Question

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?

Answer 1

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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