Force Exerted by Table on Stick at Release

In summary, a thin uniform stick of mass m, released from rest at an angle A to the vertical on a frictionless table, will experience a normal force from the table at the moment of release. The use of angular momentum is recommended for solving this problem, with torque acting on the center of mass of the rod. By finding the instantaneous center of rotation, the normal force can be calculated by considering the various forces and using the conservation of mechanical energy. The normal force is found to be 1/4 of the weight of the stick.
  • #1
No Name Required
29
0
A thin uniform stick of mass [tex]m[/tex] with its bottom end resting on a frictionless table is released from rest at an angle [tex]A[/tex] to the vertical. Find the force exerted by the table on the stick at the moment of release. Express your answer in terms of [tex]m, g[/tex] and [tex]A[/tex].

Ive been reliably informed (from the lecturer) that angular momentum is to be used but I don’t really understand how, or how it would effect the force exerted by the table.

Any help is appreciated

picture is here

http://picturerack.com/users/includes/flopt.asp?todo=3&inp=/richiec/
 
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  • #2
No Name Required said:
A thin uniform stick of mass [tex]m[/tex] with its bottom end resting on a frictionless table is released from rest at an angle [tex]A[/tex] to the vertical. Find the force exerted by the table on the stick at the moment of release. Express your answer in terms of [tex]m, g[/tex] and [tex]A[/tex].

Ive been reliably informed (from the lecturer) that angular momentum is to be used but I don’t really understand how, or how it would effect the force exerted by the table.
I don't see how angular momentum can be a factor since it is 0 at the moment of release. But torque (dL/dt) is definitely a factor.

What is the normal force at the moment of release and what is the rate of angular momentum change at that moment? Think of the torque as acting on the centre of mass of the rod.

AM
 
  • #3
AM,

I saw this problem and it looks tricky. When you say torque acts at the center of the rod, do you mean "about" the center? or are you talking about the force of gravity acting at the center? the torque depends on the refrence point right?
 
  • #4
NoName,

This is a pretty good problem!

Since the table is frictionless, what will the path of the center of mass of the stick be as it falls? So where will the center of rotation be?
 
  • #5
HackaB said:
AM,

I saw this problem and it looks tricky. When you say torque acts at the center of the rod, do you mean "about" the center? or are you talking about the force of gravity acting at the center? the torque depends on the refrence point right?
What I should have said that gravity acts at the centre of mass which is at the centre of the rod.

The torque is a little tricky. You can look at it a couple of ways. You could look at it as the gravitational force on the centre of mass about the sliding end. Or you could look at it as the normal force on the sliding end about the centre of mass. It works out to be the same.

AM
 
  • #6
Andrew Mason said:
The torque is a little tricky. You can look at it a couple of ways. You could look at it as the gravitational force on the centre of mass about the sliding end. Or you could look at it as the normal force on the sliding end about the centre of mass. It works out to be the same.
Do u mean the torque is the same measured both ways? How can you know that without knowing the normal force, which is what the question asks for?
 
  • #7
Think of the forces acting on the rod. Draw the free body diagram of the rod. So which way will the center of mass of the rod fall? .

The instantaneous center of roation is a point about which the rod will appear to purely rotate for that instant.

Remember, the instantaneous center of rotation need not lie on the rod itself and will also be at different points as the rod falls. Once you find that point, since you know that the rod appears to purley rotate about that point, find the torque due to the various forces. Also find a relation between the acceleration of the center of mass and the angular acceleration about the center of rotation and you should be able to calculate the Normal force.
 
  • #8
siddharth said:
Also find a relation between the acceleration of the center of mass and the angular acceleration about the center of rotation and you should be able to calculate the Normal force.
That's a good idea. I didn't think of that.
siddharth said:
The instantaneous center of roation is a point about which the rod will appear to purely rotate for that instant.

Remember, the instantaneous center of rotation need not lie on the rod itself and will also be at different points as the rod falls. Once you find that point, since you know that the rod appears to purley rotate about that point, find the torque due to the various forces.
Can't u use any point since the whole rod is at rest at that instant?
 
  • #9
Thanks everyone for your help! I've been sketching various stages in its falling and am carrying out several calculations. But then I had to sleep. I am true in also thinking that mechanical energy is conserved yes?
 
  • #10
No Name Required said:
Thanks everyone for your help! I've been sketching various stages in its falling and am carrying out several calculations. But then I had to sleep. I am true in also thinking that mechanical energy is conserved yes?

Yes indeed, and I just worked it out that way.

