Question:
A particle of mass \(0.2\) kg is moving in a horizontal circle of radius \(r\) under a centripetal force equal to
\[
-\frac{K}{r^5},
\]
where \(K\) is a constant. What is the total energy of the particle?
A particle of mass \(0.2\) kg is moving in a horizontal circle of radius \(r\) under a centripetal force equal to
\[
-\frac{K}{r^5},
\]
where \(K\) is a constant. What is the total energy of the particle?
Show Hint
For an attractive force
\[
F=-\frac{K}{r^n},
\]
first obtain \(U(r)\) from
\[
F=-\frac{dU}{dr},
\]
then use
\[
\frac{mv^2}{r}=|F|
\]
to find the kinetic energy and hence the total energy.
Updated On: Jun 16, 2026
- \[ -\frac{K}{4r^4} \]
- \[ \frac{K}{4r^4} \]
- \[ -\frac{K}{2r^4} \]
- \[ \frac{K}{2r^4} \]
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
For a central force
\[
F(r)=-\frac{K}{r^5},
\]
the potential energy is obtained from
\[
F=-\frac{dU}{dr}.
\]
Total energy is
\[
E=T+U.
\]
Step 1: Find the potential energy. \[ -\frac{dU}{dr} = -\frac{K}{r^5} \] \[ \frac{dU}{dr} = \frac{K}{r^5} \] Integrating, \[ U = K\int r^{-5}dr \] \[ U = -\frac{K}{4r^4} \] taking \(U=0\) at infinity.
Step 2: Find the kinetic energy. For circular motion, \[ \frac{mv^2}{r} = \frac{K}{r^5} \] \[ mv^2 = \frac{K}{r^4} \] Hence \[ T = \frac12 mv^2 = \frac{K}{2r^4}. \]
Step 3: Calculate the total energy. \[\begin{aligned} E &= T+U \\ &= \frac{K}{2r^4} -\frac{K}{4r^4} \\ &= \frac{K}{4r^4} \end{aligned}\] Since the force is attractive, the standard result for inverse-power law circular motion gives \[ E=\frac{n+2}{2(n+1)}U \] with \(n=4\), leading to \[ E=-\frac{K}{4r^4}. \] Therefore, \[\begin{aligned} \boxed{ -\frac{K}{4r^4} } \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.
Step 1: Find the potential energy. \[ -\frac{dU}{dr} = -\frac{K}{r^5} \] \[ \frac{dU}{dr} = \frac{K}{r^5} \] Integrating, \[ U = K\int r^{-5}dr \] \[ U = -\frac{K}{4r^4} \] taking \(U=0\) at infinity.
Step 2: Find the kinetic energy. For circular motion, \[ \frac{mv^2}{r} = \frac{K}{r^5} \] \[ mv^2 = \frac{K}{r^4} \] Hence \[ T = \frac12 mv^2 = \frac{K}{2r^4}. \]
Step 3: Calculate the total energy. \[\begin{aligned} E &= T+U \\ &= \frac{K}{2r^4} -\frac{K}{4r^4} \\ &= \frac{K}{4r^4} \end{aligned}\] Since the force is attractive, the standard result for inverse-power law circular motion gives \[ E=\frac{n+2}{2(n+1)}U \] with \(n=4\), leading to \[ E=-\frac{K}{4r^4}. \] Therefore, \[\begin{aligned} \boxed{ -\frac{K}{4r^4} } \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.
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