Non-Uniform Circular Motion under Central Force

Problem Statement

A particle of mass m moves under a central force $F = -\frac{k}{r^2}$. At position $r = R$, it has tangential velocity $v_0$. Find the angle between the velocity and acceleration vectors as a function of $r$.

Given Data

• Mass of particle: m
• Central force: $F = -\frac{k}{r^2}$ (attractive)
• Initial position: $r = R$
• Initial tangential velocity: $v_0$
• No initial radial velocity

Concepts & Approach

We use:

1. Central force properties — angular momentum conservation
2. Polar coordinates
3. Velocity and acceleration components in polar form
4. Dot product to find the angle

Step-by-Step Solution

Step 1: Angular Momentum Conservation

For a central force, angular momentum is conserved:

$$L = m r^2 \dfrac{d\theta}{dt}$$ At $r=R$, $v_0$ is purely tangential, so $$\left.\dfrac{d\theta}{dt}\right|_{r=R} = \dfrac{v_0}{R}$$ Thus $L = m R v_0$. Therefore at any $r$:

$m r^2 \dfrac{d\theta}{dt} = m R v_0$, so $\dfrac{d\theta}{dt} = \dfrac{R v_0}{r^2}$.

Step 2: Velocity Components in Polar Coordinates

Velocity:

$\mathbf{v} = \dfrac{dr}{dt}\,\hat{r} + r\dfrac{d\theta}{dt}\,\hat{\theta}$. So $v_r = \dfrac{dr}{dt}$ and $v_\theta = r\dfrac{d\theta}{dt} = \dfrac{R v_0}{r}$.

Step 3: Energy Conservation to Find Radial Velocity

Total energy:

$$E = \dfrac{1}{2} m \left[\left(\dfrac{dr}{dt}\right)^2 + r^2 \left(\dfrac{d\theta}{dt}\right)^2 \right] - \dfrac{k}{r}.$$ At $r=R$, $E = \dfrac{1}{2} m v_0^2 - \dfrac{k}{R}$.

At general $r$, substituting $r^2(d\theta/dt)^2 = (R v_0 / r)^2$:

$$\left(\dfrac{dr}{dt}\right)^2 = v_0^2 - \dfrac{R^2 v_0^2}{r^2} + \dfrac{2k}{m} \left( \dfrac{1}{r} - \dfrac{1}{R} \right).$$

Step 4: Acceleration Components in Polar Coordinates

General polar acceleration:

$\mathbf{a} = \left(\dfrac{d^2 r}{dt^2} - r\left(\dfrac{d\theta}{dt}\right)^2 \right) \hat{r} + \left( 2\dfrac{dr}{dt}\dfrac{d\theta}{dt} + r\dfrac{d^2 \theta}{dt^2} \right) \hat{\theta}$.

For central force motion, $\dfrac{d^2\theta}{dt^2} = 0$. From Newton:

$a_r = -\dfrac{k}{m r^2}$ and $a_\theta = 0$.

Step 5: Angle Between Velocity and Acceleration

The angle $\phi$ satisfies $\cos \phi = \dfrac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|\,|\mathbf{a}|}$. Since $a_\theta = 0$:

$$\mathbf{v} \cdot \mathbf{a} = v_r a_r = \dfrac{dr}{dt} \left( -\dfrac{k}{m r^2} \right)$$ Magnitudes: $|\mathbf{v}| = \sqrt{ \left( \dfrac{dr}{dt} \right)^2 + \dfrac{R^2 v_0^2}{r^2} }$, $|\mathbf{a}| = \dfrac{k}{m r^2}$. Hence:

$$\cos \phi = - \dfrac{ \dfrac{dr}{dt} }{ \sqrt{ \left( \dfrac{dr}{dt} \right)^2 + \dfrac{R^2 v_0^2}{r^2} } }.$$

Step 6: Express in Terms of $r$ Only

Using the expression for $(dr/dt)^2$ from Step 3:

$$\cos \phi = - \dfrac{ \sqrt{\, v_0^2 \left(1 - \dfrac{R^2}{r^2} \right) + \dfrac{2k}{m} \left( \dfrac{1}{r} - \dfrac{1}{R} \right) \,} }{ \sqrt{\, v_0^2 + \dfrac{2k}{m} \left( \dfrac{1}{r} - \dfrac{1}{R} \right) \,} }.$$

Final Answer

$$\boxed{ \displaystyle \cos \phi = - \frac{ \sqrt{\, v_0^2 \left(1 - \dfrac{R^2}{r^2} \right) + \dfrac{2k}{m} \left( \dfrac{1}{r} - \dfrac{1}{R} \right) \,} }{ \sqrt{\, v_0^2 + \dfrac{2k}{m} \left( \dfrac{1}{r} - \dfrac{1}{R} \right) \,} } }$$

Physical Interpretation

1. Negative cosine indicates $\phi > 90^\circ$ — acceleration has a component opposing velocity due to the attractive central force.
2. Energy dependence: the expression depends on kinetic and potential energy terms.
3. Special cases:
   • At $r = R$, $dr/dt = 0$, so $\phi = 90^\circ$.
   • As $r \to \infty$, the angle approaches a constant depending on energy/escape conditions.
   • For circular orbits, the angle remains $90^\circ$.
4. Dynamic behavior: $\phi$ changes as the particle moves radially.

Tags: #Physics #CentralForce #OrbitalMechanics #PolarCoordinates #NewtonianMechanics #ClassicalDynamics