Non-Uniform Circular Motion
Category : JEE Main & Advanced
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
Centripetal acceleration | Tangential acceleration | Net acceleration | Type of motion |
\[{{a}_{c}}=0\] | \[{{a}_{t}}=0\] | \[a=0\] | Uniform translatory motion |
\[{{a}_{c}}=0\] | \[{{a}_{t}}\ne 0\] | \[a={{a}_{t}}\] | Accelerated translatory motion |
\[{{a}_{c}}\ne 0\] | \[{{a}_{t}}=0\] | \[a={{a}_{c}}\] | Uniform circular motion |
\[{{a}_{c}}\ne 0\] | \[{{a}_{t}}\ne 0\] | \[a=\sqrt{a_{c}^{2}+a_{t}^{2}}\] | Non-uniform circular motion |
Note :
(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 :
i.e. \[P=\frac{dW}{dt}={{\vec{F}}_{t}}.\vec{v}\]
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