[tex]U = mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2 = mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\left( {\frac{v}{{L/2}}} \right)^2[/tex]

[tex]U = mgh + \frac{1}{2}mv^2 + \frac{{2I}}{{L^2 }}v^2 = mgh + \frac{1}{2}mv^2 + \frac{1}{6}mv^2 = mgh + \frac{2}{3}mv^2[/tex]

[tex]\frac{{dU}}{{dt}} = mg\frac{{dh}}{{dt}} + \frac{4}{3}mv\frac{{dv}}{{dt}} = - mgv + \frac{4}{3}mv\frac{{dv}}{{dt}} = \left[ {\frac{4}{3}\frac{{dv}}{{dt}} - g} \right]mv = 0[/tex]

[tex]\frac{4}{3}\frac{{dv}}{{dt}} = g[/tex]

[tex] \frac{{dv}}{{dt}} = a = \frac{3}{4}g[/tex]

[tex]F = ma = mg - N = \frac{3}{4}mg[/tex]

[tex]N = mg - \frac{3}{4}mg = \frac{1}{4}mg[/tex]
 
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  • #11
Dan, I don't think you can use v=(l/2)w in this case. Look at the motion of the point in contact with the ground. The velocity of that point is always parallel to the ground. The center of mass falls in a straight line with its velocity perpendicular to the ground. So the the center of rotation is at a point from which both these velocites are perpendicular.

So we have
[tex] mg - N = ma [/tex]
as the only forces are in vertical direction. In fact this is why the COM falls in a straight line

[tex] mg\frac{l}{2}\cos\theta = I\alpha [/tex]

The Torque due to the Normal is 0. Because the perpendicular distance from the Normal force to the center of roation is 0.


[tex] I=\frac{1}{12}ml^2 + \frac{1}{4} ml^2(\cos\theta)^2 [/tex]

From parallel axis theorem
and also

[tex] a=\frac{l}{2}(\cos\theta)\alpha [/tex]

From this N can be found.

EDIT: The angle theta is between the horizontal and the rod.
 
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  • #12
siddharth said:
Dan, I don't think you can use v=(l/2)w in this case. Look at the motion of the point in contact with the ground. The velocity of that point is always parallel to the ground. The center of mass falls in a straight line with its velocity perpendicular to the ground. So the the center of rotation is at a point from which both these velocites are perpendicular.

The kinetic energy of a translating rotating object can be written as the sum of the translational kinetic energy of the center of mass plus the rotational kinetic energy about the center of mass. As you said, the center of mass in this problem moves in a straight line, and it is the point whose acceleration is along the line of the net force. So that is the center of rotation. The point of contact is not a stationary point. If it were constrained to not move, you would have pure rotation around that point and would use the parallel axis theorem to find the moment of inertia.

If the point of contact was the center of rotation, the moment of inertia about that point would be constant. It cannot depend on orientation, i.e., it cannot be a function of theta.
 
  • #13
I am not talking about the point of contact with the ground. I am talking about the instantaneous center of rotation.

To find it, first Draw the diagram of the rod at an angle theta(with horizontal).
Draw arrows indicating the direction of velocites of the end of the rod in contact with the ground (parallel to the horizontal) and the velocity of the center of mass. Now draw two perpendiculars to the velocity arrows you have drawn. The point where they meet is the instantaneous center of rotation. About this point the body purely rotates for that instant.

In your case, you assumed that the COM is falling with a velocity v and angular velocity is w about the Center of Mass. Calculating the velocity of the point in contact with the ground we get, v(P)=-vcm(j) + l/2wsin(theta)(i) + l/2wcos(theta)(j).

But the velocity of that point is directed along the x-axis only. Therefore the y-component of the velocity is 0.

From that you get v=l/2wcos(theta) and NOT v=(l/2)w
 
  • #14
siddharth said:
I am not talking about the point of contact with the ground. I am talking about the instantaneous center of rotation.

To find it, first Draw the diagram of the rod at an angle theta(with horizontal).
Draw arrows indicating the direction of velocites of the end of the rod in contact with the ground (parallel to the horizontal) and the velocity of the center of mass. Now draw two perpendiculars to the velocity arrows you have drawn. The point where they meet is the instantaneous center of rotation. About this point the body purely rotates for that instant.

In your case, you assumed that the COM is falling with a velocity v and angular velocity is w about the Center of Mass. Calculating the velocity of the point in contact with the ground we get, v(P)=-vcm(j) + l/2wsin(theta)(i) + l/2wcos(theta)(j).

