1.5 Velocity and acceleration of particle in circular motion

1.1 Angular Variables
1.2 Kinematic equation for circular motion
1.3 Relation between angular and linear variables
1.4 Unit vectors along the radius and tangent
1.5 Velocity and acceleration of particle in circular motion

2.1 Centripetal force
2.2 Application of circular motion

3.1 Conditions for a particle in a circular motion

4.1 Conditions for a rod in a vertical circular motion

5.1 By friction
5.2 By banking of roads
5.3 By both friction and banking of road

For position vector of particle at point $P$, we can write,

$\begin{equation} \begin{aligned} {\text{Vector}} = \overrightarrow {OP} = \overrightarrow r \\ {\text{Magnitude}} = r \\ {\text{Unit vector}} = {\widehat e_r} \\\end{aligned} \end{equation} $

Note: As we know, vector can be written as,

(vector)=(magnitude).(unit vector)

Therefore, $$\begin{equation} \begin{aligned} \overrightarrow {OP} = \left( r \right).\left( {{{\widehat e}_r}} \right) \\ \overrightarrow r = r\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) \\\end{aligned} \end{equation} $$

For determining the velocity at point $P$, differentiate the above equation w.r.t. time $t$,

$$\begin{equation} \begin{aligned} \frac{{d\overrightarrow r }}{{dt}} = r\frac{{d\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right)}}{{dt}} \\ \overrightarrow v = r\left( {\frac{{d\cos \theta }}{{dt}}\widehat i + \frac{{d\sin \theta }}{{dt}}\widehat j} \right) \\ \overrightarrow v = r\left( { - \sin \theta \frac{{d\theta }}{{dt}}\widehat i + \cos \theta \frac{{d\theta }}{{dt}}\widehat j} \right) \\\end{aligned} \end{equation} $$ As $\left( {\frac{{d\theta }}{{dt}} = \omega } \right)$, $$\begin{equation} \begin{aligned} \overrightarrow v = r\frac{{d\theta }}{{dt}}\left( { - \sin \theta \widehat i + \cos \theta \widehat j} \right) \\ \overrightarrow v = r\omega \left( { - \sin \theta \widehat i + \cos \theta \widehat j} \right) \\\end{aligned} \end{equation} $$ As $\left( {{{\widehat e}_t} = - \sin \theta \widehat i + \cos \theta \widehat j} \right)$, $$\overrightarrow v = r\omega {\widehat e_t}$$

$\omega $: Angular velocity about point $O$ at any time $t$

$\overrightarrow v $: Translational velocity at point $P$ and it is along the unit vector ${\widehat e_t}$ with magnitude $r\omega $.

Now, the particle can undergo either uniform accelerated circular motion or circular motion with constant angular velocity.

As we know, $$\overrightarrow v = r\omega {\widehat e_t}$$

Differentiate the above equation wrt time $t$ we get,

$$\begin{equation} \begin{aligned} \frac{{d\overrightarrow v }}{{dt}} = r\frac{{d\omega {{\widehat e}_t}}}{{dt}} \\ \overrightarrow a = r\frac{{d\omega \left( { - \sin \theta \widehat i + \cos \theta \widehat j} \right)}}{{dt}} \\ \overrightarrow a = r\left( { - \frac{{d\omega \sin \theta }}{{dt}}\widehat i + \frac{{d\omega \cos \theta }}{{dt}}\widehat j} \right) \\\end{aligned} \end{equation} $$

The above differentiation is of the form, $$d(uv) = udv + vdu$$ So, $$\begin{equation} \begin{aligned} \overrightarrow a = r\left[ { - \left( {\omega .\frac{{d\sin \theta }}{{dt}} + \sin \theta .\frac{{d\omega }}{{dt}}} \right)\widehat i + \left( {\omega .\frac{{d\cos \theta }}{{dt}} + \cos \theta .\frac{{d\omega }}{{dt}}} \right)\widehat j} \right] \\ \overrightarrow a = r\left[ { - \left( {\omega \cos \theta \frac{{d\theta }}{{dt}} + \sin \theta .\alpha } \right)\widehat i + \left( {\omega .\left( { - \sin \theta \frac{{d\theta }}{{dt}}} \right) + \cos \theta .\alpha } \right)\widehat j} \right] \\ \overrightarrow a = r\left[ { - \left( {{\omega ^2}.\cos \theta + \sin \theta .\alpha } \right)\widehat i + \left( {{\omega ^2}.\left( { - \sin \theta } \right) + \cos \theta .\alpha } \right)\widehat j} \right] \\ \overrightarrow a = - r{\omega ^2}\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) + r\alpha \left( { - \sin \theta \widehat i + \cos \theta \widehat j} \right) \\ \overrightarrow a = - r{\omega ^2}{\widehat e_r} + r\alpha {\widehat e_t} \\ \overrightarrow a = {\overrightarrow a _c} + {\overrightarrow a _t} \\\end{aligned} \end{equation} $$

where,

$\overrightarrow a $: Net acceleration of point $P$ at any time $t$

${\overrightarrow a _c} = - r{\omega ^2}{\widehat e_r}$: is the centripetal acceleration with magnitude $r{\omega ^2}$ along the unit vector ${\widehat e_r}$.

