Moment of inertia, center of mass and vincular reaction of a rod

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  • #1
Nexus99
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Homework Statement:
A rigid rod OA of negligible thickness, mass M and length L, with linear density ## \lambda (r) = kr ## (where r is the distance from the extreme O), lies in a smooth plane and is initially stationary, hinged in its extreme O. At the time ##t = t_0## a material point of mass m = 2M, which moves on the plane with speed ## v_0 ## perpendicularly directed to the rod, hits it in the extreme A in a completely anaelastic way. Calculate: 1) the moment of inertia of the rod as a function of M and L, evaluated with respect to an axis perpendicular to the plane and passing through O. 2) The angular velocity of the system after the collision 3) The moment of inertia of the system after the impact with respect to the previous axis 4) The distance of the new center of mass after the impact from the point O 5) The magnitude of the vincular reaction which acts after the collision
Relevant Equations:
Center of mass, moment of inertia, kinetic energy
I'm struggling doing point 5, i have no idea how to solve that question. In point 1 i obtained the following result:
## I=\frac{ML^2}{2}## calculating the integral of dI, the infinitesimal moment of inertia of a small section of the rod of length dr.
2) Through the conservation of angular momentum (calculating angular momentum in point O) i obtained
## ω= \frac{4v0}{5L} ##
3) i obtained through huygens steiner:
## I=\frac{5ML^2}{2} ##
4) The center of mass before the collision is: ## x_{CM}=\frac{2L}{3} ## . After the collision:
## x_{CM}=\frac{8L}{9} ##
5) i don't know
 
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  • #2
haruspex
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Is this a translation? I've never heard of a vincular reaction. My dictionary says it pertains to a bond or ligament. The only maths connection I can find is for a vincula, a lne drawn above terms to bracket them together.
Maybe it means the force needed to keep the rod and mass together?
 
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  • #3
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Yes it is a translation i mean like constraint reaction force. It's the force needed to keep the point O stationary and keep the rod in the plane
 
  • #4
haruspex
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Yes it is a translation i mean like constraint reaction force. It's the force needed to keep the point O stationary and keep the rod in the plane
It's just the centripetal force. Since you got everything else right, I would guess you know how to find that, right?
 
  • #5
Nexus99
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It's just the centripetal force. Since you got everything else right, I would guess you know how to find that, right?
Yes, i know how to calculate this force, but i don't understant why is equal to the centripetal force. Isn't, after the collision, the constraint reaction force directed in the same direction of the tangential acceleration?

Anyway, following your suggestion:
## | \vec{R} | = 3 M \omega ^2 \frac{8}{9} L = \frac{32}{15} M \frac{v_0^2}{L} ##
 
  • #6
haruspex
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after the collision, the constraint reaction force directed in the same direction of the tangential acceleration?
During the collision there is an opposing impulse at the hinge, but once the collision is complete we just have a steady rotation around the hinge.

I don't agree with your ##\frac{32}{15}##. Check how you squared ##\omega##.
 
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  • #7
Nexus99
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During the collision there is an opposing impulse at the hinge, but once the collision is complete we just having a steady rotation around the hinge.

I don't agree with your ##\frac{32}{15}##. Check how you squared ##\omega##.
Ok, i think i got what you said.
Yes, i squared ## \omega ## wrong
real result is: ## \frac{128}{75} ##
 
  • #8
haruspex
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Ok, i think i got what you said.
Yes, i squared ## \omega ## wrong
real result is: ## \frac{128}{75} ##
Looks good. (Did I really write "we are just having"? Ouch.)
 
  • #9
Nexus99
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Looks good. (Did I really write "we are just having"? Ouch.)
Thanks for help, i hadn't even noticed the mistake 😅
Anyway, based on what you wrote in the post 6, can be asserted that the motion of the center of mass is uniform? After the collision isn't there also a tangent component of the constraint reaction that changes the tangential speed?
 
  • #10
haruspex
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After the collision isn't there also a tangent component of the constraint reaction
Only if there is friction or other stiffness in the hinge. Once the circular motion is underway, the rod could be just a string.
 
  • #11
Nexus99
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Only if there is friction or other stiffness in the hinge. Once the circular motion is underway, the rod could be just a string.
Instead if the system had been on a vertical plane we would have a tangential and a radial component of the constraint reaction to compensate for the weight force, right?
 
