30th November 2008

A uniform sphere of mass m and radius r rolls without sliding over a horizontal plane, rotating about a horizontal axle OA. in the process, the center of the sphere moves with velocity v along a circle of radius R. Find the kinetic energy of the sphere.

20 Answers

1
Rohan Ghosh ·

if an object moves with velocity v in a horizontal circle its kinetic energy = 1/2mv2

1
rahul wadhwani ·

there we would take radius of rotaion as r+R then add KE trans.+ KErot.

62
Lokesh Verma ·

There is a mistake in prashant's solution... something is wrong.. i dont know exactly what!!!

My solution is 1/2 I1ω12+1/2 I2ω22

where I1= 2/5mr2
ω1 = v/r

I2= mR2+2/5mr2
ω2 = v/R

Thus, total energy = 1/2(mr2+2/5mr2)(v/r)2 + 1/2(mR2+2/5mr2) (v/R)2

= 7/10mv2 + 1/2(2/5m(r/R)2)(v)2
= 7/10mv2 + 1/5m(r/R)2)(v)2

1
Prashant Shekhar ·

K.E.=1/2 mv2 + 1/2 I1ω12+1/2 I2ω22

ω1=v/r
ω2=v/R
I1=2/5mr2
I2=2/5mr2+mR2

Hence, KE=1/2 mv2 + 1/2 (2/5mr2)v2/r2 + 1/2(2/5mr2+mR2)v2/R2
=6/5 mv2[1+r2/6R2]

62
Lokesh Verma ·

Hint: THis is made of 2 pure rotations only

1) at point O
2) At the point where the sphere touches the ground

1
Kshitij Gupta ·

62
Lokesh Verma ·

ok.. then what is teh difference between.. a sphere .. moving along a straight line and moving in a circular orbit..

(forget rotation about the axis!)

Will the kinetic energy be the same??

1
Rohan Ghosh ·

but .. at any moment of time the velocity of the centre of mass =v

and the angular velocity = w

so the result .. but i cant understand where im going wrong!!

62
Lokesh Verma ·

But is that not for point masses?

62
Lokesh Verma ·

yes! that is true..

9
Celestine preetham ·

1/2mv2( 1 + 2r2/5R2 )

62
Lokesh Verma ·

yeah i know it is a bit hard to digest.. even when i was preparing for jeee it took me a lot of lot of time to digest this solution!! *to be true i could not solve it myself!

(It is another one from IRODOV!!!!)

1
Rohan Ghosh ·

but that will just contribute to its translational energy!! (it is moving with v so simply 1/2mv2!!)

62
Lokesh Verma ·

do u realise that there are 2 different omeagas!

one is of the ball along its axis.. another from the center!

1
Rohan Ghosh ·

if r=0 then the velocity of the bottom most point cant be 0 it will slide!!

1
Rohan Ghosh ·

as it rolliing without sliding
v=wr
translational energy =1/2mv2
rotational energy = 2/5mr2w2/2=mv2/5
hence ans ..

62
Lokesh Verma ·

There is a lot more to it..!!!

how about r .. dont u think if r is zero.. the answer will be very very differnt!

how are r and R related!

1
Rohan Ghosh ·

im getting answeras 7mv2/10

9
Celestine preetham ·

ive included translational and rotational KE wat else is there

62
Lokesh Verma ·

No dear.. there are more energies!

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