Acceleration of Particle moving in Circle

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Theorem

Let $P$ be a particle moving in a circular path $C$.

Then the acceleration of $P$ is given as:

$\mathbf a = -\dfrac {\size {\mathbf v}^2 \mathbf r} {\size {\mathbf r}^2}$

where:

$\mathbf v$ is the instantaneous velocity of $P$
$\mathbf r$ is the vector whose magnitude equals the length of the radius of $C$ and whose direction is from the center of $C$ to $P$
$\size {\, \cdot \,}$ denotes the magnitude of a vector.


Proof

We use Lagrangian mechanics to derive the result.

First we remember that $\mathbf r = \map {\mathbf r} t$ is a function of time $t$.

Let $R = \size r$ be the constant radius of the circle $C$ for clarity.

The constraint can then be written as:

$\forall t: \map f {\mathbf r} := \size {\mathbf r}^2 - R^2 = 0$

The Principle of Stationary Action then states that the first variation of the action $S$ must vanish.

$\ds \delta S = \delta \int L \rd t = 0$

The Lagrangian $L$ is given by the kinetic energy alone because the constraint force is handled implicitly by the constraint and there are no other forces acting on $P$.

$L = \dfrac 1 2 m \size {\mathbf v}^2$

To incorporate the constraint we use a scalar Lagrange multiplier $\lambda$.

Because we are dealing with functionals and not functions and the constraint is local and not global $\lambda$ is also a function of time $t$, as is $\mathbf r$:

$\lambda = \map \lambda t$

The augmented action $S^+$ is given by:

$\ds S^+ = S + \int \lambda f \d t$

We have that:

$\dfrac {\d f} {\d \mathbf r} = 2 \mathbf r$

Then:

$\ds \delta S^+ = \int \paren {-\map {\dfrac \d {\d t} } {m \mathbf v} + \map \lambda {2 \mathbf r} } \cdot \delta \mathbf r \rd t + \int f \delta \lambda \rd t = 0$

This means that both the term in the brackets as well as $f$ must vanish.

We eliminate $\lambda$ from the term in the brackets by taking its scalar product with $\mathbf r$ and simplifying $\map {\dfrac \d {\d t} } {m \mathbf v}$ to $m \mathbf a$:

$-\map {\dfrac \d {\d t} } {m \mathbf v} \cdot \mathbf r + \map \lambda {2 \mathbf r} \cdot \mathbf r = 0$

This gives:

$2 \lambda = \dfrac {m \mathbf a \cdot \mathbf r} {R^2}$

We have to replace $\mathbf a \cdot \mathbf r$.

To do this, we take the second derivative with respect to $t$ of the constraint, which must vanish too:

$\dfrac {\d^2 f} {\d t^2} = 2 \map {\dfrac \d {\d t} } {\mathbf r \cdot \mathbf v} = 2 \paren {\mathbf a \cdot \mathbf r + \size {\mathbf v}^2} = 0$

So

$2 \lambda = \dfrac {-m \size {\mathbf v}^2} {R^2}$

Plugging in $2 \lambda$ in the term with the big brackets yields:

$-m \mathbf a - \dfrac {m \size {\mathbf v}^2} {R^2} \mathbf r = 0$

Division by $m$ completes the proof.

$\blacksquare$


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