A rod initially stands vertically, then falls.

[Psi-String]

Though I solved this problem, the answer of mine is strange. Could someone check my solution, please? Thanks in advance.

Problem:
A rod with length L and mass M initially stands vertically on the ground. A small perturbation makes it start to falling down. What's the force exerted by the ground on the rod when the rod is inclined 45 degree with respect to the horizon if the contact between rod and ground is (a)rigid or (b) frictionless ?

My solution to (a):

Since the normal force N and frictional force f can't do work on the rod, the energy of the rod-earth system is conserved. Hence

$$Mg \frac{L}{2}(1-\frac{1}{\sqrt{2}})=\frac{1}{2} I \omega ^2 =\frac{1}{2} \times \frac{1}{3}ML^2 \omega^2$$

so we can know $$a_{C.M.r} =\omega^2 \frac{L}{2}=\frac{3g}{2} (1-\frac{1}{\sqrt{2}})$$

By $$\tau = I \alpha$$ ,

$$Mg(sin\frac{\pi}{4})\frac{L}{2} = (\frac{1}{3}ML^2)\alpha$$

so $$\alpha=\frac{3}{2\sqrt{2}} \frac{g}{L}$$

then $$a_{C.M.t}= \alpha \frac{L}{2} = \frac{3g}{4\sqrt{2}}$$

So now we have $$\vec{a_{C.M}}=(-\frac{3\sqrt{2}}{4}+\frac{9}{8})g \hat{i}+(\frac{3\sqrt{2}}{4}+\frac{3}{8})g \hat{j}$$

Since $$\vec{F}_{net}= M \vec{a}_{C.M}=M\vec{g}+\vec{F}_{floor}$$

we find $$\vec{F}_{floor} =Mg(-\frac{3\sqrt{2}}{4}+\frac{9}{8}) \hat{i}+Mg(\frac{3\sqrt{2}}{4}+\frac{11}{8}) \hat{j}$$
My solution to (b):

Since the contact between rod and ground is frictionless, we know that x-directional linear momenum is conserved. So the center of mass of the rod will always on y-axis.

From $$\frac{L}{2} sin\theta =y$$

we find $$\frac{L}{2} \frac{d\theta}{dt} cos\theta = v$$ ------------ Eq.1

and $$\frac{L}{2} ( \frac{d^2\theta}{dt^2} cos\theta -\frac{d\theta}{dt} \frac{d\theta}{dt} sin\theta) = \frac{L}{2}(\alpha cos \theta -\omega^2 sin\theta) = a$$ ---------Eq.2

By the conservation of energy, we have

$$Mg\frac{L}{2} (1-sin\theta) = \frac{1}{2}Mv^2+\frac{1}{2} I \omega^2=\frac{1}{2} Mv^2+\frac{1}{24} ML^2 \omega^2$$

substitute v by Eq.1 and $$\theta = \frac{\pi}{4}$$ we have $$\omega^2 =\frac{24}{5} (1-\frac{1}{\sqrt{2}}) \frac{g}{L}$$

because$$\alpha = \frac{3\sqrt{2} g}{L}$$ and by Eq.2

$$a=\frac{L}{2} ( \frac{3\sqrt{2} g}{L} \times \frac{1}{\sqrt{2}} -\frac{24}{5} (1-\frac{1}{\sqrt{2}}) \times \frac{1}{\sqrt{2}}) = (\frac{3}{2}+\frac{12}{5}-\frac{12\sqrt{2}}{5})g$$

then
$$N= mg-ma= mg(1-(\frac{3}{2}+\frac{12}{5}-\frac{12\sqrt{2}}{5})=mg(\frac{24\sqrt{2}-29}{10})$$
Thanks for help! I'll be very very appriciated!

 

[Andrew Mason]

A rod with length L and mass M initially stands vertically on the ground. A small perturbation makes it start to falling down. What's the force exerted by the ground on the rod when the rod is inclined 45 degree with respect to the horizon if the contact between rod and ground is (a)rigid or (b) frictionless ?

a) The net force on the rod: gravitational force - normal force, has to be equal to the mass of the rod x its vertical acceleration: mg - N = ma(cos45), so N = mg-ma(cos45)

$$\tau = mgL/2 = I\alpha = \frac{1}{3}mL^2\alpha = mgsin(\theta)L/2$$

$$\alpha = \frac{3gsin(45)}{2L}$$

Using $a_{cm} = \alpha L/2$,

$$a_{cm} = \frac{3g\sqrt{2}}{8}$$

so:

$$N = mg-ma(cos45) = mg - ma\sqrt{2}/2 = mg(1 - \frac{3}{16}) = 13mg/16$$

The only difference in b) is that the acceleration is not perpendicular to the rod (as it is in a)) but is vertical, so you don't have to multiply a by cos45

AM

 

[OlderDan]

a) The net force on the rod: gravitational force - normal force, has to be equal to the mass of the rod x its vertical acceleration: mg - N = ma(cos45), so N = mg-ma(cos45)

$$\tau = mgL/2 = I\alpha = \frac{1}{3}mL^2\alpha = mgsin(\theta)L/2$$

$$\alpha = \frac{3gsin(45)}{2L}$$

Using $a_{cm} = \alpha L/2$,

$$a_{cm} = \frac{3g\sqrt{2}}{8}$$

so:

$$N = mg-ma(cos45) = mg - ma\sqrt{2}/2 = mg(1 - \frac{3}{16}) = 13mg/16$$

The only difference in b) is that the acceleration is not perpendicular to the rod (as it is in a)) but is vertical, so you don't have to multiply a by cos45

AM

I can't stay to work on this right now, but I think you have left out the horizontal acceleration. In part a) the rod pivots about a stationary point, so there is angular acceleration which means the rod is accerating both horizontally and vertically. I did a calculation using centripetal acceleration of the CM that resulted in the net radial force from the pivot being F = (Mg/2)(5cosθ - 3) = .268Mg at 45°

 

[Psi-String]

a) The net force on the rod: gravitational force - normal force, has to be equal to the mass of the rod x its vertical acceleration: mg - N = ma(cos45), so N = mg-ma(cos45)

$$\tau = mgL/2 = I\alpha = \frac{1}{3}mL^2\alpha = mgsin(\theta)L/2$$

$$\alpha = \frac{3gsin(45)}{2L}$$

Using $a_{cm} = \alpha L/2$,

$$a_{cm} = \frac{3g\sqrt{2}}{8}$$

so:

$$N = mg-ma(cos45) = mg - ma\sqrt{2}/2 = mg(1 - \frac{3}{16}) = 13mg/16$$

The only difference in b) is that the acceleration is not perpendicular to the rod (as it is in a)) but is vertical, so you don't have to multiply a by cos45

AM

The $$a_{C.M}$$ you calculated is the tangent acceleration of C.M , but don't we need to know the radial acceleration of C.M ?

 

[OlderDan]

The $$a_{C.M}$$ you calculated is the tangent acceleration of C.M , but don't we need to know the radial acceleration of C.M ?

Yes you do need the centripetal acceleration. I think your method for part (a) is right, but you have made a sign mistake somewhere in the y component. I'm quite certain the result should be with a +11/8 (you added Mg instead of subtracting) and the other term negative. I think that amounts to using the wrong sign for the y acceleration component in your last step. Your final result seems unreasonable, since it says the y-component is greater than the weight of the rod. I can't imagine that happening.

 

[Psi-String]

Yes you are right! I made some mistake in the calculation.
How about (b) ? Any suggestion??

Thanks a lot!

 

[Andrew Mason]

I can't stay to work on this right now, but I think you have left out the horizontal acceleration. In part a) the rod pivots about a stationary point, so there is angular acceleration which means the rod is accerating both horizontally and vertically. I did a calculation using centripetal acceleration of the CM that resulted in the net radial force from the pivot being F = (Mg/2)(5cosθ - 3) = .268Mg at 45°

I did the calculation for the normal force. The horizontal force would have the same magnitude, but horizontally.

AM

 

[Andrew Mason]

The $$a_{C.M}$$ you calculated is the tangent acceleration of C.M , but don't we need to know the radial acceleration of C.M ?

There is no radial acceleration in a). The radius does not change.

AM

 

[Psi-String]

There is no radial acceleration in a). The radius does not change.

AM

The radius does not change, but the direction of the velocity of C.M does change, so it means that it no only has a tangent acceleration but also has a radial acceleraion to change its direction.

 

[OlderDan]

There is no radial acceleration in a). The radius does not change.

AM

That is true. dr/dt = 0 and d²r/dt² = 0. But there is a component of the acceleration of the CM in the "radial direction", the centripetal acceleration.

 

[OlderDan]

Yes you are right! I made some mistake in the calculation.
How about (b) ? Any suggestion??

Thanks a lot!

I must say you did a better job with (b) than I did the first time I looked at it some time ago.

https://www.physicsforums.com/showthread.php?t=73192

I'm not getting your answer though. I did make an algebra mistake that I think I have fixed. As of the moment, the result I have is

$$N = \frac{2}{{25}}\left[ {11 - 6\sqrt 2 } \right] Mg = .201Mg$$

Your result is considerably larger than mine at .494Mg. I thought your approach was OK at first, but I don't see where you got your equation

$$\alpha = \frac{3\sqrt{2} g}{L}$$

I think the g in that equation should in fact be N/M, and N is definitely not Mg

 

[Psi-String]

At first, I got alpha by calculating the torque about the contact point between the rod and ground. But since the contact point is moving, I think I can't calculate the torque by

$$Mg \times \frac{L}{2} \times cos\frac{\pi}{4}$$

If we calculate the torque about C.M, it seems that the torque will be

$$N \times \frac{L}{2} \times cos\frac{\pi}{4}$$

but C.M is moving too. And I believe that the alpha about contact point should be the same to the alpha about C.M.

I'm thinking that maybe we could just differentiate both side of this equation

$$\omega^2=\frac{12g}{L}(\frac{1-sin \theta}{3cos^2 \theta +1})$$

which come from $$\frac{L}{2} \frac{d\theta}{dt} cos\theta = v$$ and $$Mg\frac{L}{2} (1-sin\theta) = \frac{1}{2}Mv^2+\frac{1}{2} I \omega^2=\frac{1}{2} Mv^2+\frac{1}{24} ML^2 \omega^2$$ theta is the angle between the rod and the ground.

By this approach we do not need to calculate the torque, which invlove the factor of a moving reference point.

and I got $$\alpha =\frac{6g}{L} \times \frac{6cos\theta sin \theta)(1-sin\theta)-cos\theta(3cos^2 \theta +1)}{(3cos^2 \theta +1 )^2}$$

we can find alpha at 45 degree by setting theta=45 degree and I found

$$N=\frac{Mg}{25} (25-12\sqrt{2})$$

How do you think about this idea??

 

[OlderDan]

I'm thinking that maybe we could just differentiate both side of this equation

$$\omega^2=\frac{12g}{L}(\frac{1-sin \theta}{3cos^2 \theta +1})$$

By this approach we do not need to calculate the torque, which invlove the factor of a moving reference point.

and I got $$\alpha =\frac{6g}{L} \times \frac{6cos\theta sin \theta)(1-sin\theta)-cos\theta(3cos^2 \theta +1)}{(3cos^2 \theta +1 )^2}$$

we can find alpha at 45 degree by setting theta=45 degree and I found

$$N=\frac{Mg}{25} (25-12\sqrt{2})$$

How do you think about this idea??

I like the idea. I had used the torque about the CM, but your approach is better. If the torque about the CM is valid, then your way should prove it. The result I got earlier was the same as this except I have a 22 inside the parentheses instead of a 25. I used the derivative of ω² as you had suggested, and it did not change my result.

 

[OlderDan]

I like the idea. I had used the torque about the CM, but your approach is better. If the torque about the CM is valid, then your way should prove it. The result I got earlier was the same as this except I have a 22 inside the parentheses instead of a 25. I used the derivative of ω² as you had suggested, and it did not change my result.

While the site was away, I finally managed to work through all the algebra needed to reduce the general expression for the normal force as a fuction of angle down to the same result using either the derivative of ω² or torque about the CM to find α.

Had I known we were going to be away so long, I would have made the point sooner, but I thought your idea was worth pursuing. You should be aware of the general validity of using torque about the CM. See the annotations in your quote that follows.

