wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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


A

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B


Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upward direction as shown in figure. For horizontal motion during the time 0 to t.

vx = v0 cos 45 = v02

and x = vxt = v02.v0g = v202g.

For vertical motion,

vy = v0 sin 45 gt = v02 v0 = (12)2v0

and y = (v0 sin 45)t12gt2

= v202g - v202g = v202g(21)

The angular momentum of the particle at time t about the origin is

L = rxp = mrxv

= m(ix + jy)x(ivx + jvy)

= m(kxvy kyvx)

= mk[(v202g)v02(1 2) v202g(2 1)v02]

= kmv3022g.

Thus, the angular momentum of the particle is mv3022g in the negative Z - direction, i.e., perpendicular to the plane of motion, going into the plane.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Intuition of Angular Momentum
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon