Question

An 8 kg solid sphere has a radius of 70 mm and anangular velocity of 60 rpm at the top of a 30 degree incline. At the bottom of the incline it has an angular velocity of 15x60 rpm. Assume the sphere rolls without slipping. Find the height of the incline in meters.

The answer is 3.092 m. How do I get there?

Answer #1

for solid sphere

KE = 0.5*m*v^2 + 0.5*I*w^2

I = 2*m*r^2/5

w = v/r

gives use

KE = 0.7*m*v^2 = 0.7*m*w^2*r^2

Now,

Using energy conervation

KEi + PEi = KEf + PEf

KEi = 0.7*m*r^2*wi^2

KEf = 0.7*m*r^2*wf^2

PEi = m*g*h

PEf = 0

So,

0.7*m*r^2*wi^2 + m*g*h = 0.7*m*r^2*wf^2 + 0

h = 0.7*r^2*(wf^2 - wi^2)/g

given values are:

r = 70 mm = 0.07 m

g = 9.81 m/sec^2

wi = 60 rpm = 6.283 rad/sec

wf = 94.247 rad/sec

Using these values:

h = 0.7*0.07^2*(94.247^2 - 6.283^2)/9.81

h = 3.091 m

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