An 8.10-cm-diameter, 300 g solid sphere is released from rest at the top of a 1.60-m-long, 16.0 ? incline. It rolls, without slipping, to the bottom.
a)What is the sphere's angular velocity at the bottom of the incline?
b)What fraction of its kinetic energy is rotational?
Using Energy conservation during this motion:
KEi + PEi = KEf + PEf
KEi = Kinetic energy at the top of incline = 0 (Since starts from rest)
PEf = 0, since potential energy is zero at ground
PEi = m*g*h
h = vertical height of incline = L*sin theta = 1.60*sin 16 deg
KEf = total kinetic energy at the bottom of incline
KEf = KEtrans + KErot
KEf = 0.5*m*V^2 + 0.5*I*w^2
V = linear velocity of sphere = w*R
I = Moment of inertia of sphere = 2*m*R^2/5
w = angular velocity
So Using these values:
m*g*h = 0.5*m*(w*R)^2 + 0.5*2*m*R^2*w^2/5
g*h = w^2*R^2/2 + w^2*R^2/5
g*h = 7*w^2*R^2/10
w = sqrt (10*g*h/(7*R^2))
R = 8.10 cm/2 = 4.05 cm = 0.0405 m
Using known values:
w = sqrt ((10*9.81*1.60*sin 16 deg)/(7*0.0405^2))
w = 61.4 rad/sec
Part B.
KErot/KEtotal = (0.5*I*w^2)/(0.5*m*V^2 + 0.5*I*w^2)
Using above relations:
KErot/KEtotal = (m*w^2*R^2/5)/(0.5*m*w^2*R^2 + m*w^2*R^2/5)
KErot/KEtotal = (1/5)/(7/10)
KErot/KEtotal = 2/7
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