- #1

- 2

- 0

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

[tex]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}[/tex]

[tex]

= (\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} )

[/tex]

[tex]= \Delta K_{trans} + \Delta K_{rot} = \Delta K_{total}

[/tex]

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

[tex]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}[/tex]

[tex]

= (\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} )

[/tex]

[tex]= \Delta K_{trans} + \Delta K_{rot} = \Delta K_{total}

[/tex]

Last edited: