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When a uniform solid sphere mass m rolls from rest down an inclined plane such that its centre of mass descends through a height h, its translational kinetic energy will be

When a uniform solid sphere mass m rolls from rest down an inclined plane such that its centre of mass descends through a height h, its translational kinetic energy will be

Grade:12th pass

2 Answers

hardeep
13 Points
4 years ago
5/7mg is the answer if a solid sphere mass m rolls from rest down an inclined plane its centre if mass desend height h its tramsitional kinetic energy will be
 
Vikas TU
14149 Points
4 years ago
dear student
for a sphere rolling down an inclined plane (without slipping) the following condition exists
potential energy at top = translational kinetic energy at bottom + rotational kinetic energy at bottom
so,
PE = KEt + KEr
or
mgh = (1/2)mv^2 + (1/2)Iw^2
here
I = moment of inertia = (2/5)MR^2
and v = translational velocity = R.w
thus, we have
mgh = (1/2)M.R^2w^2 + (1/2).(2/5).MR^2.w^2
or
gh = (7/10).)R^2w^2
or the angular velocity becomes
w2 = (10/7).[gh/R^2]
we can now use this value of w2 to calculate the translational energy
Et = (1/2)mv^2 + (1/2)Iw^2 = (1/2)M.R^2w^2 
or
Et = (1/2)MR^2.[((10/7).[gh/R^2])] 
or
Et = (5/7)M.[gh]
or
E t = (5/7)Mgh
and
the rotational kinetic energy will be
Er = mgh - (5/7)Mgh
or
E r = (2/7)Mgh

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