Question
A particle is projected at time t =0 from a point P with a speed v0 at an angle of 45∘ to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point P at time t =v0g.
Answer: Option B
:
B
Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upwards direction as shown in figure. For the horizontal motion during the time 0 to t,
vx=v0cos45∘=v0√2
and x=vxt=v0√2.v0g=v20√2g
For vertical motion,
vγ=v0sin45∘−gt=v0√2−v0=1−√2√2v0
and y=(v0sin45∘)t−12gt2
y=v20√2g−v202g=v202g(√2−1)
The angular momentum of the particle at time t about the origin is
L=⃗r×⃗p=m⃗r×⃗v
=m(→ix+→jy)×(−→ivx+−→jvy)
=m(⃗kxvy−→kyvx)
=m⃗k[(v20√2g)v0√2(1−√2)−v202g(√2−1)v0√2]
=−⃗kmv302√2g
Thus, the angular momentum of the particle is mv302√2g in the negative Z-direction, i.e., perpendicular to the plane of motion, going into the plane.
Was this answer helpful ?
:
B
Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upwards direction as shown in figure. For the horizontal motion during the time 0 to t,
vx=v0cos45∘=v0√2
and x=vxt=v0√2.v0g=v20√2g
For vertical motion,
vγ=v0sin45∘−gt=v0√2−v0=1−√2√2v0
and y=(v0sin45∘)t−12gt2
y=v20√2g−v202g=v202g(√2−1)
The angular momentum of the particle at time t about the origin is
L=⃗r×⃗p=m⃗r×⃗v
=m(→ix+→jy)×(−→ivx+−→jvy)
=m(⃗kxvy−→kyvx)
=m⃗k[(v20√2g)v0√2(1−√2)−v202g(√2−1)v0√2]
=−⃗kmv302√2g
Thus, the angular momentum of the particle is mv302√2g in the negative Z-direction, i.e., perpendicular to the plane of motion, going into the plane.
Was this answer helpful ?
Submit Solution