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Here is the question:

A uniform rod of mass m and length $l$ is placed horizontally on a smooth horizontal surface. An impulse P is applied at one end perpendicular to the length of the rod. Find the velocity of centre of mass and angular velocity of the rod just after the impulse is applied.enter image description here

I know that we will conserve angular and linear momentum because no external force is being applied.

According to the solutions I have read:

Linear momentum: The initial linear momentum is $0$ and an impulse of P is being applied. Therefore, the change in linear momentum = P. $$\therefore P = m\cdot v_{cm}\\v_{cm} = \frac{P}{m}$$

But my question is, will the whole P momentum considered for linear momentum. Shouldn't the total impulse given be divided into linear momentum and angular momentum? Also, I couldn't understand the way they calculated angular momentum so please explain that as well...

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  • $\begingroup$ You can define an angular impulse in a similar way. $\endgroup$
    – Judas503
    Commented Jan 11, 2020 at 16:46
  • $\begingroup$ So will we write $\displaystyle P\frac{l}{2} = I\omega + mv_{cm}r$? $\endgroup$
    – HSB
    Commented Jan 11, 2020 at 16:52
  • $\begingroup$ Sorry, what about tangens, sinus, cosinus? $\endgroup$ Commented Oct 30, 2022 at 9:16

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No, impulse is momentum. 100% of the impulse is going to the momentum of the body afterward, and hence the movement of the center of mass.

$$ P = m \,v_{\rm cm} $$

The motion is thus $$v_{\rm cm} = \frac{P}{m} $$

So what about the rotation? Well, angular momentum is just where momentum acts in space. And the conservation of angular momentum means that the line momentum acts though is conserved.

pic

Let me elaborate on your example. Consider the point of application point 1 and the center of mass point 2 with distance $c$ between them.

The angular momentum (moment of momentum) of the impulse as seen by the center of mass is $$ L_2 = c \, P $$ (positive since $P$ acts in a counter-clockwise way relative to 1).

After the impulse is applied, the angular momentum is preserved and defines the rotational motion

$$ L_2 = I_2 \omega $$

So the motion is $$\omega = \frac{c P}{I_2} $$

Now the proof. If I transform angular momentum $L_2$ onto point 1 where the impulse acted it should be zero. This is because the impulse has zero angular momentum along the line (in space) it is acting upon.

$$ L_1 = L_2 - c P = I_2 \omega - c P = c P - c P = 0 $$ (negative since momentum acts clockwise relative to 1)

In summary,

  • The impulse carries momentum equal to $P$
  • The impusle has angular momentum zero about 1
  • The impusle has angular mometnum $c P$ about 2.
  • After the impulse is applied the body has momentum equal to $P$
  • The body has angular momentum zero about 1
  • The body has angular mometnum $c P$ about 2.

The above is a direct result of conservation of momentum, not only as a vector (magnitude and direction) but also as a line in space (passing through 2).

See also this answer expanding on the geometry of angular momentum.

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The linear momentum impulse isn’t ‘divided’ in any way.

One way to see this is that too things are being applied to the target:

  1. A linear momentum impulse. The might move some parts of the target faster, some slower, but summed result of all that will be the change in the center of mass momentum.

  2. An angular momentum impulse that comes from the applied force and moment arm (I.e. torque) and duration. That results in a change in the angular moment of the target.

These are independent results of the applied force and time, I.e. the impulse. Applied momentum and applied angular momentum are separately conserved.

What may be confusing here is the energy is not separately conserved: if you do work on the block, that does have to be divided into energy of rotation and energy of translation. But the two momentum forms are not like that: except for having similar names, they’re different.

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  • $\begingroup$ Could you be more clearer on why 'linear and angular momentum are separately conserved'... $\endgroup$
    – HSB
    Commented Jan 11, 2020 at 19:30
  • $\begingroup$ Not really. That’s just a fact of nature: they’re different things, with different conservation laws. Just like energy and momentum: different things. $\endgroup$ Commented Jan 11, 2020 at 19:32

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