At first, I got alpha by calculating the torque about the contact point between the rod and ground. But since the contact point is moving, I think I can't calculate the torque by

$$Mg \times \frac{L}{2} \times cos\frac{\pi}{4}$$

You are correct. You cannot calculate the angular accelertation about some arbitrary point that is moving. If it is moving with constant velocity, you can work in a frame where it is stationary, but when it is accelerating as it is in this problem, there is no inertial frame of reference with the point stationary. If you read the discussion in the thread I posted earlier

https://www.physicsforums.com/showthread.php?t=73192

and pay attention to the comments by siddharth you will see that he talks about a point of rotation that is momentarily at rest about which you can calculate the angular acceleration. The problem is that point is constantly changing, and it is not even a point within the object.

If we calculate the torque about C.M, it seems that the torque will be

$$N \times \frac{L}{2} \times cos\frac{\pi}{4}$$

but C.M is moving too. And I believe that the alpha about contact point should be the same to the alpha about C.M.

The bad news is the calculation about the contact point does not give the correct angular acceleration. The good news is there is only one angular acceleration and it can be found by analyzing the torque about the center of mass. The center of mass is a special point in a system of particles. That is why we can treat the translational motion as if all the forces were acting at that point. It also permits us to analyze the rotation about the CM using the torques relative to the CM. The tedious algebra I did for this problem demonstrates the validity of this approach, at least for this problem. And it is a general result.

I'm thinking that maybe we could just differentiate both side of this equation

$$\omega^2=\frac{12g}{L}(\frac{1-sin \theta}{3cos^2 \theta +1})$$

which come from $$\frac{L}{2} \frac{d\theta}{dt} cos\theta = v$$ and $$Mg\frac{L}{2} (1-sin\theta) = \frac{1}{2}Mv^2+\frac{1}{2} I \omega^2=\frac{1}{2} Mv^2+\frac{1}{24} ML^2 \omega^2$$ theta is the angle between the rod and the ground.

By this approach we do not need to calculate the torque, which invlove the factor of a moving reference point.

and I got $$\alpha =\frac{6g}{L} \times \frac{6cos\theta sin \theta)(1-sin\theta)-cos\theta(3cos^2 \theta +1)}{(3cos^2 \theta +1 )^2}$$

we can find alpha at 45 degree by setting theta=45 degree and I found

$$N=\frac{Mg}{25} (25-12\sqrt{2})$$

How do you think about this idea??

As I said before, I think it is a good idea. I used it and found that this approach confirms the validity of the torque about the CM approach.

You are not the first and you will not be the last to struggle with this concept. At least you recognized that you can't just calculate the torque and angular acceleration around any old point in a dynamic situation. You can however use the CM and it will often simplify the calculation.

One reason I wanted to work out the general solution to the problem was because in that earlier discussion I had speculated about the possibility of the rotation becoming fast enough so that the contact point would separate from the ground before the rod became horizontal. I should still work it out for an arbitray starting angle, but at least when starting from vertical I now know that the normal force never reaches zero, so contact is maintained the whole time until the rod reaches horizontal. There is actually a minimum in the normal force at 61° from the vertical.

The general result expressed in terms of the angle from the vertical is.

$$N = Mg\frac{{4 - 6\cos \theta + 3\cos ^2 \theta }}{{\left( {4 - 3\cos ^2 \theta } \right)^2 }} = Mg\frac{{7 - 6\cos \theta - 3\sin ^2 \theta }}{{\left( {1 + 3\sin ^2 \theta } \right)^2 }}$$

Since you used the angle from the horizontal, if you just swap sin for cos and vice versa it should give your result. If you compare this to your equation for α (and use the Pythagorean identity to simplify yours) you will see that it is consistent with the result you would obtain using the torque about the CM.