  • #12
haruspex
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Instead if the system had been on a vertical plane we would have a tangential and a radial component of the constraint reaction to compensate for the weight force, right?
No. If we resolve the forces from the hinge into a radial component and a component normal to that we can see that the normal component must be zero. If not, it has a moment about the mass, so must cause the rod to rotate away from the hinge!
Consider that as long as the mass is rotating fast enough about the hinge we could replace the rod with a piece of string, and there could not be forces normal to the string.
 
  • #13
Nexus99
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But if the system was on a vertical plane, wouldn't it represent a physical pendulum? And i know that, for physical pendulum, the force have a radial and normal component
Screenshot_20200728-235924_WhatsApp.jpg

the diagram of the forces drawn by my professor. In this case web have a rod that hits a mass in a vertical plane
 
  • #14
haruspex
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Whoops! You are right! I was forgetting the rod has mass, so moment of inertia. This invalidates the string analogy. The hinge reaction normal to the rod provides the torque for angular acceleration of the rod.

Don't know if there is an easier way, but I got an answer by considering the angular acceleration of the rod and the tangential accelerations of rod and mass. I introduced unknowns for the normal forces each end of the rod and the angular acceleration of the rod.
 
  • #15
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Can you post what you got?
Anyway the result i wrote in post 7, according to the book, was correct. Probably because the problem wanted to know the magnitude of the force immediately after the collision, where we a have a pure normal force
 
  • #16
haruspex
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Can you post what you got?
Anyway the result i wrote in post 7, according to the book, was correct. Probably because the problem wanted to know the magnitude of the force immediately after the collision, where we a have a pure normal force
No, it is not to do with its being right after the collision.
I called the normal forces at the ends of the rod N (hinge end) and N', both clockwise about the rod's centre. I is the MoI of the rod about its centre, ##=\frac 1{12}ML^2##.
Rotations positive clockwise, angled measured from upward vertical.
##I\alpha=(N+N')\frac L2##
##M\frac L2\alpha=Mg\sin(\theta)-N+N'##
##mL\alpha=mg\sin(\theta)-N'##
Whence
##(\frac 12M+m)L\alpha=(M+m)g\sin(\theta)-N##
##2I\alpha=NL-mL^2\alpha+mgL\sin(\theta)##
##(2I+mL^2)\alpha=NL+mgL\sin(\theta)##
##(\frac 12M+m)L(NL+mgL\sin(\theta))=((M+m)g\sin(\theta)-N)(2I+mL^2)##
##N((\frac 12M+2m)L^2+2I)=g\sin(\theta)(\frac 12MmL^2+(M+m)2I)##
Substituting for I:
##N((\frac 12M+2m)L^2+\frac 16ML^2)=g\sin(\theta)(\frac 12MmL^2+(M+m)\frac 16ML^2)##
##N((3M+12m)+M)=g\sin(\theta)(3Mm+(M+m)M)##
##N(4M+12m)=g\sin(\theta)(4Mm+M^2)##
##N=\frac{M(M+4m)}{4(M+3m)}g\sin(\theta)##.
Note that this is zero if g is zero (horizontal plane, as in the question) or M is zero (light string) or the rod is vertical.
 
  • #17
Nexus99
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No, it is not to do with its being right after the collision.
I called the normal forces at the ends of the rod N (hinge end) and N', both clockwise about the rod's centre. I is the MoI of the rod about its centre, ##=\frac 1{12}ML^2##.
Rotations positive clockwise, angled measured from upward vertical.
##I\alpha=(N+N')\frac L2##
##M\frac L2\alpha=Mg\sin(\theta)-N+N'##
##mL\alpha=mg\sin(\theta)-N'##
Whence
##(\frac 12M+m)L\alpha=(M+m)g\sin(\theta)-N##
##2I\alpha=NL-mL^2\alpha+mgL\sin(\theta)##
##(2I+mL^2)\alpha=NL+mgL\sin(\theta)##
##(\frac 12M+m)L(NL+mgL\sin(\theta))=((M+m)g\sin(\theta)-N)(2I+mL^2)##
##N((\frac 12M+2m)L^2+2I)=g\sin(\theta)(\frac 12MmL^2+(M+m)2I)##
Substituting for I:
##N((\frac 12M+2m)L^2+\frac 16ML^2)=g\sin(\theta)(\frac 12MmL^2+(M+m)\frac 16ML^2)##
##N((3M+12m)+M)=g\sin(\theta)(3Mm+(M+m)M)##
##N(4M+12m)=g\sin(\theta)(4Mm+M^2)##
##N=\frac{M(M+4m)}{4(M+3m)}g\sin(\theta)##.
Note that this is zero if g is zero (horizontal plane, as in the question) or M is zero (light string) or the rod is vertical.
Ok thanks for posting it!
 

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