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Rotational Kinematics

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A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 4.2 m down a q = 23° incline. The sphere has a mass M = 4.1 kg and a radius R = 0.28 m.
a) Of the total kinetic energy of the sphere, what fraction is translational?
b) What is the translational kinetic energy of the sphere when it reaches the bottom of the incline?
c) What is the translational speed of the sphere as it reaches the bottom of the ramp?
d) Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline?

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Solution Summary

This job also determines translational kinetic energy of the sphere.

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<br>because, total kinetic energy(K.E.):
<br>K.E. = rotational K.E. + translational K.E.
<br>rotational K.E. = (1/2)*I*w^2 (w == read as omega=> angular velocity)
<br>I = moment of inertia = (2/5)*M*R^2
<br>K.E.(R) = (1/2)*(2/5)*M*R^2*w^2 = (1/5)*M*R^2*w^2 = (1/5)*M*v^2
<br>(because, for pure rolling: R*w = v)
<br>translational K.E. = (1/2)*M*v^2
<br>therefore, total K.E.:
<br>K = (1/2)*M*v^2 + (1/5)*M*v^2 = (7/10)*M*v^2
<br>therefore, translational K.E. as a ...

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  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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