Question
Question: A particle of mass m moves under the action of a central force whose potential is given by \[V(r) = ...
A particle of mass m moves under the action of a central force whose potential is given by V(r)=Kr3,(K>0)
(i) For what energy and angular momentum will the orbit be a circle of radius a about the origin?
(ii) What is the period of this circular motion?
(iii) If the particle is slightly disturbed from this circular motion, What will be the period of small radial oscillations about r=a?
Solution
To solve this question we have to know about angular momentum. Angular momentum is what might be compared to straight energy. It is a significant amount in material science since it is a monitored amount—the all-out precise energy of a shut framework stays consistent.
Complete step by step answer:
In this question, the given data is , V(r)=Kr3.Now, we are going to derivative both sides with respect to r, we will get,
dv=3Kr2
Now, we can say,
ΔV=mΔu
⇒mdv=dv=3Kr2dr
PE of mass m,now, we can say, du/dr=3Kmr2
Therefore,
F=−drdu=∣F∣=3Kmr2
Or, 2Kmr2=rmv2
Now, for circular motion we know that, F=Fcentripetal
Therefore, v=3Kr3
When r is equal to a then, v=3Ka3
For circular motion we know that, the total energy is said to be,
E = K + U \\\
\Rightarrow E = \dfrac{1}{2}m{v^2} + m{v_r} \\\
\Rightarrow E = \dfrac{1}{2}m(3K{a^3}) + m(K{a^3}) \\\
Therefore, E=35mKa3
We know that angular momentum is equal to mvr. Which is equal to m3Ka3
L=(ma2)3Ka
Now, time for circulation motion is,