But the velocity of that point is directed along the x-axis only. Therefore the y-component of the velocity is 0.

From that you get v=l/2wcos(theta) and NOT v=(l/2)w

Then I think you have a contradiction

siddharth said:
So we have
[tex] mg - N = ma [/tex]
as the only forces are in vertical direction. In fact this is why the COM falls in a straight line

[tex] mg\frac{l}{2}\cos\theta = I\alpha [/tex]

The Torque due to the Normal is 0. Because the perpendicular distance from the Normal force to the center of roation is 0.

The torque in your equation [itex] mg\frac{l}{2}\cos\theta = I\alpha [/itex] is the torque about the contact point. The statement that the torque due to the normal is zero is only true if the contact point is the point of rotation. If there is some other point of rotation, then the normal force produces a torque.

Your assertion that the velocity of the point of contact is horizontal is clearly correct. The floor constrains the motion. Look at what that velocity is in relation to the CM.

[tex] v_{cm} = -v_{cm}j[/tex]

[tex]v_P = -v_{cm}j + \frac{L}{2}\omega sin(\theta)i + \frac{L}{2}\omega cos(\theta)j[/tex]


and their DIFFERENCE describes the relative motion between them

[tex]v_{rel} = v_P - v_{cm} = \frac{L}{2}\omega sin(\theta)i + \frac{L}{2}\omega cos(\theta)j[/tex]


The relative velocity is perpendicular to the rod, and its magnitude is l/2w, as it must be. If you were in the center of mass frame of reference, you would see the end of the rod in circular motion around you at a distance L/2, a fact that you used to figure out the horizontal velocity of the contact point.
 
  • #15
OlderDan,

But your solution is independent A, the initial angle. How is that possible? If A = 0 then N = mg, not mg/4.
 
  • #16
First of all i think the both of us are using different theta's.
Then,

OlderDan said:
The torque in your equation [itex] mg\frac{l}{2}\cos\theta = I\alpha [/itex] is the torque about the contact point.

No, The torque in the above statement is NOT about the contact point but about the Instantaneos center of Rotation. I am unable to explain this clearly without a diagram. Anyway, let me try again

Draw a straight vertical line along the normal force from the end of the rod. Draw a straight horizontal line from the center of the rod perpendicular to the direction of mg. The point where they meet is the instantaneous center of rotation (not to be confused with the contact point).

OlderDan said:
The statement that the torque due to the normal is zero is only true if the contact point is the point of rotation. If there is some other point of rotation, then the normal force produces a torque.

If you find the torque due to normal about the instantaneous center of rotation, you will find it is zero


OlderDan said:
Look at what that velocity is in relation to the CM.

[tex] v_{cm} = -v_{cm}j[/tex]

[tex]v_P = -v_{cm}j + \frac{L}{2}\omega sin(\theta)i + \frac{L}{2}\omega cos(\theta)j[/tex]


and their DIFFERENCE describes the relative motion between them

[tex]v_{rel} = v_P - v_{cm} = \frac{L}{2}\omega sin(\theta)i + \frac{L}{2}\omega cos(\theta)j[/tex]

From the Ground frame, the velocity of the contact point is only along the x-axis. (As you said, it is constrained to move on the x-axis)

Also From the ground frame the velocity of the contact point is

[tex]v_P = -v_{cm}j + \frac{L}{2}\omega sin(\theta)i + \frac{L}{2}\omega cos(\theta)j[/tex]

But the component of the velocity in the ground frame in the y-direction is zero. So from that you get vcm=l/2wcos(theta).


OlderDan said:
If you were in the center of mass frame of reference, you would see the end of the rod in circular motion around you at a distance L/2, a fact that you used to figure out the horizontal velocity of the contact point.

I used the fact that v<vector>=r<vector>w
But there v is not equal to l/2w.
 
  • #17
jdavel said:
OlderDan,

But your solution is independent A, the initial angle. How is that possible? If A = 0 then N = mg, not mg/4.

Good point! There is indeed an error in my analysis. The original premise of conservation of energy is incorrect, not the geometry. I will see if I can make the appropriate correction and still approach the problem from a work energy prespective.
 
  • #18
OlderDan said:
The original premise of conservation of energy is incorrect, not the geometry.
Why wouldn't mechanical energy be conserved?
 
  • #19
Doc Al said:
Why wouldn't mechanical energy be conserved?

