from the time the rod is released until the time it's just about to hit m. I do understand there's conservation of momentum in the collision phase. Am I correct?
The distance part is true but in some way confusing for me... it didn't occur to me that the point of contact between the axis and M doesn't move... it's just like the center of a circle in a way right?
I have no other information about the rod... the solution is w=sqrt(2g/L) if it helps...

 
Sep 13th 2012, 08:11 AM

#4  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,289

Originally Posted by assaftolko I meant there's no conservation of momentum from the time the rod is released until the time it's just about to hit m. I do understand there's conservation of momentum in the collision phase. Am I correct? 
Yes.
Originally Posted by assaftolko The distance part is true but in some way confusing for me... it didn't occur to me that the point of contact between the axis and M doesn't move... it's just like the center of a circle in a way right? 
Yes.
Originally Posted by assaftolko I have no other information about the rod... the solution is w=sqrt(2g/L) if it helps... 
Ah! What they want is for you to calculate the frequency of oscillation of the bar acting as a pendulum after the block is knocked away. the use of the word "vibration" in your original post confused me  I consider vibrations to involve a vibrating object such as a guitar string or the air inside a organ pipe. But that's not the question at all. To do this you set up an equation for the torque aplied to the bar by graviity when displaced an angle theta equal to the moment of inertia times the anglar acceleration:
T = I alpha
mg sin(theta) L/2 = I (theta double dot)
Here "theta double dot) means the angular acceleration.
For a bar pinned at one end I = mL^2/3, so:
mL^2/3 (theta double dot)  mgL/2 sin(theta) = 0
Divide through by m and note that for small theta sin(theta) = theta, so:
L^2/3 (theta double dot) gL/2 (theta) = 0
The solution is theta = A sin(wt). Now solve for w (frequency). However I don't get the same answer as you gave  I get w = sqrt(3g/2L).
Last edited by ChipB; Sep 13th 2012 at 09:11 AM.

 
Sep 13th 2012, 08:35 AM

#5  Member
Join Date: Jul 2012
Posts: 67

I found the full answer:
They write the force equation for the tangent axis, which they draw at the center of the rod, so the angle between the force vector Mg and the radial axis is theta (the small angle, not theta=20 deg).
Ft: Mgsin(theta)=Mat
gsin(theta)=L/2 * theta double dot (L/2 is the radius of the movement)
gtheta=L/2 * theta double dot
theta double dot = 2g/L * theta
and so w=sqrt(2g/L)
???????

 
Sep 13th 2012, 09:17 AM

#6  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,289

Originally Posted by assaftolko gsin(theta)=L/2 * theta double dot (L/2 is the radius of the movement) 
This treats the rod as if all its mass is concentrated at the L/2 point, so that it's moment of inertia about the axle is m (L/2)^2. This is not correct if the rod is of uniform mass distribution, as the moment of inertia for a rod pinned at one end is a bit greater: mL^2/3.

 
Sep 13th 2012, 09:26 AM

#7  Member
Join Date: Jul 2012
Posts: 67

Originally Posted by ChipB This treats the rod as if all its mass is concentrated at the L/2 point, so that it's moment of inertia about the axle is m (L/2)^2. This is not correct if the rod is of uniform mass distribution, as the moment of inertia for a rod pinned at one end is a bit greater: mL^2/3. 
but what in their way reffers to the notion of I? they just did a net force equation on a point that is in the middle of the rod, and they didnt even reffer to the force applies by the axis on the rod, just gravity.. i dont understand why they did what they did

 
Sep 13th 2012, 09:53 AM

#8  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,289

The calculation of pendulum frequency for a pendulum that's a point mass is like this:
Starting with F=ma, where F the force on the point mass in the direction perpendicular to the pendulum bar:
F = mg sin(theta)
The acceleration of the point mass moving in circular motion at distance R from the pivot is R times the second derivative of theta:
mg sin(theta) = mR(theta double dot), etc.
The result is w = sqrt(g/R). So in this problem their solution treats the rod as if it was a point mass at R=L/2.
The forces on the pivot don't matter here, as they act in a direction in line with the pendulum bar. All that matters here to determine the frequency of the the motion of the pendulum are the forces acting perpendicular to the bar, which are what cause it to oscillate back and forth.

 
Sep 13th 2012, 10:00 AM

#9  Member
Join Date: Jul 2012
Posts: 67

Originally Posted by ChipB The calculation of pendulum frequency for a pendulum that's a point mass is like this:
Starting with F=ma, where F the force on the point mass in the direction perpendicular to the pendulum bar:
F = mg sin(theta)
The acceleration of the point mass moving in circular motion at distance R from the pivot is R times the second derivative of theta:
mg sin(theta) = mR(theta double dot), etc.
The result is w = sqrt(g/R). So in this problem their solution treats the rod as if it was a point mass at R=L/2.
The forces on the pivot don't matter here, as they act in a direction in line with the pendulum bar. All that matters here to determine the frequency of the the motion of the pendulum are the forces acting perpendicular to the bar, which are what cause it to oscillate back and forth. 
How do you know in which direction does the force on the pivot act? I really didn't understand how this force behaves... and so you say that their way is correct? (before you said it's not)

 
Sep 13th 2012, 11:03 AM

#10  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,289

Originally Posted by assaftolko How do you know in which direction does the force on the pivot act? I really didn't understand how this force behaves... and so you say that their way is correct? (before you said it's not) 
The forces on the piviot are not germain to this probem. The only way it would matter is if the pivot applies a torque to the bar  but that's not what a pivot does.
My last post was to show why the frequency for a simple pendulum consisting of a point mnass on the end of the string is sqrt(g/R). I did this to try and show that their answer appears to be built on an assumption that the mass of the bar is concentrated at L/2. I do not agree with their answer, because I assume that the mass of the bar is distributed along its length (i.e., the mass is NOT concentrated all at L/2).

  Search tags for this page 
Click on a term to search for related topics.
 Thread Tools   Display Modes  Linear Mode  