Rotational Mechanics - Question

[sankalpmittal]

Homework Statement

A uniform rod of length "L" is kept vertically on a smooth horizontal surface at a point "O". If it is rotated slightly and released, it falls down on the horizontal surface. To what distance will the lower end of the rod "shift" from point "O" ?

Note : The answer in the book is given terms of "L"..

Homework Equations

Formulas of rotations. Check the wikipedia link.

The Attempt at a Solution

I can not think of any way. I tried integrating Lcosθ/2 from 0 to pi/2. I am not sure of any logic. How will the lower point shift if the rod is doing pure rotation ? Any hints will do.

Please help !

Thanks in advance...

 

[Doc Al]

Consider the forces acting on the rod.

 

[arildno]

What external forces, if any, act upon the rod in the HORIZONTAL direction?

 

[arildno]

Blaargh, you are so fast, Doc Al.

 

[sankalpmittal]

Consider the forces acting on the rod.

The force mg acting along the centre of mass perpendicular to axis of rotation.

The normal reaction from the floor will have two components: Rsinθ and Rcosθ..

Rcosθ will balance the weight and Rsinθ will cause the shifting of lower end, right ?

I assume rod is slightly rotated by angle θ.

So what should I do after this ?

 

[arildno]

The force mg acting along the centre of mass perpendicular to axis of rotation.

The normal reaction from the floor will have two components: Rsinθ and Rcosθ..

Rcosθ will balance the weight and Rsinθ will cause the shifting of lower end, right ?

I assume rod is slightly rotated by angle θ.

So what should I do after this ?

The surface is..SMOOTH.
What does that condition entail?

 

[sankalpmittal]

The surface is..SMOOTH.
What does that condition entail?

It entails that there will be no friction acting on the rod's lower end by the floor. But how will that help me ? (Thanks BTW..)

 

[Doc Al]

Apply Newton's 2nd law. What will be the translational acceleration of the rod? (In what direction?)

 

[sankalpmittal]

Apply Newton's 2nd law. What will be the translational acceleration of the rod? (In what direction?)

Rsinθ=ma , right ?

R is the reaction force from the floor to the lower end of the rod.

 

[arildno]

It entails that there will be no friction acting on the rod's lower end by the floor. But how will that help me ? (Thanks BTW..)

So, whatever force acts at the contact point, will it be:
a) Strictly horizontal
b) Strictly vertical
c) A mixture?

 

[sankalpmittal]

So, whatever force acts at the contact point, will it be:
a) Strictly horizontal
b) Strictly vertical
c) A mixture?

A mixture, right ? I already mentioned that in my previous post. I mentioned that normal reaction at contact point will have two components, horizontal which balances the weight and vertical component which causes the acceleration in lower point.

 

[Doc Al]

I mentioned that normal reaction at contact point will have two components, horizontal which balances the weight and vertical component which causes the acceleration in lower point.

What's the definition of "normal"? And how can a horizontal force balance a weight, which is a vertical force?

 

[sankalpmittal]

What's the definition of "normal"? And how can a horizontal force balance a weight, which is a vertical force?

I am not saying that horizontal force balances the weight. If I rotate the rod slightly by angle θ, then by geometry rod will make angle (90-θ) with the floor. Now, horizontal component of normal reaction, Rcos(90-θ)= Rsinθ causes acceleration, and vertical component Rsin(90-θ)=Rcosθ balances the weight of the rod. Yes now that's correct. I mistyped in my previous post. Sorry about that !

So now, what can I do after this ?

 

[Doc Al]

You seem to think that the normal force of the floor on the rod will be parallel to the rod. No.

The problem is much easier than you think. Almost a trick question.

 

[sankalpmittal]

You seem to think that the normal force of the floor on the rod will be parallel to the rod. No.

The problem is much easier than you think. Almost a trick question.

I am sorry if I misunderstood somewhere.
The normal reaction will be "just" perpendicular to the plane of floor, right ? And it will not have components right ? Then how will there be horizontal acceleration on the lower point of rod ?

 

[Doc Al]

The normal reaction will be "just" perpendicular to the plane of floor, right ?

Right!

And it will not have components right ?

Right!

Then how will there be horizontal acceleration on the lower point of rod ?

If you divide the rod into pieces, then one piece can certainly exert a horizontal force on the other. But no need to do that. Treat the rod as a whole; no external horizontal forces act on the rod.

Hint: The net force determines the acceleration of what part of the rod?

