Normal force in a curvilnear motion

In summary, the problem involves a rotating system with a bent bar and a collar. The collar starts to rotate around the vertical axis with a constant angular acceleration. The coefficient of static friction between the collar and the bent bar is given, as well as the initial distance between the collar and the spin axis. To solve the problem, the concept of pseudo force is used to take into account the non-inertial reference frame. The normal force is not just the sum of forces in the vertical direction, but also includes forces in the r and theta directions, as the collar is cylindrical in shape. The pseudo force in this case represents the sum of all other forces in the non-inertial frame, and is calculated using the acceleration of the
  • #1
EastWindBreaks
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3

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.
QQ截图20170506220030.png

Homework Equations


Scan0002.jpg


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.



Scan0001.jpg

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!
 
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  • #2
EastWindBreaks said:
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".
EastWindBreaks said:
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.
 
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  • #3
2b5b33055c23.png

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?
 
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  • #4
haruspex said:
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.

haruspex said:
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?
 
  • #5
zwierz said:
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?
 
  • #6
EastWindBreaks said:
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
 
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  • #7
zwierz said:
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.
 
  • #8
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.
 
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  • #9
zwierz said:
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.
 
  • #10
EastWindBreaks said:
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.
 
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  • #11
haruspex said:
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.
 
  • #12
EastWindBreaks said:
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.
 
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  • #13
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## .
 
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Related to Normal force in a curvilnear motion

What is normal force in a curvilinear motion?

Normal force in a curvilinear motion is the force that acts perpendicularly to a surface at the point of contact. This force is exerted by a surface on an object in order to prevent it from sinking into the surface.

What is the direction of normal force in a curvilinear motion?

The direction of normal force in a curvilinear motion is always perpendicular to the surface at the point of contact. This means that the force is at a 90 degree angle to the surface and is pointing away from the surface.

How is normal force related to centripetal force in a curvilinear motion?

In a curvilinear motion, normal force and centripetal force are equal in magnitude and opposite in direction. This means that the normal force provides the centripetal force needed to keep an object moving in a curved path.

What factors affect the normal force in a curvilinear motion?

The factors that affect the normal force in a curvilinear motion include the mass of the object, the velocity of the object, and the radius of the curve. The greater the mass or velocity of the object, the greater the normal force needed to maintain the curved path. A smaller radius also requires a greater normal force.

How does the normal force change in a banked curve?

In a banked curve, the normal force is not always perpendicular to the surface. Instead, it is tilted at an angle, providing both normal and centripetal forces. The amount of tilt depends on the speed of the object and the angle of the bank. As the speed increases, the angle of the bank must also increase in order to maintain a constant normal force and prevent the object from sliding off the surface.

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