 

[Psi-String]

My result is indeed $$\frac{Mg}{25} [22-12\sqrt{2}]$$

I typed the wrong number in last reply. Sorry for this mistake. And I didn't notice that your result is the same as mine, I'm such a stupid

Anyway, I figured out why we can calculate the torque about C.M without considering the acceleration of C.M. Since the observer is in a noninertial frame with an accleration $$a_{C.M}$$, he will observe that every point seems to experience a force $$F=ma_{C.M}$$, so for every tiny element on the rod, it experiences a force $$F= dm a_{C.M}$$, and by the definition of C.M the net torque of every element of rod about C.M will be zero. So when we want to calculate the net torque about the C.M, we just only need to consider the force "really" act on the rod, in this case, is the normal force.

Thanks for your patient to spending time answering my question, it's very nice to have someone to discuss. Thanks again.

 

[Andrew Mason]

The radius does not change, but the direction of the velocity of C.M does change, so it means that it no only has a tangent acceleration but also has a radial acceleraion to change its direction.

Yes. I was forgetting that the rod falls from the vertical so it has centripetal acceleration. This has to be supplied by the component of gravity along the length of the rod. ie. the radial components of the force of gravity and normal force add up to the centripetal force on the centre of mass of the rod:

(1)$$F_r = mg\sin(45) - N\sin(45) = m\omega^2 L/2$$

Finding the angular speed using energy as you have done:

$$mgh = \frac{1}{2}I\omega^2 = \frac{1}{6}mL^2\omega^2$$

where h is the distance the centre of mass has fallen (h=Lsin(45)/2) gives you an expression for N.

AM

 

[OlderDan]

(1)$$F_r = mg\sin(45) - N\sin(45) = m\omega^2 L/2$$

AM

The force at the pivot point is not directed along the length of the rod, which seems to be what you are saying with N sin(45°) and in a comment you made in an earlier post. There is a point where the radial component of the force at the pivot goes to zero because the radial component of gravity equals the centripetal force. I found that to be at about 53° from the vertical. At that point there is still a component perpendicular to the radial direction, with a positive vertical component and a negative horizontal component.

The horizontal component changes direction at 48° to the vertical. Clearly the horizontal component has to change direction at some point because the horizontal velocity increases from zero and then returns to zero as the rod goes from vertical to horizontal

 

[Andrew Mason]

The force at the pivot point is not directed along the length of the rod, which seems to be what you are saying with N sin(45°) and in a comment you made in an earlier post. There is a point where the radial component of the force at the pivot goes to zero because the radial component of gravity equals the centripetal force. I found that to be at about 53° from the vertical. At that point there is still a component perpendicular to the radial direction, with a positive vertical component and a negative horizontal component.

The horizontal component changes direction at 48° to the vertical. Clearly the horizontal component has to change direction at some point because the horizontal velocity increases from zero and then returns to zero as the rod goes from vertical to horizontal

Ok. One would have to add the radial component of the horizontal friction force:

(1)$$F_r = mg\sin(45) - N\sin(45) + F_f = m\omega^2 L/2$$

where:

$$\omega^2 = 6gh/L^2 = 3g\sin(45)/L$$

You can use the vertical component of the tangential acceleration to give an equation for the upward component of force (the Normal force, N). That will give you two equations with two unknowns ($F_f$ and N) which should be solveable.

AM

 

[OlderDan]

Ok. One would have to add the radial component of the horizontal friction force:

(1)$$F_r = mg\sin(45) - N\sin(45) + F_f = m\omega^2 L/2$$

where:

$$\omega^2 = 6gh/L^2 = 3g\sin(45)/L$$

You can use the vertical component of the tangential acceleration to give an equation for the upward component of force (the Normal force, N). That will give you two equations with two unknowns ($F_f$ and N) which should be solveable.

AM

Friction will take care of the additional force as long as the angle from the vertical is not too large and it is static friction. I wanted to do the problem at all angles, so I took the statement of the problem about the rigid contact at face value and imagined a stationary frictionless pivot. There is a point at which the vertical force goes to zero, so all friction would be lost. Trying to maintain a constant ratio of frictional force to a vertical normal force at large angles to the vertical would make things even more complicated. Here are the graphs I got for the force components relative to the vertical angle θ. For the stationary pivot I graphed the horizontal and vertical components of the pivot force, and then the radial and perpendicular components. There is also a graph of the net pivot force magnitude, and finally a graph of the normal force for the case of frictionless contact. The graphing program only likes radians, and the force units are Mg.