 

[sankalpmittal]

If you divide the rod into pieces, then one piece can certainly exert a horizontal force on the other. But no need to do that. Treat the rod as a whole; no external horizontal forces act on the rod.

Hint: The net force determines the acceleration of what part of the rod?

Sorry, but net external force on the rod seems to be producing a "couple" on the rod when the rod gets slightly rotated and had unstable equilibrium, and nothing else.

 

[Doc Al]

Sorry, but net external force on the rod seems to be producing a "couple" on the rod when the rod gets slightly rotated and had unstable equilibrium, and nothing else.

So you think the net external force is zero?

 

[arildno]

The normal force to a surface is simply so large that the object the surface is in contact with does not pentrate into that surface.

Thus, the normal force is NOT, necessarily, equal to the normally directed component of gravity.

 

[sankalpmittal]

The normal force to a surface is simply so large that the object the surface is in contact with does not pentrate into that surface.

Thus, the normal force is NOT, necessarily, equal to the normally directed component of gravity.

If Normal Reaction were not equal to force of gravity on centre of mass of the rod, there will be either net force up or net force down. In former case, rod will jump up and in latter case, it will penetrate in the floor. That is why it has to equal force of gravity.

Doc Al, I can not see any net external force on the rod acting. I can only see net external torque.

 

[Doc Al]

If Normal Reaction were not equal to force of gravity on centre of mass of the rod, there will be either net force up or net force down.

That is true.

In former case, rod will jump up and in latter case, it will penetrate in the floor.

Not quite. Having a net force just means that there will be an acceleration.

That is why it has to equal force of gravity.

Doc Al, I can not see any net external force on the rod acting. I can only see net external torque.

So it just sits there? (Rotating in space?)

What does Newton's 2nd law tell you? ƩF = ma.

 

[sankalpmittal]

That is true.


Not quite. Having a net force just means that there will be an acceleration.


So it just sits there? (Rotating in space?)

What does Newton's 2nd law tell you? ƩF = ma.

ƩF=0, because you said that there will be no net external horizontal force on the rod and vertical mg is equal to normal reaction.

If net external horizontal force is 0, it seems that there will be no shifting of lower point of the rod in contact with floor from point "O".

 

[Doc Al]

ƩF=0,

No, there is a net force on the rod. It falls!

because you said that there will be no net external horizontal force on the rod

That's true.

and vertical mg is equal to normal reaction.

That's not true.

If net external horizontal force is 0, it seems that there will be no shifting of lower point of the rod in contact with floor from point "O".

That is an incorrect conclusion.

For an extended body, ƩF = ma refers to the acceleration of what point of the body?

 

[sankalpmittal]

No, there is a net force on the rod. It falls!


That's true.


That's not true.


That is an incorrect conclusion.

For an extended body, ƩF = ma refers to the acceleration of what point of the body?

Alright, so that means mg>R , where R is the normal reaction. So downward acceleration in the centre of mass of the rod will be mg-R=ma. But how will that laterally shift the lower point of the rod from contact point "O" ?

 

[Doc Al]

Alright, so that means mg>R , where R is the normal reaction. So downward acceleration in the centre of mass of the rod will be mg-R=ma.

Good. There is no horizontal net force, so the center of mass will not accelerate horizontally. But there is downward vertical net force, so the center of mass will accelerate downward.

But how will that laterally shift the lower point of the rod from contact point "O" ?

If you choose to analyze just the lower portion of the rod, then there is a horizontal force on that piece. But that is internal to the rod, so it has no effect on the motion of the center of mass of the rod.

But no need to do that analysis, since just realizing that the friction free surface implies that the center of mass can only move downward is all you need.

 

[arildno]

If Normal Reaction were not equal to force of gravity on centre of mass of the rod, there will be either net force up or net force down. In former case, rod will jump up and in latter case, it will penetrate in the floor. That is why it has to equal force of gravity.

Incorrect. The proper condition for the normal force is NOT a dynamic condition at all, but a tricky kinematic condition(s).
At all contact points "i", we have the kinematic conditions:
\vec{v}_{rel,c,i}\cdot\vec{n}_{i}\geq{0}
, where v_{rel,c,i} is the relative velocity of the object to the surface at point "i", and n_i is the outward normal vector at the surface at point "i"

It is not possible to deduce, in generality, from this, that the normal force has to be equal to gravity's normal component. That is where your error lies, in thinking the kinematic condition is equivalent to that particular dynamic condition.

 

[arildno]

The lateral shifts of the other parts of the rods are induced by internal stresses in the rod, so that points at one side of C.M will accelerate to the left (and downwards), while on the other side of C.M, the points will accelerate to the right (and downwards).

 

[ehild]

Alright, so that means mg>R , where R is the normal reaction. So downward acceleration in the centre of mass of the rod will be mg-R=ma. But how will that laterally shift the lower point of the rod from contact point "O" ?

As soon as the rod is rotated off from its vertical position the normal force from the ground causes torque with respect to the CM. The rod will rotate around the CM while the CM falls. You have two equations, one for the vertical acceleration of the CM and one for the angular acceleration β about the CM. displacement of the lover end of the rod and the hight of the CM are related through the rotation angle θ.

 

[sankalpmittal]

As soon as the rod is rotated off from its vertical position the normal force from the ground causes torque with respect to the CM. The rod will rotate around the CM while the CM falls. You have two equations, one for the vertical acceleration of the CM and one for the angular acceleration β about the CM. displacement of the lover end of the rod and the hight of the CM are related through the rotation angle θ.

ehild,

That's excellent hint but I found the method given by Doc Al and arildno simpler. As it's not necessary that normal reaction has to equal force mg acting along centre of mass of the rod, CM of the rod may or may not fall vertically i.e. may or may not have linear acceleration vertically. But anyway, the two forces acting on the rod "will" produce couple on it. By symmetry of rotation, the centre of mass will land on the distance L/2 from its original position "horizontally" right. But the net external force acting horizontally on CM is 0, and it was initially at rest, therefore it must not change its position horizontally. To compensate for this change, the lower point of the rod must shift L/2 horizontally due left so that CM again comes in its original position.

Thanks everyone (arildno, Doc Al and ehid) !

Edit:
Sorry. Normal reaction does not pass through CM, and so there will be net force on the CM downward. Will it be mg or mg-R ? I think former is correct. Also if a rod is pivoted at the end and it is left swing in a semicircle, will there be linear acceleration in its centre of mass ?

 

[Doc Al]

Normal reaction does not pass through CM, and so there will be net force on the CM downward. Will it be mg or mg-R ? I think former is correct.

What determines the acceleration of the CM is the net force on the object. It doesn't matter where the individual forces are applied.

Also if a rod is pivoted at the end and it is left swing in a semicircle, will there be linear acceleration in its centre of mass ?

Not sure exactly what you have in mind, but if the rod swings in a semicircle the CM will have an acceleration.

 

[BBAI BBAI]

As there is no acceleration of centre of mass the CM moves in a straight line its falls downwards so the end point moves L/2 Distance.

 

[Doc Al]

As there is no acceleration of centre of mass the CM moves in a straight line its falls downwards so the end point moves L/2 Distance.

The center of mass is accelerated.

 

[ehild]

Edit:
Sorry. Normal reaction does not pass through CM, and so there will be net force on the CM downward. Will it be mg or mg-R ? I think former is correct. Also if a rod is pivoted at the end and it is left swing in a semicircle, will there be linear acceleration in its centre of mass ?

Sankalpmittal,

The rod is an extended body. You can imagine it as a system of point masses, connected to each other by internal forces.

To make it simpler, assume a system of two point masses m1 and m2, their position vectors r1 and r2. The external forces, F1 and F2 act on m1 and m2, respectively, and the internal force f12 is exerted by m2 on m1 and f21 = -f12 is exerted by m1 on m2.

You can write up Newton's II for each mass:

m1a1=F1+f12

m2a2=F2-f12

Add up the equations:

m1a1+m2a2=F1+F2 . (*) The internal forces cancel.

The position vector of the CM is defined as R_CM= (m1r1+m2r2)/(m1+m2). So (*) can be written in terms of the acceleration of the CM and the total mass M=m1+m2 of the system. : Ma_CM=F1+F2.

You can get the equivalent formula for a system of any number of point masses.

The CM of the extended body moves as if all the external forces acted on it. No matter at what point the external forces act: The CM will be accelerated by the sum of forces.


The rod in the problem is acted on by gravity and the normal force (reaction force R). Gravity acts at each mass element, but the resultant of the gravitational forces acts at the CM. The normal force acts from the ground, at the lower end of the rod. The CM is accelerated by the net force mg-R.

ehild

 

[sankalpmittal]

Sankalpmittal,

The rod is an extended body. You can imagine it as a system of point masses, connected to each other by internal forces.

To make it simpler, assume a system of two point masses m1 and m2, their position vectors r1 and r2. ...

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ehild

Thanks a lot everyone which includes ehild ! I get it now.

It was all the concept of the net force on centre of mass of the rod system.