*How strange is the lot of us mortals! Each of us is here for a brief sojourn; for what purpose we know not, though some times sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our happiness depends.*

– Albert Einstein

*L*and mass

*M*is hinged without friction at the end B. A small block of clay of mass

*m*moving horizontally with velocity

*v*srikes the end B of the rod and gets stuck to it. The angular velocity of the system about A just after the collision is

(b) 3*mv/*(*mL* + *ML*)

(c) (*mv*+ *ML*)*/* 3*mL*

(d) *mv/*(3*mL* + *ML*)

**3**

*mv/*(3

*mL*+

*ML*)

The angular momentum of the system about the point A just before collision is the angular momentum of the block of clay which is equal to *mvL*. (Note that the perpendicular distance of the line of action of the linear momentum (*mv*) of the block of clay from the point A is *L*).

The angular momentum of the system about the point A just after the collision is *Iω* where *I* is the total moment of inertia of the rod and clay and *ω *is the angular velocity of the system immediately after the collision.

We have *I = ML*^{2}/3 + *mL*^{2}

Equating the angular momenta before and after collision, we have

*mvL = * (*ML*^{2}/3 + *mL*^{2})* ω*

Therefore* ω = *3*mv/*(3*mL* + *ML*).

(2) A particle is projected at an angle of 60º with the horizontal with linear momentum of magnitude *p*. The horizontal range of this projectile is *R*. Just before the projectile strikes the ground at A, what is the magnitude of its angular momentum *about **an axis perpendicular the plane of motion and passing through* *the point of projection*? Neglect air resistance.

(a) *pR*

(b) *pR*/2

(c) (√3) *pR* /2

(d) (√3) *pR*

(e)** **Zero

This is a very simple question. The magnitude of the linear momentum of the projectile just before it strikes the ground will be equal to *p*. The magnitude of the angular momentum at the moment will be *p*×Lever arm = *p*×ON = *p*×*R *sin 60º = (√3) *pR* /2.

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