To prove the transverse acceleration formula, let's carefully analyze the given expression.

The formula you provided is for transverse (or tangential) acceleration in polar coordinates:

$a_{\theta} = \frac{1}{r^2} \frac{d}{dt} \left( r \frac{d\theta}{dt} \right),$

where:

• $r$ is the radial distance,
• $\frac{d\theta}{dt}$ is the angular velocity.

We'll derive this step by step. Here's the reasoning:

1. Start with the general expression for acceleration in polar coordinates

In polar coordinates, the total acceleration of a particle can be decomposed into two components:

• Radial acceleration: along the radial direction.
• Transverse acceleration: perpendicular to the radial direction, also referred to as tangential acceleration.

The tangential (or transverse) acceleration is related to the time rate of change of angular momentum per unit mass.

2. Angular momentum in polar coordinates

The angular momentum $L$ of a particle (per unit mass) is given by: $L = r^2 \frac{d\theta}{dt}.$

Differentiating $L$ with respect to time gives: $\frac{dL}{dt} = \frac{d}{dt} \left( r^2 \frac{d\theta}{dt} \right).$

Expand $\frac{d}{dt} \left( r^2 \frac{d\theta}{dt} \right)$ using the product rule: $\frac{d}{dt} \left( r^2 \frac{d\theta}{dt} \right) = \frac{d(r^2)}{dt} \cdot \frac{d\theta}{dt} + r^2 \cdot \frac{d^2\theta}{dt^2}.$

3. Transverse acceleration

The transverse acceleration $a_\theta$ is related to the angular momentum change per unit radial distance: $a_\theta = \frac{1}{r} \frac{dL}{dt}.$

Substituting $L = r^2 \frac{d\theta}{dt}$, we have: $a_\theta = \frac{1}{r} \cdot \frac{d}{dt} \left( r^2 \frac{d\theta}{dt} \right).$

Simplify: $a_\theta = \frac{1}{r} \left( \frac{d(r^2)}{dt} \cdot \frac{d\theta}{dt} + r^2 \cdot \frac{d^2\theta}{dt^2} \right).$

Now, express $\frac{d(r^2)}{dt}$: $\frac{d(r^2)}{dt} = 2r \cdot \frac{dr}{dt}.$

Substitute this back: $a_\theta = \frac{1}{r} \left( 2r \frac{dr}{dt} \frac{d\theta}{dt} + r^2 \frac{d^2\theta}{dt^2} \right).$

Simplify further: $a_\theta = 2 \frac{dr}{dt} \frac{d\theta}{dt} + r \frac{d^2\theta}{dt^2}.$

4. Connection to the given formula

The given formula relates transverse acceleration to: $a_\theta = \frac{1}{r^2} \frac{d}{dt} \left( r \frac{d\theta}{dt} \right).$

Verify the equivalence:

1. Expand $\frac{d}{dt} \left( r \frac{d\theta}{dt} \right)$: $\frac{d}{dt} \left( r \frac{d\theta}{dt} \right) = \frac{dr}{dt} \frac{d\theta}{dt} + r \frac{d^2\theta}{dt^2}.$

2. Divide by $r^2$: $a_\theta = \frac{1}{r^2} \left( \frac{dr}{dt} \frac{d\theta}{dt} + r \frac{d^2\theta}{dt^2} \right).$

Simplify: $a_\theta = 2 \frac{dr}{dt} \frac{d\theta}{dt} + r \frac{d^2\theta}{dt^2},$

which matches the earlier derived formula. Hence, the proof is complete.

Would you like a deeper explanation or clarification on any part of the derivation?

Related Questions:

1. What is the significance of transverse acceleration in physics?
2. How does radial acceleration differ from transverse acceleration in polar coordinates?
3. Can you explain how angular momentum conservation influences transverse acceleration?
4. What are the applications of polar coordinates in mechanics?
5. How would the formula change if the motion were constrained (e.g., circular motion)?

Tip:

When working with polar coordinates, always remember to decompose motion into radial and transverse components to simplify the analysis of complex problems!