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Question: A particle of mass $m$ is moving in a horizontal circle of radius $r$ under a centripetal force give...

A particle of mass mm is moving in a horizontal circle of radius rr under a centripetal force given by (−Kr2)\left(\frac{-K}{r^2}\right), where KK is a constant. Then

A

the total energy of the particle is (−K2r)\left(\frac{-K}{2r}\right)

B

the kinetic energy of the particle is (Kr)\left(\frac{K}{r}\right)

C

the potential energy of the particle is (K2r)\left(\frac{K}{2r}\right)

D

the kinetic energy of the particle is (−Kr)\left(\frac{-K}{r}\right)

Answer

the total energy of the particle is (−K2r)\left(\frac{-K}{2r}\right)

Explanation

Solution

The particle is moving in a horizontal circle of radius rr under a centripetal force given by F=(−Kr2)F = \left(\frac{-K}{r^2}\right). The negative sign indicates that the force is attractive, directed towards the center of the circle. The magnitude of the centripetal force is ∣F∣=Kr2|F| = \frac{K}{r^2}.

For a particle moving in a circle of radius rr with speed vv, the centripetal force required is mv2r\frac{mv^2}{r}.
Thus, we have:
mv2r=Kr2\frac{mv^2}{r} = \frac{K}{r^2}

From this equation, we can find the kinetic energy of the particle. The kinetic energy is KE=12mv2KE = \frac{1}{2}mv^2.
Multiplying the equation by r2\frac{r}{2}:
12mv2=12Kr2r=K2r\frac{1}{2}mv^2 = \frac{1}{2} \frac{K}{r^2} r = \frac{K}{2r}
So, the kinetic energy of the particle is KE=K2rKE = \frac{K}{2r}.

Next, we need to find the potential energy associated with the force F(r)=−Kr2F(r) = -\frac{K}{r^2}. The force is conservative, and the potential energy U(r)U(r) is related to the force by F(r)=−dUdrF(r) = -\frac{dU}{dr}.
So, −Kr2=−dUdr-\frac{K}{r^2} = -\frac{dU}{dr}.
dUdr=Kr2\frac{dU}{dr} = \frac{K}{r^2}.

To find U(r)U(r), we integrate with respect to rr:
U(r)=∫Kr2dr=K∫r−2dr=K(r−1−1)+C=−Kr+CU(r) = \int \frac{K}{r^2} dr = K \int r^{-2} dr = K \left(\frac{r^{-1}}{-1}\right) + C = -\frac{K}{r} + C.
We usually choose the potential energy to be zero at infinity, i.e., U(∞)=0U(\infty) = 0.
0=−K∞+C  ⟹  0=0+C  ⟹  C=00 = -\frac{K}{\infty} + C \implies 0 = 0 + C \implies C = 0.
So, the potential energy of the particle is U(r)=−KrU(r) = -\frac{K}{r}.

The total energy EE of the particle is the sum of its kinetic energy and potential energy:
E=KE+U(r)E = KE + U(r).
E=K2r+(−Kr)=K2r−Kr=K−2K2r=−K2rE = \frac{K}{2r} + \left(-\frac{K}{r}\right) = \frac{K}{2r} - \frac{K}{r} = \frac{K - 2K}{2r} = \frac{-K}{2r}.

Therefore, the total energy of the particle is (−K2r)\left(\frac{-K}{2r}\right).