CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The potential energy of mass 1 kg at a distance r m from a fixed point O is given by V(r)=9r22 J. With the particle moving in a circular orbit of radius 1 m about the point O if v is the speed of the particle and L is the magnitude of its angular momentum about O, then which of the following statements is/are true?

Open in App
Solution

Given, potential energy of the particle at a distance r in a conservative field, V(r)=9r22

We know that force acting on the particle, F=dVdr

F=ddr(9r22)

F=9r (towards centre)


At r=1 m

F=9×1=9 N

From the equilibrium of force in the circular motion, we know
centripetal force will be equal to the force F,

mv2r=9 (sign removed as sign shows direction )

Here, r=1 m and mass of particle, m=1 kg
and v is the velocity of the particle as shown in figure.
Now,

1×v21=9

v=3 m/s

Also, we know thwt angular momentum,
L=r×p.......(1)
where,
p is linear momentum of the particle which is, p=mv.......(2)

From equaltion (1) and (2) we have,
L=r×mv
L=mr×v

So, magnitude of angular momentum,
L=mrvsinθ
Here, angle between r and v is 90
L=1×1×3×sin90

L=3 kg-m/s

Thus options (b) and (c) are correct answers.
Why this question?
This question test your concept of conservative force relation and its realtion with potential energy. Also useful in understanding of application the angular momentum.
It was asked in JEE advanced 2018.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Circular Motion: A Need for Speed
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon