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InterviewSolution
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Figure shows a student, again sitting on a stool that can rotate freely about a vertical axis. The student, initially at rest, is holding a bicycle wheel whose rim is loaded with lead and whose rotational inertia I_(wh) about its central axis is 1.2kg*m^(2). The wheel is rotating at an angular speed omega_(wh) of 3.9 rev/s, as seen from overhead, the rotation is counterclockwise. The axis of the wheel is vertical, and the angular Momentum vecL_(wh) of the wheel points vertically upward. The student now inverts the wheel so that, as seen from overhead, it is rotating clockwise. Its angular momentum is now -vecL_(wh). The inversion results in the student, the stool, and the wheel's center rotating together as a composite rigid body about the stool's rotation axis, with rotational inertia I_(b)=6.8kg*m^(2). With what angular speed omega_(b) and in what direction does the composite body rotate after the inversion of the wheel? |
| Answer» <html><body><p></p>Solution :1. The angular speed `omega_(b)` we seek is related to the final angular <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a> `vecL_(b)` of the composite body about the stool.s <a href="https://interviewquestions.tuteehub.com/tag/rotation-11281" style="font-weight:bold;" target="_blank" title="Click to know more about ROTATION">ROTATION</a> <a href="https://interviewquestions.tuteehub.com/tag/axis-889931" style="font-weight:bold;" target="_blank" title="Click to know more about AXIS">AXIS</a> by `(L=Iomega)`. <br/> 2. The initial angular speed `omega_(wh)` of the wheel is related to the angular momentum `vecL_(wh)` of the wheel.s rotation about its center by the same equation. <br/> 3. The vector <a href="https://interviewquestions.tuteehub.com/tag/addition-367641" style="font-weight:bold;" target="_blank" title="Click to know more about ADDITION">ADDITION</a> of `vecL_(b)andvecL_(wh)` gives the total angular momentum `L_("tot")` of the system of the student, stool, and wheel. <br/> 4. As the wheel is inverted, no net external torque acts on that system to change `vecL_("tot")` about any vertical axis. So, the system.s total angular momentum is conserved about any vertical axis, including the rotation axis through the stool. <br/> Calculations: The conservation of `vecL_("tot")` is represented with vectors in Fig. We can also write this conservation in terms of components along a vertical axis as <br/> `L_(bf)+L_(wh.f)=L_(b,i)+L_(wh,i)`. <br/> Where i and f refer to the initial state and the final state. Because inversion of the wheel inverted the angular momentum vector of the wheel.s rotation, we substitute `-L_(wh,i)" for "L_(wh,f)`. Then, if we set `L_(b,i)=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>`. <br/> `L_(b,f)=2L_(wh,i)`. <br/> Using Equation, we next substitute `I_(b)omega_(b)" for "L_(b.f)andI_(wh)omega_(wh)" for "L_(wh,i)` and solve for `omega_(b)`, finding <br/>`omega_(b)=(2I_(wh))/I_(b)omega_(wh)` <br/> = `((2)(1.2kg*m^(2))(3.9rev//s))/(6.8kg*m^(2))=1.4rev//s`.</body></html> | |