The negative sign suggest that the centripetal acceleration is towards the center $O$. Centripetal acceleration is due to the circular motion.

${\overrightarrow a _t} = r\alpha {\widehat e_t}$: is the tangential acceleration with magnitude $r\alpha $ along the unit vector ${\widehat e_t}$ at point $P$.

Tangential acceleration is due to the angular acceleration in the circular motion.

Therefore, magnitude of net acceleration at any time $t$ is,

$$\begin{equation} \begin{aligned} \left| {\overrightarrow a } \right| = \sqrt {{{\left( {{{\overrightarrow a }_c}} \right)}^2} + {{\left( {{{\overrightarrow a }_t}} \right)}^2}} \\ \left| {\overrightarrow a } \right| = \sqrt {{{\left( { - r{\omega ^2}{{\widehat e}_r}} \right)}^2} + {{\left( {r\alpha {{\widehat e}_t}} \right)}^2}} \\\end{aligned} \end{equation} $$

As, $\left| {{{\widehat e}_r}} \right| = \left| {{{\widehat e}_t}} \right| = 1$,

$$\left| {\overrightarrow a } \right| = \sqrt {{{\left( {r{\omega ^2}} \right)}^2} + {{\left( {r\alpha } \right)}^2}} $$

Assume the particle is moving in a circular path with constant angular velocity $\omega $. So,

$$\frac{{d\omega }}{{dt}} = 0$$

Also, the velocity of particle is given by, $$\overrightarrow v = r\omega {\widehat e_t}$$

Differentiating the above equation wrt time $t$ we get,

$$\begin{equation} \begin{aligned} \frac{{d\overrightarrow v }}{{dt}} = r\frac{{d\omega {{\widehat e}_t}}}{{dt}} \\ \overrightarrow a = r\frac{{d\omega \left( { - \sin \theta \widehat i + \cos \theta \widehat j} \right)}}{{dt}} \\ \overrightarrow a = r\left( { - \frac{{d\omega \sin \theta }}{{dt}}\widehat i + \frac{{d\omega \cos \theta }}{{dt}}\widehat j} \right) \\ \overrightarrow a = r\left[ { - \left( {\omega .\frac{{d\sin \theta }}{{dt}} + \sin \theta .\frac{{d\omega }}{{dt}}} \right)\widehat i + \left( {\omega .\frac{{d\cos \theta }}{{dt}} + \cos \theta .\frac{{d\omega }}{{dt}}} \right)\widehat j} \right] \\\end{aligned} \end{equation} $$

As $\left( {\frac{{d\omega }}{{dt}} = 0} \right)$, So, $$\begin{equation} \begin{aligned} \overrightarrow a = r\left[ { - \omega \cos \theta \frac{{d\theta }}{{dt}}\widehat i - \omega \sin \theta \frac{{d\theta }}{{dt}}\widehat j} \right] \\ \overrightarrow a = - r\omega \frac{{d\theta }}{{dt}}\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) \\ \overrightarrow a = - r{\omega ^2}\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) \\ \overrightarrow a = - r{\omega ^2}{\widehat e_r} \\ \overrightarrow a = {\overrightarrow a _c} \\\end{aligned} \end{equation} $$

Since the particle is in the circular motion with constant velocity. So, the particle has only centripetal acceleration $\left( {{{\overrightarrow a }_c}} \right)$ towards the centre $O$ and the translational acceleration $\left( {{{\overrightarrow a }_t}} \right)$ is zero.

Therefore, the magnitude of net acceleration at any time $t$ is,

$$\begin{equation} \begin{aligned} \left| {\overrightarrow a } \right| = \left| {{{\overrightarrow a }_c}} \right| \\ \left| {\overrightarrow a } \right| = r{\omega ^2} \\\end{aligned} \end{equation} $$

Therefore, the particle in circular motion with constant angular velocity will have only centripetal acceleration.

From the above two cases we understand that circular motion is always an accelerated motion.

Question 2. A particle in rest starts to move in a uniformly acceleration circular motion of radius $2\ m$ with an angular acceleration of $1\;rad/{s^2}$. Find the centripetal, translation and net acceleration at time t=3 sec.

Solution: As we know, $\alpha = 1\;rad/{s^2},\;{\omega _0} = 0$

So, angular velocity at $t=2\sec$, $$\begin{equation} \begin{aligned} \omega = {\omega _0} + \alpha t \\ \omega = 0 + 1(2) \\ \omega = 2\;rad/s \\\end{aligned} \end{equation} $$

Centripetal acceleration: ${a_c} = r{\omega ^2} = 2{(2)^2} = 8\;m/{s^2}$

Translational acceleration: ${a_t} = r\alpha = 2(1) = 2\;m/{s^2}$

Net acceleration: $$\begin{equation} \begin{aligned} \overrightarrow a = {\overrightarrow a _c} + {\overrightarrow a _t} \\ \left| {\overrightarrow a } \right| = \sqrt {a_c^2 + a_t^2} \\ \left| {\overrightarrow a } \right| = \sqrt {{8^2} + {2^2}} \\ \left| {\overrightarrow a } \right| = 2\sqrt {17} \;m/{s^2} \\\end{aligned} \end{equation} $$