A solid sphere of mass M and radius R is placed on a smooth horizontal surface . It is given a horizontal impulse J at a height h above centre of mass and sphere starts rolling then, the value of h and speed of centre of mass are

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2015-05-18T19:32:18+05:30
After applying horizontal impulse J horizontal velocity of the sphere becomes v
as initial linear momentum of the system was 0 so
J = Mv
so v = J/M
now angular impulse about the C.M.
= hMv = Iω  ( I = moment of inertia , ω = angular velocity)
⇒hMv = 2/5 MR² ×ω
so ω = 5vh/2R²
now while rolling
v = ωR
⇒v = 5vh/2R
⇒h = 2R/5
2015-05-19T13:33:36+05:30

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Sphere of mass M, radius R.  Initial velocity = 0.  Initial angular velocity = 0.
Initial linear and angular momenta = 0.
There is no friction on the surface.  So there is no torque applied by friction.

The horizontal impulse J = Force * Δt = Δp = change in linear momentum
=> J = m v,  where v = velocity of sphere after impulse is given.
p = m v

This is according to the conservation of linear momentum
=>  v = J / m        -- (1)

Since the line of impulse does not pass through C, center of mass of sphere, there is a torque, angular momentum, or moment of force.  Hence, the sphere rotates.

As the ball starts rolling and does not slip on the smooth surface,
v = R ω            =>  ω = v / R =  J / (m R)    --- (2)
where ω = angular velocity of the sphere about the Center of sphere C.

The moment of Inertia I of the solid sphere about its cm C = 2/5 m R²

Angular momentum after impulse = angular impulse given, as L initial =0, by applying the law of conservation of angular momentum:

r = perpendicular distance from C onto = h
L = h * p  = h m v = h m R ω = h m R J / (m R) = h J
=> L = h J      --- (3)

For a solid sphere angular momentum about center C  =  I ω
=   2/5 m R² ω = 2/5 m R^2 J / (m R)  = 2/5 * R J    --- (4)

compare (3) and (4)
L = h J = 2/5 * R J

=>  h = 2/5  * R
and we have already, v = J / m        and    ω = J / (m R)
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Important info:

If the impulse is given above the center of mass C and below height  of  2/5 * R, then it moves linearly faster than Rω.    If the impulse is given at a height above 2/5 * R above CM, then the ball has higher ω than v/R,  ie., v < ω R.

So the ball slips and rolls.

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