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

A particle of mass m=1 kg is moving along the line y=x+2 with speed 2 m/s. The angular momentum of the particle about origin O is

A
L=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
L=22(^k)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
L=22(^k)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
L=2(^k)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C L=22(^k)
Given, M=1 kg,v=2 m/s
Comparing y=x+2 with y=mx+c
{where m=tanθ= slope, c= intercept on y-axis}
m=1,c=2
tanθ=1,θ=45


Hence, the velocity in vector form
v=2cosθ^i+2sinθ^j
=2cos45^i+2cos45^j
=2^i+2^j
As we know, angular momentum L=M(r×v) remains constant on any point of the line, since the perpendicular distance of the line from the origin is constant.
For position vector (r), let us assume particle is at point (0,2)
Then, r=(00)^i+(20)^j=2^j
Hence, L=M(r×v)
L=1×[2^j×(2^i+2^j)]
L=22(^k)
[^j×^i=^k,^j×^j=0]

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon