Question

prove that linear acceleration = angular acceleration * distance of the particles from axis​

Answer 1

Consider a particle moving along the circumference of a circle of radius $\sf{r,}$ so $\sf{r}$ is the distance of the particle from the axis.

When the particle traverses an angular displacement of $\theta$ (in radian), then the distance travelled by the particle $\sf{s}$ is given by,

$\sf{\longrightarrow \dfrac{s}{2\pi r}=\dfrac{\theta}{2\pi}}$

$\sf{\longrightarrow s=r\theta}$

Differentiating wrt time $\sf{t,}$

$\sf{\longrightarrow\dfrac{ds}{dt}=\dfrac{d}{dt}[r\theta]}$

As $\sf{r}$ is constant wrt $\sf{t,}$

$\sf{\longrightarrow\dfrac{ds}{dt}=r\cdot\dfrac{d\theta}{dt}}$

$\sf{\longrightarrow v=r\omega}$

where $\sf{v}$ is linear velocity and $\omega$ is angular velocity.

Again differentiating wrt time $\sf{t,}$

$\sf{\longrightarrow\dfrac{dv}{dt}=\dfrac{d}{dt}[r\omega]}$

$\sf{\longrightarrow\dfrac{dv}{dt}=r\cdot\dfrac{d\omega}{dt}}$

$\sf{\longrightarrow\underline{\underline{a=r\alpha}}}$

where $\sf{a}$ is linear acceleration and $\sf{\alpha}$ is angular acceleration.

Hence the Proof!