Question

A mass $m$ moves in a horizontal smooth plane in uniform circular motion with angular frequency $\Omega$. The centripetal force is provided by a spring whose force constant is $k$. A very small radial impulse is given to the mass. Then:

• A. Angular momentum of the mass remain conserved about centre of circle.
• B. Linear momentum of the mass is conserved
• C. Angular frequency of radial oscillations is equal to $\sqrt{2\omega^2 + \frac{k}{m}}$
• D. Angular frequency of radial oscillations is equal to $\sqrt{3\omega^2 + \frac{k}{m}}$

Answer 1

Answer: (A), (D)

Angular momentum of the mass remain conserved about centre of circle. Angular frequency of radial oscillations is equal to $\sqrt{3\omega^2 + \frac{k}{m}}$

Answer 2

The initial motion is uniform circular motion with angular frequency $\Omega$ and radius $r_0$. The centripetal force is provided by the spring. Let the natural length of the spring be $r_n$. The tension in the spring is $T = k(r_0 - r_n)$.

The centripetal force is $F_c = m r_0 \Omega^2$.

In uniform circular motion, $k(r_0 - r_n) = m r_0 \Omega^2$. (1)

A very small radial impulse is given to the mass.

(A) Angular momentum of the mass remain conserved about centre of circle.

The force exerted by the spring is always directed towards the center of the circle (or away from it). This is a central force. For a central force, the torque about the center is zero ($\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times F \hat{r} = 0$). Since the net torque about the center is zero, the angular momentum about the center is conserved. The radial impulse changes the radial velocity but does not change the angular velocity instantaneously, nor does it introduce a tangential force component. The subsequent motion is under the influence of the central spring force. Therefore, the angular momentum about the center remains conserved. So, option (A) is correct.

(B) Linear momentum of the mass is conserved.

In circular motion, the velocity vector is continuously changing direction, so linear momentum is not conserved. The spring force is acting on the mass, which changes its linear momentum. Therefore, linear momentum is not conserved. So, option (B) is incorrect.

(C) and (D) Angular frequency of radial oscillations.

Let the mass move in polar coordinates $(r, \theta)$. The velocity is $\vec{v} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}$.

The angular momentum about the center is $L = m r^2 \dot{\theta}$. Since angular momentum is conserved, $L = L_0$, where $L_0$ is the initial angular momentum during uniform circular motion.

$L_0 = m r_0^2 \Omega$.

So, $m r^2 \dot{\theta} = m r_0^2 \Omega$, which gives $\dot{\theta} = \frac{r_0^2 \Omega}{r^2}$.

The potential energy of the spring is $V(r) = \frac{1}{2} k (r - r_n)^2$.

The total energy of the system is $E = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) + V(r)$.

Using conservation of angular momentum, $E = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \left(\frac{L}{m r^2}\right)^2 + \frac{1}{2} k (r - r_n)^2$.

$E = \frac{1}{2} m \dot{r}^2 + \frac{L^2}{2m r^2} + \frac{1}{2} k (r - r_n)^2$.

The radial motion is equivalent to a particle of mass $m$ moving in an effective potential $V_{eff}(r) = \frac{L^2}{2m r^2} + \frac{1}{2} k (r - r_n)^2$.

The equation of radial motion is $m \ddot{r} = - \frac{dV_{eff}}{dr}$.

$\frac{dV_{eff}}{dr} = \frac{d}{dr} \left(\frac{L^2}{2m r^2}\right) + \frac{d}{dr} \left(\frac{1}{2} k (r - r_n)^2\right) = - \frac{L^2}{m r^3} + k(r - r_n)$.

So, $m \ddot{r} = \frac{L^2}{m r^3} - k(r - r_n)$.

The equilibrium radius $r_0$ is where $\ddot{r} = 0$.

$\frac{L_0^2}{m r_0^3} - k(r_0 - r_n) = 0$.

Substituting $L_0 = m r_0^2 \Omega$, we get $\frac{(m r_0^2 \Omega)^2}{m r_0^3} - k(r_0 - r_n) = 0$, which is $m r_0 \Omega^2 = k(r_0 - r_n)$, consistent with (1).

To find the frequency of small radial oscillations about $r_0$, let $r = r_0 + x$, where $x$ is a small displacement.

The effective potential for small $x$ can be expanded around $r_0$:

$V_{eff}(r_0 + x) \approx V_{eff}(r_0) + V_{eff}'(r_0) x + \frac{1}{2} V_{eff}''(r_0) x^2 + \dots$

Since $r_0$ is an equilibrium point, $V_{eff}'(r_0) = 0$.

The potential for oscillations is approximately $\frac{1}{2} V_{eff}''(r_0) x^2$.

This corresponds to a simple harmonic oscillator with effective force constant $k_{eff} = V_{eff}''(r_0)$.

The angular frequency of oscillations is $\omega_{osc} = \sqrt{\frac{k_{eff}}{m}} = \sqrt{\frac{V_{eff}''(r_0)}{m}}$.

Let's calculate $V_{eff}''(r)$.

$V_{eff}'(r) = - \frac{L^2}{m} r^{-3} + k(r - r_n)$.

$V_{eff}''(r) = - \frac{L^2}{m} (-3 r^{-4}) + k = \frac{3L^2}{m r^4} + k$.

Now evaluate $V_{eff}''(r)$ at $r = r_0$.

$V_{eff}''(r_0) = \frac{3L_0^2}{m r_0^4} + k$.

Substitute $L_0 = m r_0^2 \Omega$:

$V_{eff}''(r_0) = \frac{3(m r_0^2 \Omega)^2}{m r_0^4} + k = \frac{3m^2 r_0^4 \Omega^2}{m r_0^4} + k = 3m \Omega^2 + k$.

The effective force constant for radial oscillations is $k_{eff} = 3m \Omega^2 + k$.

The angular frequency of radial oscillations is $\omega_{osc} = \sqrt{\frac{k_{eff}}{m}} = \sqrt{\frac{3m \Omega^2 + k}{m}} = \sqrt{3 \Omega^2 + \frac{k}{m}}$.

Comparing this with the given options:

(C) Angular frequency of radial oscillations is equal to $\sqrt{2\omega^2 + \frac{k}{m}}$. Incorrect. (Note: The question uses $\Omega$ for angular frequency of circular motion, but options use $\omega$. Assuming $\omega = \Omega$.)

(D) Angular frequency of radial oscillations is equal to $\sqrt{3\omega^2 + \frac{k}{m}}$. Correct, assuming $\omega = \Omega$.

Final check of the options based on the derived result $\omega_{osc} = \sqrt{3 \Omega^2 + \frac{k}{m}}$.

Option (A) is correct.

Option (B) is incorrect.

Option (C) uses $\sqrt{2\omega^2 + \frac{k}{m}}$. If $\omega = \Omega$, this is $\sqrt{2\Omega^2 + \frac{k}{m}}$, which is not our result.

Option (D) uses $\sqrt{3\omega^2 + \frac{k}{m}}$. If $\omega = \Omega$, this is $\sqrt{3\Omega^2 + \frac{k}{m}}$, which matches our result.