Hey guys. If a body is spinning around the z - axis and an element of that body detaches on its own and flies off tangentially, does the rest of the rotating body experience a change in angular speed??? My thought is that the angular speed remains constant, just like velocity of a man holding a bowling ball on a skateboard remains constant before and after he drops it.
According to the conservation of angular momentum, the body must experience a change in its angular velocity because its moment of inertia changes(mr^2) as there is no external torque.
Thanks for the reply.....so how do you describe the new angular velocity. I understand that the body's moment of inertia decreases, but im still not sure how to use conservation of angular momentum to solve for the final angular velocity. If the element mass flies off at the same linear velocity that it was going at in the first place, doesnt that mean that nothing changes?
If the initial angular velocity was w and moment of inertia of body is I, then inital angular momentum of system is Iw. Now, let's consider the moment of inertia for the element to be I1 and that of the remaining body be I2. If their angular velocities are w1 and w2, then: \[Iw=I1w1+I2w2\] The element that flies off will also rotate with some angular velocity w1
Even if the element flies off with same speed, the angular velocity of the body left off will still change because its moment of inertia has changed.
im not sure what ur telling me, so would the bodies angular speed increase or decrease. lets say that elemnt doesnt have any rotational speed as it flies off, (just tangential!) that element has the same momentum before and after it breaks off .....started as rw angular velocity, then turned to v linear velocity, which are equivalent in magnitude.............
See, if the element flies off, the moment of inertia of the body decreases(because the mass decreases_). So to maintain constant angular momentum, the body now rotates with a faster angular velocity. Makes sense? Ok, consider that the element does not rotate. So, its angular velocity is zero. Hence its angular momentum is also zero. Then, in the above formula, w2=-0 So it becomes: \[Iw=I1w1\] Got it?
Hey, either we have a miscommunication with the problem, or your confusing what is going on. The element at the beginning of the problem was rotating as a part of the entire body around the axis of rotation, therefore it has angular momentum with respect to the point of rotation the element keeps the same angular momentum before and after it flies off the rotating body as given by the cross product of angular momentum (m(r x v)). that breaks down to mvR, which was what the angular velocity was before it broke off. * **A particle can have angular momentum with respect to a point even if its not rotating around that point, m(r x v), always**....... So if the elements' angular momentum with respect to the point of rotation is constant no matter how far that thing flies, then shouldnt the remaining bodies' angular momentum stay the same as it was before (oringinal angular velocity of the entire combined body) Clarify if you can.
Hey,you are right! I had almost forgotten that: **A particle can have angular momentum with respect to a point even if its not rotating around that point, m(r x v), always**....... So,the element that flies off has same angular momentum as before. So, the angular momentum of the body does not change. If the moment of inertia of the element is I' and that of the remaining part is I, \[(I+I')w=Iw1+I'w\] \[rightarrowIw=Iw1\] Thus,w=w1. Thanks, you taught me something new today! By the way, where are you from?
Oh cool, I really hope this is actually true. I mean, it should be, based on this logic........ You might wanna double check the idea with a physics teacher or something, but I think we have it. I'm from MA u? i use these videos a lot. 8.01 and 8.02, they really help.
Yeah,Walter lewin rocks! MA means? I m from India
oh cool, USA, massachusetts
So, you aspire to study at MIT?
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