Because work is being done on the rod by the normal force. Just because the point of contact of the force with the rod is not moving does not mean the object as a whole is not moving. We know the center of mass will move. I need to add a term that accounts for the F(h).dh
 
  • #20
Maybe I'm missing something (besides a few marbles), but why not treat this as a straightforward dynamics problem?
From Newton's 2nd law:
[tex]mg - N = ma[/tex]

[tex]N (L/2) \sin A = I \alpha[/tex]

The rotational inertia about the cm is:
[tex]I = 1/12 m L^2[/tex]

The constraint of the contact point slipping along the floor gives:
[tex]a = \alpha (L/2)\sin A [/tex]

Solving the above, I get:
[tex]N = mg/(1 + 3\sin^2 A)[/tex]
 
  • #21
OlderDan said:
Because work is being done on the rod by the normal force. Just because the point of contact of the force with the rod is not moving does not mean the object as a whole is not moving. We know the center of mass will move. I need to add a term that accounts for the F(h).dh
While you can certainly apply Newton's law, the fact that the contact point does not move (in the direction of the force) tells you that the normal force does no work. Energy should be conserved.
 
  • #22
siddharth said:
I am not talking about the point of contact with the ground. I am talking about the instantaneous center of rotation.

To find it, first Draw the diagram of the rod at an angle theta(with horizontal).
Draw arrows indicating the direction of velocites of the end of the rod in contact with the ground (parallel to the horizontal) and the velocity of the center of mass. Now draw two perpendiculars to the velocity arrows you have drawn. The point where they meet is the instantaneous center of rotation. About this point the body purely rotates for that instant.

In your case, you assumed that the COM is falling with a velocity v and angular velocity is w about the Center of Mass. Calculating the velocity of the point in contact with the ground we get, v(P)=-vcm(j) + l/2wsin(theta)(i) + l/2wcos(theta)(j).

But the velocity of that point is directed along the x-axis only. Therefore the y-component of the velocity is 0.

From that you get v=l/2wcos(theta) and NOT v=(l/2)w

OK.. I'm starting to understand and come around to your point of view Your relatiohship between v and theta looks correct. Mine is definitely wrong. I'm going to keep working on this from the energy conservation perspective. I'm even starting to believe in that again.
 
  • #23
Doc Al said:
Maybe I'm missing something (besides a few marbles), but why not treat this as a straightforward dynamics problem?
From Newton's 2nd law:
[tex]mg - N = ma[/tex]

[tex]N (L/2) \sin A = I \alpha[/tex]

The rotational inertia about the cm is:
[tex]I = 1/12 m L^2[/tex]

The constraint of the contact point slipping along the floor gives:
[tex]a = \alpha (L/2)\sin A [/tex]

Solving the above, I get:
[tex]N = mg/(1 + 3\sin^2 A)[/tex]

A is the angle rod makes with the vertical? If so, that's the same answer you get if you do it my way.
 
  • #24
siddharth said:
A is the angle rod makes with the vertical? If so, that's the same answer you get if you do it my way.
Yes, A is given as the angle with the vertical. And, yes, there several ways to skin this cat. :smile:
 
  • #25
Doc Al said:
Maybe I'm missing something (besides a few marbles), but why not treat this as a straightforward dynamics problem?
From Newton's 2nd law:
[tex]mg - N = ma[/tex]

[tex]N (L/2) \sin A = I \alpha[/tex]

The rotational inertia about the cm is:
[tex]I = 1/12 m L^2[/tex]

The constraint of the contact point slipping along the floor gives:
[tex]a = \alpha (L/2)\sin A [/tex]

Solving the above, I get:
[tex]N = mg/(1 + 3\sin^2 A)[/tex]

I think I have finally caught up to you and siddharth on this thing. You are right of course about the conservation of energy. I guess I had a serious brain burp, or a senior moment on that one. siddharth was right all along on the velocity issue, though it took me to his latest discription to get in touch with his instantaneous center of rotation notion. A picture is helpful in this regard, so I drew one up and am posting it here. Please check it siddharth and make sure I have interpreted it correctly. The diagram does seem to fit with the case you were making and the final result.

I did manage to work through the energy approach and got the same result once I fixed the velocity to angular velocity relationship. It does not add anything to what has already been done, except that in the course of doing it I believe I have extended the problem to the case of what happens to the normal force after the rod is in motion. It is not a closed form solution because there is a term that depends on the angular (or linear if you want to put it in those terms) velocity. It is posted here from the F = ma and [itex] \tau = I\alpha[/itex] point of view that you followed, and includes the calculation of the relationship between a and [itex]\alpha[/itex] that leads to the velocity dependence of the normal force.

Newton's second law

[tex] h = \frac{L}{2}\cos A [/tex]

[tex] \frac{{dh}}{{dt}} = - \frac{L}{2}\omega \sin A = v [/tex]

[tex] \frac{{d^2 h}}{{dt^2 }} = - \frac{L}{2}\left( {\alpha \sin A + \omega ^2 \cos A} \right) = a [/tex]

[tex] N - mg = ma = - m\frac{L}{2}\left( {\alpha \sin A + \omega ^2 \cos A} \right) [/tex]

[tex] N - mg = - \frac{{mL\alpha }}{2}\sin A - \frac{{mL\omega ^2 }}{2}\cos A [/tex]

Torque

[tex] N\frac{L}{2}\sin A = I\alpha {\rm{ ; }}I = \frac{{mL^2 }}{{12}} [/tex]

[tex]
\frac{{N\sin A}}{2} = \frac{{I\alpha }}{L} = \frac{{mL\alpha }}{{12}} = \frac{1}{6}\frac{{mL\alpha }}{2} [/tex]

[tex] \frac{{mL\alpha }}{2} = 3N\sin A [/tex]

Substituting

[tex] N - mg = - \frac{{mL\alpha }}{2}\sin A - \frac{{mL\omega ^2 }}{2}\cos A [/tex]

[tex] N = mg - 3N\sin ^2 A - \frac{{mL\omega ^2 }}{2}\cos A [/tex]

[tex] N\left( {1 + 3\sin ^2 A} \right) = mg - \frac{{mL\omega ^2 }}{2}\cos A [/tex]

[tex] N = \frac{{2mg - mL\omega ^2 \cos A}}{{2 + 6\sin ^2 A}} [/tex]

For the stated condition that the Normal force at the time of release is to be found, this result reduces to your result because at that moment [itex]\omega[/itex] is zero.

I think an interpretation of this result is that if you started with an initial [itex]\omega[/itex] you might get to a point where the rotation creates separation from the floor before the rod hits flat. I'm wondering if it might happen even if you release the rod from rest. I dropped a pencil a few times, and it looks to me like there might be separation before it hits, but it is hard to tell. Someone else might want to explore that and get a closed form solution, but I'm done with this one for a while.
 

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  • #26
That is a very nicely drawn picture OlderDan! thanks for posting it. Has it been assumed that the right end stays in contact with the table throughout the whole motion? It seems like that would be true, but how do we know for sure?
 
  • #27
HackaB said:
That is a very nicely drawn picture OlderDan! thanks for posting it. Has it been assumed that the right end stays in contact with the table throughout the whole motion? It seems like that would be true, but how do we know for sure?

The original problem is only asking for the normal force when the rod is first released, so it is not an issue for the problem. The wording of the problem is suggestive that things change after the rod starts moving, and I believe I have shown that to be the case. I suspect that it might not maintain contact throughout, at least if started high enough, but I certainly have not demonstrated that, and I'm not sure.
 
  • #28
wow! its amazing how tricky a simple enough looking question can get! It seems as if the answer posted by several of you is correct by a majority vote. I have learned a lot from this thread. Thankyou. I will be getting the solution to this next week. I will post what my lecturer has to say about it.

Thanks again
 

1. What is the force exerted by a table on a stick at release?

The force exerted by a table on a stick at release refers to the amount of force that the table exerts on the stick when it is released from the table's surface. This force is often referred to as the normal force and is perpendicular to the surface of the table.

2. How is the force exerted by a table on a stick at release calculated?

The force exerted by a table on a stick at release can be calculated using Newton's Second Law of Motion, which states that force is equal to mass multiplied by acceleration (F=ma). In this case, the mass of the stick and the acceleration due to gravity can be used to calculate the force.

3. Does the force exerted by a table on a stick at release change with the weight of the stick?

Yes, the force exerted by a table on a stick at release will change with the weight of the stick. The greater the weight of the stick, the greater the force exerted by the table on the stick at release will be.

4. What factors can affect the force exerted by a table on a stick at release?

The force exerted by a table on a stick at release can be affected by various factors such as the weight and mass of the stick, the type of surface the table is on, and the angle at which the stick is released from the table.

5. Is the force exerted by a table on a stick at release the same as the force of gravity?

No, the force exerted by a table on a stick at release and the force of gravity are not the same. The force of gravity is the force that pulls objects towards the center of the Earth, while the force exerted by a table on a stick at release is the force that the table exerts on the stick in a perpendicular direction.

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