Normal force in a curvilnear motion

[EastWindBreaks]

Homework Statement

The system shown is initially at rest when the bent bar starts to rotate about the vertical axis AB with constant angular acceleration a 0 = 3 rad/ s2 . The coefficient of static friction between the collar of mass m = 2 kg and the bent bar is f.Ls = 0.35, and the collar is initially d = 70 em from the spin axis AB.

1)Assuming the motion starts at t= 0, determine the time at which the collar starts to slip relative to the bent bar.

Homework Equations

3. The Attempt at a Solution
Hello everyone! I am studying for my final, and this problem is from the textbook.
here is my thought process when I tried to solve this problem:

rotating frame is an non-inertial reference frame, so i have to use centrifugal force instead of centripetal force. The only force that is going to make the mass to slip is the centrifugal force pointing in the direction of r, since the mass is a collar which it can only slip in the r direction.

Below is my attempt on this problem, its solved but i still have some questions. Initially, I thought the normal force in this case is just mg, i fixed it after checking with the solution, however, i still don't understand why force in theta direction would affect the normal force, to my knowledge, normal force is the sum of the forces in the vertical direction, but the force in the theta direction is perpendicular to the z axis, i am so confused.

another version of solution I found online :

[let angular velocity at any time be w
so w = 3 * t rad / sec
centrifugal force = m* w^2 * r = 2 * 9*t^2 * 0.7 = 12.6 t^2

acceleration of mass = 0.7 * 3 = 2.1 m/sec^2

so pseudo force of block due to acceleration = 2.1 * 2 = 4.2 N

weight of block = m*g = 19.6 N

the pseudo force and the weight are perpendicular to each other ..

so the net normal reaction will ne sqrt ( 4.2^2 + 19.6^2 ) = 20.04495 N

maximum friction force = 0.35 * 20.04495 = 7.015732 N

12.6 t^2 = 7.015732

so time t = 0.746193 secs]

so, pseudo force is the same force in the theta direction? I thought pseudo force is same as centrifugal force? or is he using the wrong term here? Please help~Heaven Bless You!

 

[haruspex]

normal force is the sum of the forces in the vertical direction,

No. The normal force between two surfaces is the minimum magnitude action and reaction necssary to prevent interpenetration. One can deduce that it acts normally to the plane of contact, hence the term "normal".

thought pseudo force is same as centrifugal force

It depends on the reference frame. If you take your frame to be rotating around the axis at a constant speed, which happens at the instant under consideration to match the speed of the rod's rotation, and centred at the mass, then the only pseudo force is the centrifugal one. But typically one would take a non-inertial frame moving with the mass. In that frame, it is not accelerating at all, so the pseudo force must represent all of the acceleration of the mass.

 

[zwierz]

Here $\boldsymbol T$ is a reaction force. $m\boldsymbol a=\boldsymbol T+m\boldsymbol g;\quad \boldsymbol a=-d\dot\theta^2\boldsymbol e_r+d\ddot\theta \boldsymbol e_\theta;\quad \ddot\theta=\alpha_0.$
Further $\boldsymbol T=\boldsymbol T_n+T_{friction}\boldsymbol e_r,\quad \boldsymbol T_n=T_\theta\boldsymbol e_\theta+T_z\boldsymbol e_z;\quad |T_{friction}|=\mu_S|\boldsymbol T_n|$.
So why are pseudo forces needed?

 

[EastWindBreaks]

No. The normal force between two surfaces is the minimum magnitude action and reaction necssary to prevent interpenetration. One can deduce that it acts normally to the plane of contact, hence the term "normal".

Thank you for helping me again! I see, so because its a collar on a bar, its different from the typical "a box on a ground" scenario, in such scenario, the normal force is just the sum of the forces in the vertical direction, but we have a collar here, so there are more than 2 directions where the forces can act, because the contacting surface is a cylinder.

It depends on the reference frame. If you take your frame to be rotating around the axis at a constant speed, which happens at the instant under consideration to match the speed of the rod's rotation, and centred at the mass, then the only pseudo force is the centrifugal one. But typically one would take a non-inertial frame moving with the mass. In that frame, it is not accelerating at all, so the pseudo force must represent all of the acceleration of the mass.

" If you take your frame to be rotating around the axis at a constant speed", is it a inertial frame?
just to confirm, there are only two types of frame of reference inertial and non-inertial, correct?
So, in non-inertial frame, the object appears to be stationary, so pseudo force= the sum of all other forces= the force in theta direction+ the force in the r direction?

from the second version solution, he used angular acceleration* distance from the center=acceleration of the mass, then he got pseudo force from this acceleration, only acceleration in the theta( tangential) direction can be calculated using angular acceleration * distance from the center, under this logic,pseudo force in this case is just the force in the theta (tangential) direction? or maybe he should have used the term " force in the theta direction" instead of " pseudo force"? because you said that pseudo force represent all of the other forces in a non-inertial frame, which the second version of solution is, correct?

 

[EastWindBreaks]

View attachment 199656
Here $\boldsymbol T$ is a reaction force. $m\boldsymbol a=\boldsymbol T+m\boldsymbol g;\quad \boldsymbol a=-d\dot\theta^2\boldsymbol e_r+d\ddot\theta \boldsymbol e_\theta;\quad \ddot\theta=\alpha_0.$
Further $\boldsymbol T=\boldsymbol T_n+T_{friction}\boldsymbol e_r,\quad \boldsymbol T_n=T_\theta\boldsymbol e_\theta+T_z\boldsymbol e_z;\quad |T_{friction}|=\mu_S|\boldsymbol T_n|$.
So why are pseudo forces needed?

you had your total normal force = the normal force in the z-direction + the normal force in the r direction, i thought it should be z direction + theta direction?

