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?

Figure shows a student, again sitting on a stool that can rotate freel

1.	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 1.2kgm^(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)=6.8kgm^(2). With what angular speed omega_(b) and in what direction does the composite body rotate after the inversion of the wheel?
Answer» Solution :1. The angular speed `omega_(b)` we seek is related to the final angular MOMENTUM `vecL_(b)` of the composite body about the stool.s ROTATION AXIS by `(L=Iomega)`. 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. 3. The vector ADDITION of `vecL_(b)andvecL_(wh)` gives the total angular momentum `L_("tot")` of the system of the student, stool, and wheel. 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. 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 `L_(bf)+L_(wh.f)=L_(b,i)+L_(wh,i)`. 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)=0`. `L_(b,f)=2L_(wh,i)`. 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 `omega_(b)=(2I_(wh))/I_(b)omega_(wh)` = `((2)(1.2kgm^(2))(3.9rev//s))/(6.8kgm^(2))=1.4rev//s`.

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