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zucchini
 3 years ago
The top of a wheel of mass M and radius R is connected to a spring (at its equilibrium length) with spring constant k, as shown in Fig. 6.33. Assume that all the mass of the wheel is at its center. If the wheel rolls without slipping, what is the frequency of (small) oscillations? The top of a wheel of mass M and radius R is connected to a spring (at its equilibrium length) with spring constant k, as shown in Fig. 6.33. Assume that all the mass of the wheel is at its center. If the wheel rolls without slipping, what is the frequency of (small) oscillations? @MIT 8.01 Physics …
zucchini
 3 years ago
The top of a wheel of mass M and radius R is connected to a spring (at its equilibrium length) with spring constant k, as shown in Fig. 6.33. Assume that all the mass of the wheel is at its center. If the wheel rolls without slipping, what is the frequency of (small) oscillations? The top of a wheel of mass M and radius R is connected to a spring (at its equilibrium length) with spring constant k, as shown in Fig. 6.33. Assume that all the mass of the wheel is at its center. If the wheel rolls without slipping, what is the frequency of (small) oscillations? @MIT 8.01 Physics …

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arijit.mech
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1320493173057:dw actually moi about the instantaneous center of the wheel is given by 0.5mr^2+m(2r)^2 =4.5mr^2 now d/dt(KE+PE+RE)=0 KE=kinetic energy=.5mv^2 PE=potential energy=.5KX^2 RE=ROTATIONAL ENERGY=.5IW^2

VincentLyon.Fr
 3 years ago
Best ResponseYou've already chosen the best response.0Although this is 5 months old, it's still worth it correcting the few mistakes in the answer. KE : correct RE : zero because the problem states that we have a pointmass at the centre of the wheel. PE : the displacement of the end of the spring is twice that of the centre (no slipping) => PE = 1/2k(2x)²
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