Force applied to dumbbell to CM and edge

[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.

 

[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.

 

[Doc Al]

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.

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.

 

[A.T.]

BUT how can a force applied for a certain time give something more energy

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

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.

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

 

[PJC01]

I can understand your confusion about the concept of energy in this scenario. The key thing to remember is that energy is a conserved quantity, meaning it cannot be created or destroyed, only transferred or transformed. In the case of the dumbbell, the energy is being transferred between different forms - from potential energy (due to the applied force) to kinetic energy (motion).

In the first case, when the force is applied to the center of mass, the dumbbell only has translational motion and no rotational motion. This means that all the energy from the force is transferred to the dumbbell's translational kinetic energy. In the second case, when the force is applied to the edge, the dumbbell has both translational and rotational motion. This means that the energy from the force is divided between the two forms - some goes towards translational kinetic energy and some towards rotational kinetic energy.

Now, to address your question about how a force applied to a different position can give more energy - this is due to the concept of torque. Torque is a measure of how much force is causing an object to rotate about a certain point. In the second case, when the force is applied to the edge, there is a greater distance between the point of application and the center of mass, resulting in a larger torque. This larger torque requires more energy to overcome and thus, the dumbbell ends up with more energy in the form of rotational kinetic energy.

To provide some mathematical proof, we can use the equation for work: W = Fd, where W is work, F is force, and d is distance. In the first case, the force F is applied directly to the center of mass, so d is equal to 0. This means that no work is done on the dumbbell and therefore, no energy is transferred to it. In the second case, the force F is applied at a distance d from the center of mass, resulting in a non-zero value for W and thus, a transfer of energy to the dumbbell.

To further illustrate this concept, let's take your example of flicking a pen. When you flick the pen at its center of mass, you are essentially applying a force at a distance of 0 from the center of mass. This means that all the energy from the flick is transferred to the pen's translational motion, resulting in it moving quickly across the table. However, when you flick the pen at the end