So, How to Find x - Rod Standing on Frictionless Table (Rotation)

[FedEx]

A rod standing on frictionless table(Rotation)

Homework Statement

A rod of length l is standing on a friction less surface.
A slight impulse is given to the rod and hence the rod starts falling. Find the torque,angular acceleration ,angular velocity, the normal force by the ground and the distance through which the end part of the rod slides on the ground when the rod makes an angle of $$\theta$$ withe the vertical

Homework Equations

Law of conservation of energy and equations of moment of forces.

The Attempt at a Solution

We can apply the laws of conservation of energy

$$mgh = mg\frac{l}{2}(1-cos\theta) + something$$

I don't know what this something would be. It may be the sum of the rotational energy and something. It can't just be rotational energy cause the rod is also sliding on the surface.

From this we would get angular velocity. Differentiating it would give us angular acceleration. Now we can apply the equation of moments. Let x be the distance by which the rod slides.

$$N\frac{l}{2}cos\theta = I\alpha$$

The above is about the centra of mass.

$$Mg\frac{l}2}sin\theta = I_{0}\beta$$

The above is about the point which touches the ground(ie the endpoint)

What we have to do is to find a relation between alpha and beta and the acc which we get from the law of conservation of energy.

Moreover these are just three equations while we have four variables ( alpha, beta, N and x) and we have no equation containing x.

 

[FedEx]

In the first and second equation the h and the x should not be there. I am trying to edit but vain

 

[Doc Al]

We can apply the laws of conservation of energy

$$mgh = mg\frac{l}{2}(1-cos\theta) + something$$

I don't know what this something would be. It may be the sum of the rotational energy and something. It can't just be rotational energy cause the rod is also sliding on the surface.

The total KE of the rod.

Hint: What path does the center of mass take?

 

[FedEx]

The total KE of the rod.

Hint: What path does the center of mass take?

Does total mean rotational + translational. If yes than what would be the relation between linear velocity and angular velocity. I don't think that it would be
$$v = \frac{l}{2}\omega$$

The path should be a parabola. But still that is not helpful. Let's suppose that initially the co ordinate of the endpoint of the rod is (0,0) then the position of the centre of the mass of the rod once it becomes horizontal would be less than l/2 cause the rod would have slided a certain distance in the -ve x direction.

though i can apply

$$\frac{l}{2} - x = vt$$

$$\frac{l}{2} = \frac{1}{2}gt^{2}$$

Hence

$$\sqrt{\frac{g}{l}}(\frac{l}{2} - x) = v_{x}$$

And what else can i apply?

Please help.

We have obtained the above equation but we have also added a variable (ie V_x)

We definitely assume V_x to be constant

Now at $$\theta$$

$$v_y = \sqrt{2gh}$$

$$h = \frac{l}{2} - xtan\theta - \frac{l}{2}cos\theta + x(cosec\theta)(cos\theta)$$

This is weird.

 

[Doc Al]

Does total mean rotational + translational.

Yes. Translational KE of the center of mass plus rotational KE about the center of mass.

If yes than what would be the relation between linear velocity and angular velocity. I don't think that it would be
$$v = \frac{l}{2}\omega$$

Use a bit of geometry. Find the position of the rod's center as a function of angle.

The path should be a parabola.

No, it's simpler than that. Hint: Consider the forces acting on the rod.

 

[FedEx]

Then there is just one way left. A straight line. And in that case then the center of mass finally would be at (0,0).

I can definitely apply geometry once i know the point about which i have to take the velocities. I have done various problems but not the ones in which the point about which the object is rotating is sliding.

Just elaborate a few important points eg how to determine x at a certain angle, the relation between $$\alpha & \beta$$

 

[Doc Al]

Then there is just one way left. A straight line. And in that case then the center of mass finally would be at (0,0).

Right. The center of mass falls straight down since there is no horizontal force on the rod.

I can definitely apply geometry once i know the point about which i have to take the velocities. I have done various problems but not the ones in which the point about which the object is rotating is sliding.

Just elaborate a few important points eg how to determine x at a certain angle, the relation between $$\alpha & \beta$$

What's the relationship between the position of the center of mass and the angle of the rod?

 

[FedEx]

$$y_{cm} = \frac{l}{2}cos\theta$$

Yup that's better. Now a final relation between alpha and beta is needed. But as you have said that the path is a straight line i feel that alpha and beta are the same.

I must say thing. The way you help people is ingenious. Thats because you help in such a way that one feels thayt he or she has done the problem on his own. Cause you do not give direct solutions but you give deep insights. Its like " Dont offer a fish to the hungry, but teach him how to fish"

 

[FedEx]

Hey Doc

You have really helped me in this problem. And i am about to finish it only if you could just comment on my last post.

 

[Doc Al]

$$y_{cm} = \frac{l}{2}cos\theta$$

Good.

Yup that's better. Now a final relation between alpha and beta is needed. But as you have said that the path is a straight line i feel that alpha and beta are the same.

What do you mean by "alpha" and "beta"?

I must say thing. The way you help people is ingenious. Thats because you help in such a way that one feels thayt he or she has done the problem on his own. Cause you do not give direct solutions but you give deep insights. Its like " Dont offer a fish to the hungry, but teach him how to fish"

Thank you. There is a method to my madness.

 

[Doc Al]

OK, I see how you use alpha and beta. They are the same.

$$N\frac{l}{2}cos\theta = I\alpha$$

The above is about the centra of mass.

This is OK.

$$Mg\frac{l}2}sin\theta = I_{0}\beta$$

The above is about the point which touches the ground(ie the endpoint)

This relationship is not necessarily true. The point of contact accelerates.

 

[FedEx]

The point of contact accelerates. Let's say instead of beta we use

$$Mg\frac{l}2}sin\theta = I_{0}\sqrt{{\beta}^2 + ({\alpha}\frac{l}{2})^2 - 2{\beta}{\alpha}{\frac{l}{2}}cos{\theta}}$$

 

[Doc Al]

$$N\frac{l}{2}cos\theta = I\alpha$$

The above is about the centra of mass.

That should be sinθ, not cosθ.

The point of contact accelerates. Let's say instead of beta we use

$$Mg\frac{l}2}sin\theta = I_{0}\sqrt{{\beta}^2 + ({\alpha}\frac{l}{2})^2 - 2{\beta}{\alpha}{\frac{l}{2}}cos{\theta}}$$

Why not just forget about that point? The torque about the center of mass is the one you need.

 

[FedEx]

Bingo.

Forget the end point. In the first post i have written 3 equations forget the last one.

Now instead of something we can use

$$\frac{1}{2}I(\omega)^2 + \frac{1}{2}m(2g)(cos\theta)\frac{l}{2}$$

And we will diff it so that we will get an equation in alpha.

Now we have got 3 variables and 3 equations. Sum solved.

Thank you.