JEE Main & Advanced Physics Two Dimensional Motion Non-Uniform Circular Motion

If the speed of the particle in a horizontal circular motion changes with respect to time, then its motion is said to be non-uniform circular motion.

Consider a particle describing a circular path of radius r with centre at O. Let at an instant the particle be at P and \[\overrightarrow{\upsilon }\] be its linear velocity and \[\overrightarrow{\omega }\] be its angular velocity.

Then,           \[\vec{\upsilon }=\vec{\omega }\times \vec{r}\]               ...(i)           

Differentiating both sides of w.r.t. time t we have

\[\frac{\overset{\to }{\mathop{d\upsilon }}\,}{dt}=\frac{\overset{\to }{\mathop{d\omega }}\,}{dt}\,\times \vec{r}\,+\,\vec{\omega }\,\times \,\frac{\overset{\to }{\mathop{dr}}\,}{dt}\]        ...(ii)

Here,  \[\frac{\overset{\to }{\mathop{dv}}\,}{dt}=\vec{a},\,\,\](Resultant acceleration)

\[\vec{a}=\vec{\alpha }\,\times \,\vec{r}\,\,\,\,\,+\,\,\,\,\vec{\omega }\,\times \,\vec{\upsilon }\]            

\[\frac{\overset{\to }{\mathop{d\omega }}\,}{dt}=\vec{\alpha }\,\,\](Angular acceleration)  

\[\vec{a}=\,\,\,\,\,\,\,{{\vec{a}}_{t}}\,\,\,\,\,+\,\,\,\,\,{{\vec{a}}_{c}}\]                ...(iii)                

\[\frac{\overset{\to }{\mathop{dr}}\,}{dt}=\vec{\upsilon }\,\] (Linear velocity)

Thus the resultant acceleration of the particle at P has two component accelerations

(1) Tangential acceleration : \[\overrightarrow{{{a}_{t}}}=\overrightarrow{\alpha }\times \overrightarrow{\,r}\] 

It acts along the tangent to the circular path at P in the plane of circular path.

According to right hand rule since \[\vec{\alpha }\] and \[\vec{r}\] are perpendicular to each other, therefore, the magnitude of tangential acceleration is given by

\[|{{\overrightarrow{a}}_{t}}|\,=\,|\overrightarrow{\alpha }\,\times \,\overrightarrow{r}|\,=\,\alpha \,r\,\sin \,{{90}^{o}}\,=\alpha \,r.\]

(2) Centripetal (Radial) acceleration : \[\overrightarrow{{{a}_{c}}}=\overrightarrow{\omega }\times \overrightarrow{v}\]

It is also called centripetal acceleration of the particle at P. It acts along the radius of the particle at P.

According to right hand rule since \[\overrightarrow{\omega }\] and \[\overrightarrow{\upsilon }\] are perpendicular to each other, therefore, the magnitude of centripetal acceleration is given by

\[|{{\vec{a}}_{c}}|\,=\,|\vec{\omega }\,\times \,\vec{\upsilon }|\,=\,\omega \,\upsilon \,\sin \,{{90}^{o}}\]= \[\omega \,\upsilon \,=\,\omega (\omega \,r)\,=\,{{\omega }^{2}}r={{\upsilon }^{2}}/r\]      

Tangential and centripetal acceleration

Note :

• Here at governs the magnitude of \[\overrightarrow{v}\] while \[{{\overrightarrow{a}}_{c}}\] its direction of motion.

(3) Force : In non-uniform circular motion the particle simultaneously possesses two forces

Centripetal force : \[{{F}_{c}}=m{{a}_{c}}=\frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}\]

Tangential force : \[{{F}_{t}}=m{{a}_{t}}\] Net force : \[{{F}_{\text{net}}}=ma\]= \[m\sqrt{a_{c}^{2}+a_{t}^{2}}\]

Note : 

• In non-uniform circular motion work done by centripetal force will be zero since \[{{\vec{F}}_{c}}\,\bot \,\vec{v}\]
• In non uniform circular motion work done by tangential force will not be zero since \[{{F}_{t}}\ne 0\]  
• Rate of work done by net force in non-uniform circular motion = rate of work done by tangential force                

i.e. \[P=\frac{dW}{dt}={{\vec{F}}_{t}}.\vec{v}\]