A physics student stands at the exact center of a frictionless circular platform that…

PHYSICS · Multi-part · CBSE · Physics · System of Particles and Rotational Motion · System of Particles and Rotational Motion

Problem

A physics student stands at the exact center of a frictionless circular platform that can rotate about a vertical axle through its center. The combined moment of inertia of the student and platform about the axle is Is=5.0 kg m2I_s = 5.0\,\text{kg\,m}^2. The student holds a bicycle wheel of moment of inertia Iw=0.70 kg m2I_w = 0.70\,\text{kg\,m}^2 whose rim is spinning at ωw=25 rad s−1\omega_w = 25\,\text{rad\,s}^{-1} about its own axle. The wheel’s axle is initially aligned vertically, and the student–platform system is initially at rest. The student quickly flips the wheel over so that it keeps spinning at nearly the same magnitude but in the opposite direction (i.e., the wheel’s spin angular momentum about the vertical axis is reversed). The flip takes 0.40 s0.40\,\text{s} to complete. During the flip, a small friction torque of magnitude τf=0.060 N m\tau_f = 0.060\,\text{N\,m} acts on the platform due to the axle bearing. Throughout, neglect air resistance. (a) Taking the student, platform, and wheel as the system, explain whether total angular momentum about the vertical axis can be treated as conserved during the flip. In your answer, account for the effect of the bearing friction torque. (b) Derive an expression for the angular speed Ω\Omega of the student–platform system immediately after the wheel has been flipped, including the effect of the bearing friction torque. (c) Calculate the numerical value of Ω\Omega using the data provided. (d) Suppose the bearing torque were an order of magnitude larger but all other conditions unchanged. Would angular momentum conservation still be a valid approximation for this event? Justify your answer quantitatively.

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