Question

A solid sphere of mass m and radius R is placed on a smooth horizontal surface. A sudden blow is given horizontally to the sphere at a height h = 4R/5 above the center line. If I is the impulse of the blow then find

(a) the minimum time after which the highest point B will touch the ground.
(b) the displacement of the center of mass during this interval.

Answer 1

mass =  m,    Radius: R  .  Sphere is solid.
h =  4R/5 above the horizontal center line of the sphere.     blow horizontally.

I = impulse = F * Δ t = Δp  as per the conservation of linear momentum.

Linear Momentum of sphere after impulse  = m v = p = I

As h > 2 R / 5 ,  The ball probably slips along with rolling.   So, perhaps v ≠ R ω.  If the horizontal impulse is given at a height 2/5 R, for a solid sphere, then it rolls without slipping.

Moment of inertia MI about cm  = 2/5 * m R²
Angular momentum = MI * ω  = 2/5 m R² ω

  L = h * p = Angular impulse given, as initial angular momentum = 0
=>   h * p = 4/5 R * m v = 4/5 m v R   =  2/5 m R² ω
=>  ω = 2 v / R     and    v = R ω / 2        ---- (2)

So    ω =  2 I / (m R)

We need to find the Time duration for the ball to rotate by π radians, 180 deg.:
     = t =  π / ω = π m R / (2 I)

So the horizontal (translational, linear) displacement
      = s = v * t = I / m  * π m R / (2 I) =  π/2 * R

Also,  the Energy K.E. given by the impulse:

   ½ m v²  + ½ I ω² =  ½ m v² + ½ * 2/5 * m R² * ω²
                           =  m [v² / 2 + R² ω² /5]
                           =  m R²  ω²  [1/8 + 1/5 ]
          K. E.         =  13/40 * m R² ω²
                          =  13/10 * m v²

Answer 2

Answer:

amarbel is an example of