# Force applied to dumbbell to CM and edge

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1. Nov 24, 2014

### Malabeh

Today in class we learned about how if a force F is applied to a dumbbell on its center of mass and then separately applied to the edge where it gets its maximum torque, the dumbbell will have the same CM velocity after. It makes sense when you think about it in terms of momentum. F times T gives the impulse and so it is the same in both cases. I get confused about its energy. The second case has more energy because it is rotating and translating.....makes sense, BUT how can a force applied for a certain time give something more energy when it is applied to a different position? Also, in my mind I imagine the force being applied to the dumbbell in case 2 and it just rotates around that point because there is maximum torque. Obviously my thoughts are flawed, would anyone care to explain why? If you can provide mathematical proof that'd be great!!

EDIT: If I flick my pen at its center of mass, it goes across the table really fast, but if I flick it at the end, it spins and only has a minor transnational velocity. That seems counter intuitive to what my teacher said.

2. Nov 25, 2014

### jbriggs444

Change in momentum is given by impulse. Impulse is the integral of force over time. Change in energy is given by work. Work is the integral of force over distance.

The end of the dumbbell moves relative to the center of mass. If you apply the same force for the same time at the end of the dumbbell rather than at the center, you will have imparted the same impulse. But you will have imparted more energy because of that additional movement.

3. Nov 25, 2014

### Staff: Mentor

It's not so easy to flick the end of the pen with the same force as you can flick the center of mass, since the pen starts spinning and the point of contact moves away from you more quickly. When flicking the center you are probably exerting a greater impulse--more force for a greater time.

4. Nov 26, 2014

### A.T.

Because energy gain depends on the distance the force is applied over, not the duration it is applied for.

How can you be sure you applied the same force over the same time in both cases?