Does Hitting an Object at the Corner Affect Its Speed Compared to the Center?

In summary: Yes, the kinetic energy is equal. The difference in the amount of energy used to produce the rotation is due to the fact that the object has to apply the force through a greater distance to get the same rotation.
  • #1
deluksic
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I've just watched this vid about rotations and torque:

http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-21/

and they say that if u hit or pull something for a corner for example.. it will have the same speed like if u hit it in center of mass... BUT , how can it be possible if u have some impulse acted in a center of mass and the body gets some velocity but no rotation. Now if u do the same but in a corner or some edge, it will gain the SAME velocity AND ROTATION... that means that the body is getting more energy if u hit it at corner than if u hit it at center of mass...

can u tell me what is wrong with my thinking...
 
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  • #2
There's nothing wrong with your thinking. The same force produces the same acceleration of the center of mass regardless of where on the body it is applied. However, if you apply the force away from the center, that same force will also produce a rotation about the center of mass. This does require more energy. That same force must act through a greater distance--thus do more work--to produce the rotation. Just as you suspect, it requires more energy to both rotate and translate the object.
 
  • #3
In fact, Rotational energy is derived from Linear translation energy. Think of adding all 1/2mv^2 in the rigid body (those closest to CM have 0J, furthest particles carry max. KE) and you could get the formula 1/2 I w^2. So he's talking about that, at a single position, the particle on the object gains SAME velocity as you hit the center. But it is only instantaneous, what you really observe is rotation only.

BTW, if you do hit the object at its corner, it won't have translational motion and rotation. Try that with a pen, hit its corner and it's going to rotate about its CM
 
  • #4
lswtech said:
In fact, Rotational energy is derived from Linear translation energy. Think of adding all 1/2mv^2 in the rigid body (those closest to CM have 0J, furthest particles carry max. KE) and you could get the formula 1/2 I w^2. So he's talking about that, at a single position, the particle on the object gains SAME velocity as you hit the center. But it is only instantaneous, what you really observe is rotation only.

BTW, if you do hit the object at its corner, it won't have translational motion and rotation. Try that with a pen, hit its corner and it's going to rotate about its CM
If a single force applied to an object could produce only rotation about its center of mass but no translation, that would violate Newton's 2nd law. (Try hitting the pen on a frictionless surface.)
 
  • #5
I'd reply to your question a bit differently...

"BUT , how can it be possible if u have some impulse acted in a center of mass and the body gets some velocity but no rotation. Now if u do the same but in a corner or some edge, it will gain the SAME velocity AND ROTATION... that means that the body is getting more energy if u hit it at corner than if u hit it at center of mass..."

Your thinking is fine as Doc Al noted...and I agree with his answer.

But I took your question to mean that if a fixed impulse is applied, some force over some period of time, once at the center of mass and a second time an equal impulse away from the center of mass.

In the first all the acceleration is linear...the body accelerates and moves in a straight line without rotation. In the second case, as you imply, there is both linear and rotational momentum...two forms of energy is consumed...so it is logical for the linear component to be a bit less with the difference in energy used for rotational movement...when the impulse is fixed...
 
  • #6
but if impulse (F*t) give u some energy (does it depend on mass?), it has to be converted into CM velocity and rotational velocity, so velocity of CM does depend on where u apply that impulse.

**didnt saw the last reply... so it is true that it does depend on where is force being applyed? CM is actually getting less velocity in second case than in first?
 
  • #7
Impulse is just FΔt, so the same impulse produces the same change in velocity of the center of mass regardless of where it is applied on the body. Of course, applying that same impulse off-center requires greater energy.
 
  • #8
Yes you are correct. It should have translation also. I was thinking about fixed axis motions only. It should be restated that angular momentum about CM as well as linear momentum are both conserved
 
  • #9
guys u haven't answered my q...

if force is applying in center of mass then kinetic energy is equal to (mv2)/2, and if it is away from center of mass then kinetic energy is equal to (Iw2)/2 + (mv2)/2 and if in both cases linear velocity is the same then u get extra energy...
 
  • #10
deluksic said:
guys u haven't answered my q...

if force is applying in center of mass then kinetic energy is equal to (mv2)/2, and if it is away from center of mass then kinetic energy is equal to (Iw2)/2 + (mv2)/2 and if in both cases linear velocity is the same then u get extra energy...
I answered that already. When applying the force off center do you end up with greater total KE? Yes. But that just means that you had to do more work when applying that force. The greater energy does not come for free.
 
  • #11
hmm i have mistake in my expression instead of "force" i meant impulse and impulse in both cases is the same... but anyways u do get extra KE right? if that's right then it actually does matter where do you apply the impulse, and linear velocity does change if u change the position of impulse...

i need all this to make simulation of balls colliding with walls (linear, and curved) and other balls... u can download it here:

https://sites.google.com/site/deluksicgames/contact-us -third download
 
  • #12
deluksic said:
hmm i have mistake in my expression instead of "force" i meant impulse and impulse in both cases is the same... but anyways u do get extra KE right?
Yes, you'll end up with 'extra' rotational KE because to exert that same impulse off-center requires more work.
if that's right then it actually does matter where do you apply the impulse, and linear velocity does change...
No. The same impulse gives the same change in linear velocity.

Don't confuse the impulse delivered by a force, FΔt = Δmv, with the mechanical energy delivered by a force, FΔx = ΔKE.
 
  • #13
Good simulations.

I think I got your point now. If you have a small sphere, with velocity v, hitting a rod with the same mass at one end elastically, the result is not the intuitive case-- "the sphere stops and the rod moves forward with velocity v" but something more complicated. You need to consider the velocity of a particle at one end and do Coefficient of Restitution formula for that particle, and set up impulse(momentum equations) for both Linear and Angular about O.
 
  • #14
thanks!

well not only for that but also for friction and other parts of simulation...

actually i have a script that adds a force on a body which is determined by x,y(position where the force is applyed) direction and magitude... after one step of simulation the computer calculate new speed(rotational and linear) for that body...

well it does work but something isn't right with coservation of energy...

can u just tell me: how do i calculate the reaction force if ball hits a wall with some speed (lets say on 90°, totally elastic)?
i tried that by using impulse: F = -(m*v) (my equation doesn't have time cause it is one step of simulation)
but this just stopped the ball.. i need elastic collision (if u noticed there is change elasticity and if u set it to 1 then it does bounce but not quite right)
 
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  • #16
thanks... this is just what i need!
 
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  • #18
GT1 said:
Does more force really needed to move the box when the force applied at an angle?
Doesn't it depend on the coefficient of friction?

http://books.google.com/books?id=oV... Programmers&hl=iw&pg=PA42#v=onepage&q&f=true
Yikes. They fail to consider that a force applied at an angle reduces the normal force between box and ground and thus the friction.

I hope that bit of analysis is not typical of the book. The parts I glanced at when I saw the book--years ago--looked OK to me. I obviously didn't look at that bit. As always, Caveat Lector!
 
  • #19
GT1 said:
Does more force really needed to move the box when the force applied at an angle?
Doesn't it depend on the coefficient of friction?
To answer your question: Yes. When applied at an angle, the force needed to overcome friction depends on the angle and the coefficient of friction.
 
  • #20
deluksic said:
and they say that if u hit or pull something for a corner for example.. it will have the same speed like if u hit it in center of mass.
What the lecture states is if the impulse (force x time) is the same. If the force is off center, then over time the point of application of force must move at a faster speed (a faster rate of accelertion) in order to maintain the same force, so that over time, force x distance is greater if the point of application of force is off center, so therefore more work is done, more energy added to the rod, and the linear speed of the rod is the same if the same impulse is applied to the center or the end of the rod, (but the acceleration, speed, and work done is greater if the impulse is applied to the end of the rod).

Going back to your point about a rod being "hit", assume that instead of an impulse, it's an elastic collision, with some moving object colliding into a non-moving rod in space. All of the initial energy = 1/2 m v2 of the moving object, and after collision, some of that energy is transferred to the rod. If it's an off center collisoin, then some of that energy is converted into angular energy, and the linear speed of the rod will be less in that case. In order for the impulse to be the same in both cases, in the case of the off center collision, the moving object would have to have more mass and/or more speed before the collision, so that it's total energy is greater by the amount of angular energy that will be added to the rod via an elastic (no energy lost) collision.
 
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1. What is rotational physics and why is it important?

Rotational physics is a branch of physics that deals with the motion of objects that are rotating or spinning. It is important because it helps explain many natural phenomena, such as the movement of planets and the behavior of gyroscopes. It also has practical applications in engineering and technology, such as in the design of machines and vehicles.

2. How does rotational physics differ from linear physics?

Rotational physics differs from linear physics in that it studies the motion of objects around a fixed axis, rather than in a straight line. This requires different mathematical equations and concepts, such as torque and angular velocity, to describe and analyze rotational motion.

3. What are some real-world examples of rotational physics?

Some examples of rotational physics in everyday life include the rotation of the Earth on its axis, the spinning of a top or a merry-go-round, the movement of gears in a machine, and the motion of a thrown frisbee or a spinning basketball.

4. How is rotational motion affected by mass and distance from the axis of rotation?

Mass affects rotational motion through its moment of inertia, which is a measure of an object's resistance to changes in its rotational motion. Objects with a larger moment of inertia will require more torque to rotate. Distance from the axis of rotation also affects rotational motion, as objects farther from the axis will have a larger moment of inertia and will require more torque to rotate.

5. What is the role of angular momentum in rotational physics?

Angular momentum is a conserved quantity in rotational physics, meaning that it remains constant unless acted upon by an external torque. It plays a crucial role in understanding the behavior of rotating objects, as it is a measure of an object's resistance to changes in its rotational motion. It also helps explain phenomena such as the conservation of angular momentum in spinning objects and the stability of spinning tops and gyroscopes.

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