
A particle of mass m moves under the action of a central force whose potential is given by \[V(r) = K{r^3},(K > 0)\]
(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 \[r = a?\]
Answer
552.3k+ views
Hint: 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 , \[V(r) = K{r^3}\].Now, we are going to derivative both sides with respect to r, we will get,
\[dv = 3K{r^2}\]
Now, we can say,
\[\Delta V = \dfrac{{\Delta u}}{m}\]
\[ \Rightarrow \dfrac{{dv}}{m} = dv = 3K{r^2}dr\]
PE of mass $m$,now, we can say, \[du/dr = 3Km{r^2}\]
Therefore,
\[F = - \dfrac{{du}}{{dr}} = \left| F \right| = 3Km{r^2}\]
Or, \[2Km{r^2} = \dfrac{{m{v^2}}}{r}\]
Now, for circular motion we know that, \[F = {F_{centripetal}}\]
Therefore, \[v = \sqrt {3K{r^3}} \]
When r is equal to a then, \[v = \sqrt {3K{a^3}} \]
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, \[E = \dfrac{5}{3}mK{a^3}\]
We know that angular momentum is equal to mvr. Which is equal to \[m\sqrt {3K{a^3}} \]
\[L = (m{a^2})\sqrt {3Ka} \]
Now, time for circulation motion is,
\[T = \dfrac{{2\pi r}}{v} \\
\Rightarrow T= \dfrac{{2\pi (a)}}{{\sqrt {3K{a^3}} }} \\
\therefore T= \dfrac{{2\pi }}{{\sqrt {3Ka} }}\]
This is the right answer.
Note:we have to know that, Proper MKS or SI units for precise energy are kilogram meters squared every second (kg-m2/sec). For a given article or framework segregated from outside powers, the all out rakish energy is a steady, a reality that is known as the law of preservation of precise force.
Complete step by step answer:
In this question, the given data is , \[V(r) = K{r^3}\].Now, we are going to derivative both sides with respect to r, we will get,
\[dv = 3K{r^2}\]
Now, we can say,
\[\Delta V = \dfrac{{\Delta u}}{m}\]
\[ \Rightarrow \dfrac{{dv}}{m} = dv = 3K{r^2}dr\]
PE of mass $m$,now, we can say, \[du/dr = 3Km{r^2}\]
Therefore,
\[F = - \dfrac{{du}}{{dr}} = \left| F \right| = 3Km{r^2}\]
Or, \[2Km{r^2} = \dfrac{{m{v^2}}}{r}\]
Now, for circular motion we know that, \[F = {F_{centripetal}}\]
Therefore, \[v = \sqrt {3K{r^3}} \]
When r is equal to a then, \[v = \sqrt {3K{a^3}} \]
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, \[E = \dfrac{5}{3}mK{a^3}\]
We know that angular momentum is equal to mvr. Which is equal to \[m\sqrt {3K{a^3}} \]
\[L = (m{a^2})\sqrt {3Ka} \]
Now, time for circulation motion is,
\[T = \dfrac{{2\pi r}}{v} \\
\Rightarrow T= \dfrac{{2\pi (a)}}{{\sqrt {3K{a^3}} }} \\
\therefore T= \dfrac{{2\pi }}{{\sqrt {3Ka} }}\]
This is the right answer.
Note:we have to know that, Proper MKS or SI units for precise energy are kilogram meters squared every second (kg-m2/sec). For a given article or framework segregated from outside powers, the all out rakish energy is a steady, a reality that is known as the law of preservation of precise force.
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