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```
A uniform sphere of radius r starts rolling down without slipping from the top of another fixed sphere of radius R.find the angular velocity of sphere of radius r at the instant when it leaves contact with surface of fixed sphere.
A uniform sphere of radius r starts rolling down without slipping from the top of another fixed sphere of radius R.find the angular velocity of sphere of radius r at the instant when it leaves contact with surface of fixed sphere.

```
6 years ago

rake
73 Points
```							In the equilibrium, the centripetal
force must be equal to the down word force.

So,

mv2/(R+r) = mg
cos?

When the body is rolling without
slipping, in accordance to principle of conservation of energy, potential
energy must be in the form of translational energy plus rotational kinetic
energy.

So,

mgh = ½ mv2 + ½ I?2

Or,

mg(R+r) (1-cos?) = ½ mv2+(1/5 )mv2

= (7/10) mv2

Or,

(10/7) mg (1-cos?) = mg cos?

Again,

mv2 = (10/7) mg (R+r)
(1-cos?)

10/7 = 17/7 cos?

Or, cos? = 10/17

The velocity of sphere radius r at the instant when it leaves contact
with surface of fixed sphere will be,

v = v[g(R+r) cos?] =v[(10/17){g(R+r)}]

So the angular velocity will be,

? = v/r

=v[(10/17r2){g(R+r)}]

From the above observation, we
conclude that, the angular velocity of sphere radius r at the instant when it leaves contact with surface of fixed
sphere would bev[(10/17r2){g(R+r)}] .
```
6 years ago
618 Points
```							Dear student,Please find the answr to your question. In the equilibrium, the centripetal force must be equal to the down word force.So,mv2/(R+r) = mg cos?When the body is rolling without slipping, in accordance to principle of conservation of energy, potential energy must be in the form of translational energy plus rotational kinetic energy.So,mgh = ½ mv2 + ½ I?2Or,mg(R+r) (1-cos?) = ½ mv2+(1/5 )mv2= (7/10) mv2Or,(10/7) mg (1-cos?) = mg cos? Again,mv2 = (10/7) mg (R+r) (1-cos?)10/7 = 17/7 cos?Or, cos? = 10/17The velocity of sphere radius r at the instant when it leaves contact with surface of fixed sphere will be,v = v[g(R+r) cos?] =v[(10/17){g(R+r)}]So the angular velocity will be,? = v/r=v[(10/17r2){g(R+r)}]From the above observation, we conclude that, the angular velocity of sphere radius r at the instant when it leaves contact with surface of fixed sphere would be v[(10/17r2){g(R+r)}] Thanks and regards,Kushagra
```
5 months ago
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