A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the bottom of an inclined plane of height h=3.5m, as shown in the figure. The spherical shell rolls without slipping. Find wif the shell comes to rest at the top of the inclined plane. (Take g-9.81 m/s2, Ispherical shell = 2/3 MR2 ).Express your answer using one decimal place.

47.8rad/s

Explanation:

For energy to be conserved.

The potential energy sustain by the object would be equal to K.E

P.E = m× g× h = 2 × 9.81× 3.5= 68.67J

Now K.E = 1/2 × I × (w1^2 - w0^2)

I = 2/3 × M × R2

= 2/3 × 2 × (0.23)^2= 0.0705

Hence

W1 = final angular velocity

Wo = initial angular velocity

From P.E = K.E we have;

68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)

(w1^2 - w0^2) = 1948.09

W1^2 = 1948.09 + (18.3^2)

W1^2=2282.98

W1 = √2282.98

=47.78rad/s

= 47.8rad/s to 1 decimal place.

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank