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

• Nexus99
In summary, the conversation discusses a problem a student is having with a math equation. The student provides a summary of the content, explains the mistake they made, and offers a solution.
Nexus99
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

Last edited:
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?

berkeman and Nexus99
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

Nexus99 said:
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?

haruspex said:
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?

## | \vec{R} | = 3 M \omega ^2 \frac{8}{9} L = \frac{32}{15} M \frac{v_0^2}{L} ##

Nexus99 said:
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##.

Last edited:
haruspex said:
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} ##

Nexus99 said:
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.)

haruspex said:
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?

Nexus99 said:
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.

haruspex said:
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?

Nexus99 said:
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.

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

the diagram of the forces drawn by my professor. In this case web have a rod that hits a mass in a vertical plane

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.

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

Nexus99 said:
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.

haruspex said:
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!

## 1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is a physical property that depends on the object's mass and distribution of mass around its axis of rotation.

## 2. How is moment of inertia calculated?

Moment of inertia is calculated by multiplying the mass of an object by the square of its distance from the axis of rotation. This distance is also known as the object's radius of gyration.

## 3. What is center of mass?

The center of mass is the point at which an object's mass is evenly distributed in all directions. It is the balance point of an object and is often located at or near its geometric center.

## 4. How is center of mass determined?

The center of mass can be determined by finding the weighted average of an object's mass distribution. This can be done by dividing the total mass by the sum of the individual masses multiplied by their respective distances from a reference point.

## 5. What is vincular reaction of a rod?

Vincular reaction, also known as constraint force, is the force exerted by a fixed support on a rod or beam. It is perpendicular to the surface of the support and prevents the rod from moving or rotating in certain directions.

• Introductory Physics Homework Help
Replies
335
Views
9K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
745
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
469
• Introductory Physics Homework Help
Replies
7
Views
828