A rod that falls while rotating from the end of a table

[Kim Gi Hyuk]

Homework Statement

A rod of mass m and length L is inclined by theta_o ( thera_o > 0 ) when the center of the rod is in contact with the end of the table, and the initial angular velocity is omega_o = 0. At this time, the rod was slightly pushed by an amount d in the longitudinal direction. The acceleration due to gravity is g, and the moment of inertia of the rod about the point of contact is I.

1. When the rod rotates without slipping at the point of contact and leaves the table, find the inclination theta_f and the angular velocity omega_f.

2. If the rod rotates slightly at the point of contact and then slips, find the angle theta_s at which the slip occurs.

3. When leaving situation 2, find the inclination theta_f and the angular velocity omega_f.

Relevant Equations

4. m dv/dt = mg + R + F
where R is normal force and F is frictional force.

5. I d/dt omega = d mg sin theta
where I is rotational inertia moment about contact point.

I tried to solve the problem like this.

4-1. m d/dt (v_1) = mg sin(theta) - F
4-2. m d/dt (v_2) = R - mg cos(theta)

If there is no slip, v_1 = v_2 = 0, so
F= mg sin(theta), R= mg cos(theta).

In the rotational motion equation, we intergrate

5-1. I omega^2 = 2dmg ( sin theta_f - theta_o )

But I don't know when it leaves the point of contact or what happens if there is slippage.

Ref. European Journal of Physics 16(4):172-176, July 1995.

 

[haruspex]

How are you defining $v_1, v_2$? Those look like constants, but you have written derivatives of them.
The problem statement implies there are no linear velocities to consider. You only have a linear displacement and, until contact is lost, an angular velocity about the contact point.

Not sure I understand part 2. No friction coefficient is given, so it seems like you are to take that as zero. Can that lead to a nonzero angle before slipping? Seems unlikely, but I haven’t checked yet.

I do not have access to the journal item you linked.

 

[Kim Gi Hyuk]

Thank for your comments.

v_1 is the velocity along the rod (-F), and v_2 is the velocity in the direction of normal force(R). As you said, both velocities are 0 when maintainig contact.

"You only have a linear displacement and, until contact is lost, an angular velocity about the contact point."

I mean 'linear displacement d' is the initial condition(and v_1 = v_2 =0), and then the rod starts rotating with torque.

For that reference, you can also search by this title.

Tumbling toast, Murphy's Law and the fundamental constants​

 

[Kim Gi Hyuk]

 

[haruspex]

Thank for your comments.

v_1 is the velocity along the rod (-F), and v_2 is the velocity in the direction of normal force(R).

Ok, but without friction (part 2) F=0, so eqn 4-1 tells you that as soon as $\theta>0$ it will start to slip. It seems as though the question should have specified a static friction coefficient.

As you said, both velocities are 0 when maintainig contact.

Not exactly. Remember, the point of contact is not at the mass centre, so the mass centre has acceleration normal to the rod.
Thinking more about part 1, that is not making sense to me either. If it cannot slip then I do not see how the normal force can become zero until the rod is vertical.
If the only forces are mg and some force along the rod then the mass centre has a tangential acceleration $g\cos(\theta)$, and hence an angular acceleration. But without a normal force there is no torque about the mass centre.

Edit:
In the real world, it would first start to slip. That increases the distance between the mass centre and the contact point. That makes it possible to lose contact completely without requiring any continuing angular acceleration.

Tumbling toast, Murphy's Law and the fundamental constants

I can find it, but I cannot access it. It is paywalled.

 

[Tom.G]

These may be of some help:
https://oikofuge.com/matthews-tumbling-toast/

https://dspace.mit.edu/bitstream/handle/1721.1/146968/5.0121905.pdf?sequence=1

Cheers,
Tom

 

[haruspex]

These may be of some help:
https://oikofuge.com/matthews-tumbling-toast/

The outline given there agrees with my "real world" edit to post #5:
"With any angle of tipping beyond zero (horizontal), gravity also produces a force that tries to slide the toast farther over the edge of the table. This is initially opposed by friction with the table edge, but eventually translates into a sliding motion. Gravity continues to accelerate the rate of rotation until the combination of sliding and rotation lifts the trailing part of the toast away from the table edge. "
I can't see that it helps to understand how we are supposed to handle the idealised version in post #1.

 

[Kim Gi Hyuk]

Ok, but without friction (part 2) F=0, so eqn 4-1 tells you that as soon as $\theta>0$ it will start to slip. It seems as though the question should have specified a static friction coefficient.

Not exactly. Remember, the point of contact is not at the mass centre, so the mass centre has acceleration normal to the rod.
Thinking more about part 1, that is not making sense to me either. If it cannot slip then I do not see how the normal force can become zero until the rod is vertical.
If the only forces are mg and some force along the rod then the mass centre has a tangential acceleration $g\cos(\theta)$, and hence an angular acceleration. But without a normal force there is no torque about the mass centre.

Edit:
In the real world, it would first start to slip. That increases the distance between the mass centre and the contact point. That makes it possible to lose contact completely without requiring any continuing angular acceleration.

I can find it, but I cannot access it. It is paywalled.

 

[Kim Gi Hyuk]

Not exactly. Remember, the point of contact is not at the mass centre, so the mass centre has acceleration normal to the rod.
Thinking more about part 1, that is not making sense to me either. If it cannot slip then I do not see how the normal force can become zero until the rod is vertical.

Ok,
(1) I confused 'contact point' and 'center of mass'.
For the center of mass, I think, is as follows in the rotaional coordinate system,
4-1-1 md omega^2 = mg sin theta - F
4-2-1 md d/dt(omega) = R - mg cos theta

(2) If no slip, normal force can become zero only when the rod is vertical. However, I do not know when the normal force becomes 0 when slippage occurs.

 

[haruspex]

The article linked in post #8 is the same as in post #6. I do not see there any discussion of question 1 in post #1.
The article does address the Question 2 case, but takes the angle at which slip occurs, $\phi$, to be determined by experiment:
"where the value of the critical overhang parameter η₀ and slip angle φ at which detachment takes place may be determined experimentally".
If you take the coefficient of friction to be known you should be able to determine these.
But this is incorrect:

If there is no slip, v_1 = v_2 = 0, so
F= mg sin(theta)

You are overlooking centripetal acceleration.

 

[Kim Gi Hyuk]

The outline given there agrees with my "real world" edit to post #5:
"With any angle of tipping beyond zero (horizontal), gravity also produces a force that tries to slide the toast farther over the edge of the table. This is initially opposed by friction with the table edge, but eventually translates into a sliding motion. Gravity continues to accelerate the rate of rotation until the combination of sliding and rotation lifts the trailing part of the toast away from the table edge. "
I can't see that it helps to understand how we are supposed to handle the idealised version in post #1.

Thanks for your concise and meaningful expanation!

 

[Kim Gi Hyuk]

The article linked in post #8 is the same as in post #6. I do not see there any discussion of question 1 in post #1.
The article does address the Question 2 case, but takes the angle at which slip occurs, $\phi$, to be determined by experiment:
"where the value of the critical overhang parameter η₀ and slip angle φ at which detachment takes place may be determined experimentally".
If you take the coefficient of friction to be known you should be able to determine these.
But this is incorrect:

You are overlooking centripetal acceleration.

Thanks for your comments! I will try to apply the equation for center of mass motion in a rotating coordinate system.