 

[zwierz]

you had your total normal force to be the normal force in the z-direction+ the normal force in the r direction,

oh no! please see my formulas carefully

 

[EastWindBreaks]

oh no! please see my formulas carefully

oh my bad, I wasn't being careful, which frame of reference are you using? your equation ma= T+ mg, its a condensed form of summing the forces in 3 directions, correct? and Thank you for helping.

 

[zwierz]

The equation ma=T+mg is written relative to an inertial frame. The collar experiences only two forces: the force T from the rod and the force mg from the Earth.
It is a one of the core concepts of Newton mechanics: forces show up from interactions between bodies. No bodies -- no forces.
Further we employ additional hypotheses about the force T. For example we break it into normal and tangent components and impose a relation between these two components. Another additional hypotheses are also possible but the equation ma=T+mg holds in any case.

 

[EastWindBreaks]

The equation ma=T+mg is written relative to an inertial frame. The collar experiences only two forces: the force T from the rod and the force mg from the Earth.
It is a one of the core concepts of Newton mechanics: forces show up from interactions between bodies. No bodies -- no forces.
Further we employ additional hypotheses about the force T. For example we break it into normal and tangent components and impose a relation between these two components. Another additional hypotheses are also possible but the equation ma=T+mg holds in any case.

Awesome, your approach is very straightforward, i just learned about non-inertial frame and pseudo forces, they seem more confusing to me.

 

[haruspex]

So, in non-inertial frame, the object appears to be stationary,

An inertial frame is one that is neither accelerating nor rotating. Any other frame is a non-inertial frame.
Personally I prefer to use inertial frames, though I have seen cases where using a non-inertial frame is simpler.
Typically, if you are going to use a non-inertial frame it will be the reference frame of the object, so any acceleration which the object is undergoing has to be represented by a pseudo force.
But there is nothing to stop you from using a non-inertial frame in which the object still has some acceleration - other than the confusion which may arise. In this case, the pseudo force represents the acceleration of the frame only.

 

[EastWindBreaks]

a non-inertial frame in which the object still has some acceleration - other than the confusion which may arise. In this case, the pseudo force represents the acceleration of the frame only.

just to be sure, which of my listed approaches are you referring to? the second one? correct? since it involves pseudo forces and other forces.

 

[haruspex]

just to be sure, which of my listed approaches are you referring to? the second one? correct? since it involves pseudo forces and other forces.

It was a general statement, but it does apply to your approach where you considered the rotation (at current rate) as centrifugal but did not account for the acceleration in that rate. You could either add another pseudo force to account for it (thereby keeping the force sum as zero) or allow that the force sum is not zero and leads to the tangential acceleration.

See ifthis makes it clear
Inertial Frame approach:

Sum of real forces = m(centripetal acceleration + tangential acceleration)​

Frame of mass:

sum of forces = real forces + centrifugal pseudo force + tangential pseudo force = 0​

(Note that a pseudo force is opposite in sign to the acceleration it represents.)
Hybrid frame, rotating steadily at current speed of bar:

sum of forces = real forces + centrifugal pseudo force = m(tangential acceleration)​

You can see that the three equations are completely equivalent.

 

[zwierz]

The concept of non inertial frames and pseudo forces is as mathematically trivial as physically confusing. Let $Oxyz$ be an inertial frame and a non inertial frame $A\xi\eta\zeta$ somehow moves relative $Oxyz$ . Angular velocity and angular acceleration of the frame $A\xi\eta\zeta$ denote by $\boldsymbol\omega,\boldsymbol\epsilon$ respectively.

Consider a mass point $M$ of mass $m$ that experiences a true force $\boldsymbol F$. Then relative $Oxyz$ we have $m\boldsymbol a=\boldsymbol F$.
On the other hand we know that $\boldsymbol a=\boldsymbol a_r+\boldsymbol a_t+\boldsymbol a_c$, where
$\boldsymbol a_r$ is acceleration of the point $M$ relative to $A\xi\eta\zeta$;
$\boldsymbol a_t=\boldsymbol a_A+\boldsymbol\omega\times(\boldsymbol\omega\times\boldsymbol{AM})+\boldsymbol\epsilon\times\boldsymbol{AM}$ is the acceleration of transport;
$\boldsymbol a_c=2\boldsymbol\omega\times \boldsymbol v_r$ is the Coriolis acceleration.

Now rewrite the Second Newton law as follows
$$m\boldsymbol a_r=\boldsymbol F+\boldsymbol F_c+\boldsymbol F_t,\quad \boldsymbol F_c=-m\boldsymbol a_c,\quad \boldsymbol F_t=-m\boldsymbol a_t.\qquad(*)$$
The terms $\boldsymbol F_c,\boldsymbol F_t$ are so called fictitious forces. This trivial trick is justified by the fact that the "forces" $\boldsymbol F_c,\boldsymbol F_t$ do not depend on relative acceleration and equation (*) can be treated as the Newton second law relative the frame $A\xi\eta\zeta$ .