Question

A disc of mass 3kg and radius 0.5m is placed on a horizontal frictionless table. With the help of a cue we gives a horizontal impulse of 15Ns to the disc at topmost point then velocity of centre of mass will be

Answer 1

Answer:

Given: A uniform solid sphere is placed on a smooth horizontal surface. An impulse I is given horizontally to the sphere at a height h=

5

4

R above the centre line. m and R are mass and radius of sphere respectively.

To find:

(a) The angular velocity of sphere & linear velocity of centre of mass of the sphere after impulse.

(b) The minimum time after which the highest point B will touch the ground.

(c) The displacement of the centre of mass during this interval.

Solution:

As per the given criteria,

Mass = m, Radius: R, Sphere is solid.

h=

5

4

R above the horizontal center line of the sphere

We know

impulse = I=F×Δt=Δp as per the conservation of linear momentum.

Linear Momentum of sphere after impulse =mv=p=I

As h>

5

2

R , The ball probably slips along with rolling. So, perhaps v



=Rω. If the horizontal impulse is given at a height

5

2

R, for a solid sphere, then it rolls without slipping.

Moment of inertia MI about center of mass, I

cm

=

5

2

×mR

2

Angular momentum, L=I

cm

ω=

5

2

mR

2

ω

L=h×p= Angular impulse given, as initial angular momentum = 0

⟹>h×p=

5

4

Rmv=

5

2

mR

2

ω

⟹ω=

R

2v

...(ii)

So ω=

mR

2I

We need to find the Time duration for the ball to rotate by π radians, 180 deg.:

=t=

ω

π

=

2I

πωR

So the horizontal (translational, linear) displacement

=s=vt=

m

I

×

2I

πmR

=

2

π

×R

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

KE=

2

1

mv

2

+

2

1

Iω

2

⟹KE=

2

1

mv

2

+

2

1

×

5

2

mR

2

×ω

2

⟹KE=

2

1

m(

2

ωR

)

2

+

5

1

mR

2

ω

2

⟹KE=

40

5+8

mR

2

ω

2

⟹KE=

40

13

mR

2

(

R

2v

)

2

⟹KE=

10

13

mv

2