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Rotational Dynamics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsRotational Dynamics4 Marks
Question
Obtain an expression for the acceleration of a particle performing circular motion. Explain its two components.
OR
For a particle performing uniform circular motion, $\vec{v}=\vec{\omega} \times \vec{r}$. Obtain an expression for the linear acceleration of a particle performing non-uniform circular motion.
OR
In circular motion, assuming $\vec{v}=\vec{\omega} \times \vec{r}$, obtain an expression for the resultant acceleration of a particle in terms of tangential and radial components.
✓

Answer

Consider a particle moving along a circular path of constant radius $r$. If its motion is nonuniform, then its angular speed $\omega$ and linear speed $v$ both change with time.
Therefore, in general, the particle has both angular acceleration $\vec{\alpha}=\frac{d \vec{\omega}}{d t}$ and tangential acceleration $\overrightarrow{a_{\mathrm{t}}}$. $\vec{\alpha}$ has the direction of $d \vec{\omega}$, which is in the direction of $\vec{\omega}$ if $\omega$ is increasing and opposite to $\vec{\omega}$ if $\omega$ is decreasing.
At any instant, the linear velocity $\vec{v}$, angular velocity $\vec{\omega}$ and radius vector $\vec{r}$ are related by
$
\vec{v}=\vec{\omega} \times \vec{r}\ . . . {(1)}
$
The linear acceleration of the particle is
$
\begin{aligned}
\vec{a} & =\frac{d \vec{v}}{d t}\ . . . {(2)} \\
\therefore \vec{a} & =\frac{d}{d t}(\vec{\omega} \times \vec{r})=\frac{d \vec{\omega}}{d t} \times \vec{r}+\vec{\omega} \times \frac{\overrightarrow{d r}}{d t} \\
& =\vec{\alpha} \times \vec{r}+\vec{\omega} \times \vec{v} \quad\left(\because \frac{d \vec{r}}{d t}=\vec{v}\right)\ . . . {(3)}
\end{aligned}\
$
$\vec{\alpha} \times \vec{r}$ is tangential to the circular path and is in the direction of $\vec{v}$ if $\vec{\alpha}$ is in the direction of $\vec{\omega}$, and it is opposite to $\vec{v}$ if $\vec{\alpha}$ is opposite to $\vec{\omega}$. Thus, $\vec{\alpha} \times \vec{r}$ is the tangential acceleration, $\overrightarrow{a_t}$.
$
\overrightarrow{a_t}=\vec{\alpha} \times \vec{r}\ . . . {(4)}
$
In magnitude, $a_{\mathrm{t}}=\alpha r$ since $\vec{\alpha}$ is perpendicular to $\vec{r}$.
Also, $\vec{\omega} \times \vec{v}$ is along the radius towards the centre of the circle, i.e., opposite to $\vec{r}$, i.e., along $-\vec{r}$; this
acceleration is called the radial or centripetal acceleration $\overrightarrow{a_{\mathbf{r}}}$.
$\overrightarrow{a_{\mathrm{r}}}=\vec{\omega} \times \vec{v} \ldots(5)$
In magnitude, $a_r=w v$
since $\vec{\omega}$ is perpendicular to $\vec{v}$.
$
\therefore \vec{a}=\overrightarrow{a_{\mathrm{t}}}+\overrightarrow{a_{\mathrm{r}}} \ . . . {(6)}
$
This is the required expression.

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