A ball (thin spherical shell) of mass m and radius R and a very small lead marbl

Question

A ball (thin spherical shell) of mass m and radius R and a very small lead marble (much smaller radius than the ball, so it can be treated as a point particle) of mass M roll smoothly down an incline. The angle of the incline is 0. Note that their initial and final contact points have the same difference in height, h. Rotational inertia of a spherical shell is Icon = 2/3mR2 = betamR2. Which object will come first to the bottom of the incline and why? Find the difference in their speeds at the bottom of the incline, in terms of m, M.h. theta, g, and numerical constants as needed. Find the difference in time when they reach the bottom, in terms of m, M, h, theta, g, and numerical constants as needed. Answers: marble, Delta upsilon = 2gh(1 - 1/ 1 + beta) = 0.319 gh, delta t = 2h/gsin2theta( 1 + beta - 1) = 0.41 h/gsin2theta I tried working this problem several different ways but can't figure out how to derive out to get these provided solutions, please help?

Explanation / Answer

a) the marble will reach the bottom first because all its energy isspent in converting to kinetic energy while the sphere will losesome of its KE in rolling... b) For the marble, we can just conserve energy to calculate the timeof descent. lets equate potential energy to kinetic energy: mgh=1/2mv2v=2gh we know by resolving the weight of the marble into axes, thatma=mgsin a=gsin v=u+at, 2gh=0+gsint t=2gh/gsin For the thin spherical shell, I=mR2 Its PE=mgh is converted to rotational as well as translationalvelocity mgh=1/2mv2+1/2mR22 2gh=v2+R22Butv=r 2gh=v2+R2v2/R2 v2=2gh/(1+) v=2gh/(1+) so v=2gh-2gh/(1+)          =2gh[1-1/(1+)], which is the requiredexpression... part c is simple, use a=v/t to calculate t (same as we calculatedit for the marble, and subtract)

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