Questioning My Understanding: Is My Answer Wrong?

[kuruman]

My understanding is that: In an inertial frame, centripetal force is the cause of rotation, and when viewed from a non-inertial frame of reference, centrifugal force is the cause of rotation.

I agree with the first half of this statement but not the second. Say I am in an inertial frame looking at you standing on a rotating platform at some distance from the rotation axis. I see you rotate and conclude that the force of static friction provides the centripetal acceleration needed to keep you going in a circle. As far as you are concerned, you are not rotating but you experience a centrifugal force which is canceled by the force of friction at your feet, so that you remain at rest in the non-inertial frame. As far as you are concerned, you are not accelerating and the centrifugal force you experience is not the cause of any rotation in your frame.

 

[haruspex]

In an inertial frame, centripetal force is the cause of rotation,

Yes.

when viewed from a non-inertial frame of reference, centrifugal force is the cause of rotation.

No, mostly, but see the example at the end below.

Presumably you mean specifically the frame of reference of the rotating object. In that frame there is no rotation, or motion of any sort, by definition, but there may be unbalanced forces which ought to be causing an acceleration according to Newton's laws. To fix this up, we invent centrifugal force, etc., to represent the rotation of the frame relative to an inertial frame. These fictitious forces are added to the real applied forces.
True, you could take any non-inertial frame, independently of the object, but then things get more complicated. We may observe that the object does rotate about some axis in that frame, but not in a way that seems to match the applied forces.
In this case, we have both centrifugal and centripetal forces:
- centrifugal force (and Coriolis, and Euler, as appropriate), computed from the acceleration of our chosen non-inertial frame, is added to the real applied forces, and,
- of the resultant, the component normal to the observed velocity produces the centripetal force, resulting in the observed centripetal acceleration.

In the case where the observer is being whirled around an axis on the end of a cable, the centrifugal force in the observer's frame points away from the axis and balances the tension in the cable. This results in the zero acceleration the observer records.
In an inertial frame, the tension provides the centripetal force, which points towards the axis.
But now consider Bob, rotating on the spot, observing Alice, standing still nearby (on a frictionless surface, say). To Bob, Alice appears to be rotating around him, so accelerating towards him, yet not subject to any horizontal forces. In this case, the "centrifugal" force on Alice in Bob's reference frame, when we do the cross product calculation, turns out be pointing towards Bob, and is providing the centripetal acceleration Bob perceives.

 

[Richard R Richard]

Let's do some formulas
The problem says that a small imbalance is used to cause the rod to overturn, we can assume that the initial angular and linear momentum wasn't significantly altered.
As already said, if there is no friction, the ground cannot push the rod by horizontal reaction force, because there is no horizontal component of weigth, then there is no change in the linear moment of the center of mass, therefore its horizontal position remains constant, response b. End of problem, but we can calculate 3 results from different situations .
An energy analysis shows
$$ \dfrac {L} {2} mg = \dfrac12mv_{CM} ^ 2 + \dfrac12I_{CM} \omega_{CM} ^ 2 $$
An instant before hitting the floor it is verified that if N> 0 during the entire fall then
$$ \omega_{CM} = \dfrac {v_{CM}} {\frac L2} $$
Since $I_{CM} = \dfrac {1} {12} mL ^ 2$
Resulting that $$ v_{CMx} = 0 $$ and that $$ v_{CMy} = \sqrt {\dfrac {3Lg} {4}} $$
If $\mu_s$ is large enough that the fulcrum does not move then the energy balance system becomes
$$ \dfrac {L} {2} mg = \dfrac12mv_{CM} ^ 2 + \dfrac12I_{ext} \omega_{CM} ^ 2 $$
As now the instantaneous center of rotation changes to the lower end of the rod
Since $I_{ext} = \dfrac {1} {3} mL ^ 2$
Then $$ v_{CMx} = 0 $$ and that $$ v_{CMy} = \sqrt {\dfrac {3Lg} {7}} $$ the position of the CM on impact is $L /2$ away from position $x_o$
But always at some point, , if the rod is free for move
$$ mg \sin \theta> \mu mg \cos \theta \quad \to \quad \mu <\tan \theta $$
and at that moment the support point slides losing kinetic energy, the calculation of the final position of the CM becomes complex depending on the value of the work of the friction force which is a variable force in time and the length it slides as well.

 

[haruspex]

if N> 0 during the entire fall

Is there a trivial way to prove that?
It always bothers me with questions about falling and rotating rods that the question setter seems to overlook the possibility that the rod becomes airborne. I believe there is no such in the present case, but it can occur in other scenarios, such as a rod that had been standing against a wall.

 

[Richard R Richard]

Is there a trivial way to prove that?

Is there a time during fall when the modulus of the angular velocity of rotation multiplied by the radius is greater than the rate of fall of the center of mass? In this case no, in fact the angular acceleration is the result of a bond between the rod and the ground.

 

[haruspex]

Is there a time during fall when the modulus of the angular velocity of rotation multiplied by the radius is greater than the rate of fall of the center of mass?

When the rod length 2r makes angle theta to the vertical, if the base is remaining in contact with the ground then the vertical velocity of the mass centre is $v=\omega r\sin(\theta)$, so for θ <π/2 yes, $r\omega>v$.
At first, most of the KE will be in the rotation. The concern then is whether the angular momentum may become so great that the downward velocity can't catch up.