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 …
MIT 8.01 Physics I Classical Mechanics, Fall 1999
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
|dw: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
Although 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 point-mass 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