Independent translational momentum

[nomism]

Hi everyone,

I was watching an MIT OpenCourseWare video where the lecturer describes the problem of a rulers motion on a frictionless surface after being subjected to a 'short' impulse orthogonal to the long axis of the ruler.

(sorry as a new user I can't post images, youtube 'MIT torque lecture 21' for a problem description at 14:45 mins)

He goes on to show that the resulting translational velocity is independent of the distance between the centre of mass and the line of action of impulse
despite the fact that resulting angular velocity is dependent on that same distance.

The resultant translational direction seems obvious, but I have trouble reconciling the translational speed with the work energy theorem in the accompanying class notes (below) as I thought that equal impulses would input equal amounts of energy into the system

i.e. by my thinking rotational momentum would decrease and translational momentum would increase as the impulse is applied closer to the centre of mass

Please could someone explain this to me

$$W_{o,f}^{total} = \int_{o}^{f} \vec{F}_{ext}^{total}\cdot d\vec{r} + \int_{o}^{f} \vec{\tau }_{cm}^{total}\cdot d\vec{r}$$
$$= (\frac{1}{2}mV_{cm,f}^{2} + \frac{1}{2}I_{cm}\omega_{cm,f}^{2} )- (\frac{1}{2}mV_{cm,f}^{2} + \frac{1}{2}I_{cm}\omega_{cm,i}^{2} )$$
$$= \Delta K_{trans} + \Delta K_{rot} = \Delta K_{total}$$

 

[jfy4]

if $\tau$ is torque, the units don't match... I don't think that can be. What is $\tau$ in this equation?

 

[PriceMike]

damnet that is tough

 

[nomism]

Thanks for the replies so far. The tau symbol is torque but I made a mistake in the second integral, it should have been with respect to the angle d theta. The units should match now.

I also forget to include the brackets to show that this is the z component of tau (i.e. the magnitude of the torque vector orthogonal to the centre of mass)

(sorry about not being able to post links but the set of notes the equation was taken from can be found by googling
'Module 27: Rigid Body Dynamics: Rotation and Translation about a Fixed Axis')

$$W_{o,f}^{total} = \int_{o}^{f} \vec{F}_{ext}^{total}\cdot d\vec{r} + \int_{o}^{f} (\vec{\tau }_{cm}^{total})_{z} \cdot d\theta$$
$$= (\frac{1}{2}mV_{cm,f}^{2} + \frac{1}{2}I_{cm}\omega_{cm,f}^{2} )- (\frac{1}{2}mV_{cm,f}^{2} + \frac{1}{2}I_{cm}\omega_{cm,i}^{2} )$$
$$= \Delta K_{trans} + \Delta K_{rot} = \Delta K_{rot}$$

Perhaps my problem is that the inputs are in fact different. I.e. a force at a distance also creates a torque and the creation of said torque does not reduce the force therefore the creation of an angular impulse does not affect the translational impulse. Does this sound reasonable to everyone?

I also can't think of an object that would impart a constant impulse kinetically. The closest I could think of is a hockey puck sliding across the surface and imparting it's kinetic energy to the ruler. However I can imagine that point of impacts distance from the centre of mass would have an effect on the pucks speed after the impact as well as the rulers.

Has anybody ever seen an experiment like this?

 

[PJC01]

I can explain this phenomenon using the principles of conservation of momentum and energy. The key concept here is that of independent translational momentum, which means that the translational velocity of an object is not affected by the distribution of its mass. This is because the translational motion of an object is determined by its total momentum, which is the product of its mass and velocity. As long as the total momentum remains constant, the translational velocity will also remain constant.

In the case of the ruler on a frictionless surface, the impulse applied at a point on the ruler will result in a change in both translational and rotational velocities. However, due to the independent translational momentum, the resulting translational velocity will be the same regardless of where the impulse is applied on the ruler.

Now, let's consider the work-energy theorem. This theorem states that the change in kinetic energy of a system is equal to the work done on the system. In this case, the work done by the external force (impulse) will be equal to the change in kinetic energy of the system, which includes both translational and rotational components. However, since the translational velocity remains constant, the change in translational kinetic energy will be zero. This means that all of the work done by the impulse will be used to change the rotational kinetic energy of the system.

To sum up, the independent translational momentum of an object allows for its translational velocity to remain constant despite changes in its rotational velocity. This is due to the conservation of momentum and the work-energy theorem, which explain the distribution of energy within the system. I hope this helps to clarify any confusion you may have had regarding this problem.