A uniform solid 2.5 kg sphere has a radius of 0.35 m. it begins at rest, at the

Question

A uniform solid 2.5 kg sphere has a radius of 0.35 m. it begins at rest, at the top of a straight ramp, where the bottom of the sphere is 5.2 m above the floor. It then rolls down the ramp. Note that the sphere rotates about its center of mass and there is negligible friction (just enough for rotation but no significant losses of mechanical energy occur). Complete the following table based upon this problem. Note that the two points being considered arc the top of the ramp, and halfway down the ramp. You are not being asked (in part a) anything about what happens at the bottom of the ramp! Show your work and equations used.

Explanation / Answer

at the top of the ramp

PE1=m*g*(h+r) = 135.975 J

linear KE1 = 0

rotational KE1 = 0

total mechanical energy 1= 135.98 J

half of the ramp

PE2 = m*g*(0.5h+r)= 72.275

total mechanical energy-PE2 = translation + rot KE =0.5*m*V^2 + 0.5*I*^2 = 63.7

for rolling V = r

0.5*(m + I/r2 )V2 = 63.7 J ; I=2mr2/3 = 0.204167 kgm2

V= 5.53m/s

translational KE2 = 0.5*2.5*V2 = 38.226 J

rotational KE2 = 0.5*I*2 = 25.484 J

total energy2 = 135.98 J

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