Question

a particle of mass m is projected with the velocity u find its angular momentum at the highest point with respect to origin​

Answer 1

A particle of mass m is projected with a velocity v, making an angle of 30° with the horizontal. What is the magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h?

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6 ANSWERS



Mike Wilkes, former Research Physicist at Air Force Research Laboratory, Albuquerque, NM (1992-2005)

Answered May 19, 2018

No doubt David Shaffer's answer is correct, which is why I was the first to upvote it. A picture and a few more details helped me understand why. The situation for projectile motion when the projectile reaches its maximum height hh at point M above the point of projection P is illustrated in the following Figure:



The initial horizontal and vertical velocity components at P are vx0=vcos30∘=v3–√/2vx0=vcos⁡30∘=v3/2 and vy0=vsin30∘=v/2vy0=vsin⁡30∘=v/2, respectively, where vvis the magnitude of the initial velocity vector vv at P. These are vHvH and vVvV, respectively, in David Shaffer's answer. The crucial point to realize is that the projectile's vertical velocity component is zero at the maximum height h,h, so it has only a horizontal component of velocity there (equal to its initial horizontal velocity component vx0vx0, since there are no forces in the horizontal direction).

The linear momentum of the projectile at point M is p=mvMp=mvM, where vMvM is the velocity vector at M. The angular momentum about P of the projectile at point M is the vector cross product L=r×p=mr×vML=r×p=mr×vM, where rr is the position vector of M with respect to P. The magnitude of this cross product is

L=mrvMsinθ,L=mrvMsin⁡θ,

where θθ is the angle between the vectors rrand vMvM, and rr and vMvM are the magnitudes of these vectors. Since vM=vx=vx0=v3–√/2vM=vx=vx0=v3/2 and sinθ=h/rsin⁡θ=h/r(from the geometry), this reduces to

L=mr(v3–√2)(hr)=mvh3–√2.(1)(1)L=mr(v32)(hr)=mvh32.

The only force assumed to be acting is the conservative gravitational force, so we have an energy conservation principle: the sum of kinetic energy and gravitational potential energy is a constant, hence must be the same at any two points of the trajectory. Choosing the horizontal plane through P to be the reference surface for zero gravitational potential energy, we see that at point P we have only the kinetic energy (1/2)mv2(1/2)mv2, while at point M we have the sum of kinetic and potential energies: (1/2)mv2M+mgh(1/2)mvM2+mgh. Setting these equal, and cancelling the common factor of mm yields

12v2=12v2M+gh.12v2=12vM2+gh.

In terms of components, v2=v2x0+v2y0v2=vx02+vy02, and v2M=v2x0vM2=vx02, so making these substitutions yields

12(v2x0+v2y0)=12v2x0+gh.12(vx02+vy02)=12vx02+gh.

The v2x0vx02 terms cancel, and we can solve the remaining equation for hh to obtain

h=12gv2y0.h=12gvy02.

Since vy0=v/2vy0=v/2, this can be written as

h=12g(v2)2=v28g.h=12g(v2)2=v28g.

Substituting this expression for hh in equation (1) yields the final result for the magnitude of the angular momentum:

L=mv33–√16g.

hope this s helps
please mark my answer as a brainlist one pleaseeee