B Rolling Motion Test: Take the Challenge and Justify Your Answers

[kuruman]

This is not a zero sum game. For a given impulse, the body does not acquire rotational kinetic energy at the expense of translational kinetic energy. The amount of rotational KE depends on the distance of the CM from the point of impact . The ensuing angular velocity about the CM is subject to angular momentum conservation just like the ensuing linear velocity is subject to linear momentum conservation. The two are independent of each other.

 

[erobz]

I personally don't think its irrational to be taken by surprise by this as a lay person (like me). It seems like you are getting something for nothing because "position" is often just arbitrary. I think, how can an arbitrary choice have such obvious consequences, and then you find out it does and it's a bit shocking. I guess the key here is that the position @kuruman is talking about isn't really arbitrary, its relative to a very special position in physics on an extended body, and that is what makes the difference.

This is even further off topic, but I remember thinking how a lever actually multiplies the force? Right away conservation of energy pops into your head, and you start reconciling forces and displacements. Then I thought: "But if the system is static there is no displacement, yet I still have a multiplied force?" It seems strange, and I'm not sure I still understand that, but you realize there are important consequences of position and it's not always obvious why...

I don't know where you are at in your education, but you probably have time to figure some of this stuff out. I wouldn't worry too much.

 

[erobz]

Spoiler: [USER=192687]@kuruman[/USER] - tricycle problem

Does the final problem have to do with Newtons third law, where the net force from your foot is canceled by the force your butt exerts on the seat?

I showed the wife and kids this strangeness. They had fun! Thanks!

 

[Chenkel]

This is not a zero sum game. For a given impulse, the body does not acquire rotational kinetic energy at the expense of translational kinetic energy. The amount of rotational KE depends on the distance of the CM from the point of impact . The ensuing angular velocity about the CM is subject to angular momentum conservation just like the ensuing linear velocity is subject to linear momentum conservation. The two are independent of each other.

Let's take an example of a rod that weights 1kg, and has a length of 2 meters, $M = 1$, $L=2$, $I_{cm}=\frac 1 {12} M L^2 = \frac 1 3$ if a force of 10N is applied a distance $R = 1$ from the center of mass for 100ms, the change in angular momentum is $\tau \Delta t = RF\Delta t = (1)(10)(.1) = 1$, the angular velocity is $\dot \theta = \frac {L_{cm}} {I_{cm}} = (1)(3) = 3$ radians per second. The change in linear momentum is $F\Delta t = (10)(.1)=1$, so the velocity is $\frac P M = 1$ meters per second. If you apply the same force to the cm you will have no rotation, but you will have the same linear velocity. So in the first case:
$$KE = \frac 1 2 Mv^2 + \frac 1 2 I_{cm}{\dot \theta}^2 = \frac 1 2 + \frac 1 2 \frac 1 3 3^2 = 2$$In the second case$$KE = \frac 1 2 Mv^2 = \frac 1 2$$In this example we're not expending translating KE for rotational KE, but I'm wondering what the reasoning is for the result, we apply the same force at a different location and apparently we do more work with the same impulse? What's going on?

 

[jbriggs444]

does anyone know why there seems to be more energy in the body with the same impulse in the off center force example?

The answer has been given a number of times already [in another recent thread]

Energy goes as force times displacement. Momentum goes as force times time. The two need not be proportional. With an off-center hit there is less resistance at the point of impact, so more velocity at the point of impact, so more displacement at the point of impact, so more energy transferred for a given impulse.

 

[Chenkel]

The answer has been given a number of times already [in another recent thread]

Energy goes as force times displacement. Momentum goes as force times time. The two need not be proportional. With an off-center hit there is less resistance at the point of impact, so more velocity at the point of impact, so more displacement at the point of impact, so more energy transferred for a given impulse.

So I can take my constant impulse and potentially create a more kinetic energy from said impulse depending on how I apply it? I'm guessing the same impulse will create more energy if the distance traveled is larger, so that would take more effort on my part, the person applying the force. The puzzle is starting to come together. Thank you.

 

[malawi_glenn]

So I can take my constant impulse and potentially create a more kinetic energy from said impulse depending on how I apply it? I'm guessing the same impulse will create more energy if the distance traveled is larger, so that would take more effort on my part, the person applying the force. The puzzle is starting to come together. Thank you.

Same force, but applied to CM vs (almost at the) tip of the rod. The time of impact is also the same, hence same impulse. When force is applied to the tip of the rod, the displacement is longer, thus more work is done.

 

[Orodruin]

So I can take my constant impulse and potentially create a more kinetic energy from said impulse depending on how I apply it? I'm guessing the same impulse will create more energy if the distance traveled is larger, so that would take more effort on my part, the person applying the force. The puzzle is starting to come together. Thank you.

Simply put: Impulse is not energy. It is therefore not a given that a particular impulse should correspond to a particular energy.

The details have been discussed above.