# NEET Physics Two Dimensional Motion Non Uniform Circular Motion

Non Uniform Circular Motion

Category : NEET

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 $|{{\vec{a}}_{c}}|\,=\,|\vec{\omega }\,\times \,\vec{\upsilon }|\,=\,\omega \,\upsilon \,\sin \,{{90}^{o}}=\omega \,\upsilon \,=\,\omega (\omega \,r)\,=\,{{\omega }^{2}}r={{\upsilon }^{2}}/r$ (3) Tangential and centripetal acceleration in different motions

 Centripetal acceleration Tangential acceleration Net acceleration Type of motion ac = 0 at = 0 a = 0 Uniform translatory motion ac = 0 at ¹ 0 a = at Accelerated translatory motion ac ¹ 0 at = 0 a = ac Uniform circular motion ac ¹ 0 at ¹ 0 $a=\sqrt{a_{c}^{2}+a_{t}^{2}}$ Non-uniform circular motion
Note: q Here at governs the magnitude of $\overrightarrow{v}$ while ${{\overrightarrow{a}}_{c}}$ its direction of motion. (4) 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 of force will not be zero since ${{F}_{t}}\ne 0$ · Rate of work done by net force in non-uniform circular = rate of work done by tangential force i.e. $P=\frac{dW}{dt}={{\vec{F}}_{t}}.\vec{v}$ Sample problems based on non-uniform circular motion Problem 142. The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered. It is given as $\operatorname{K}.E. = a{{s}^{2}}$ where $a$ is a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 2002] (a) $2a\frac{{{s}^{2}}}{R}$    (b) $2as{{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)}^{1/2}}$        (c) $2as$               (d) $2a\ \frac{{{R}^{2}}}{s}$ Solution: (b) In non-uniform circular motion two forces will work on a particle ${{F}_{c}}\,\text{and }{{F}_{t}}$ So the net force ${{F}_{Net}}=\sqrt{F_{c}^{2}+F_{t}^{2}}$        ?. (i) Centripetal force ${{F}_{c}}=\frac{m{{v}^{2}}}{R}$$=\frac{2a{{s}^{2}}}{R}$      ?. (ii)      [As kinetic energy $\frac{1}{2}m{{v}^{2}}=a{{s}^{2}}$ given] Again from: $\frac{1}{2}m{{v}^{2}}=a{{s}^{2}}\Rightarrow {{v}^{2}}=\frac{2a{{s}^{2}}}{m}\Rightarrow v=s\sqrt{\frac{2a}{m}}$ Tangential acceleration ${{a}_{t}}=\frac{dv}{dt}=\frac{dv}{ds}\,.\,\frac{ds}{dt}\,\,\,\Rightarrow \,\,\,{{a}_{t}}=\frac{d}{ds}\left[ s\sqrt{\frac{2a}{m}} \right]\,.v$ ${{a}_{t}}=v\sqrt{\frac{2a}{m}}=s\sqrt{\frac{2a}{m}}\,\sqrt{\frac{2a}{m}}=\frac{2as}{m}$ and ${{F}_{t}}=m{{a}_{t}}=2as$                  ?. (iii) Now substituting value of ${{\operatorname{F}}_{c}}\,\,and {{F}_{t}}$ in equation (i) $\therefore {{F}_{Net}}=\sqrt{{{\left( \frac{2a{{s}^{2}}}{R} \right)}^{2}}+{{\left( 2as \right)}^{2}}}=2as\,{{\left[ 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right]}^{1/2}}$ Problem 143. A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration ${{a}_{c}}$ is varying with time $t$ as ${{a}_{c}}={{k}^{2}}r{{t}^{2}}$, where $k$ is a constant. The power delivered to the particle by the forces acting on it is [IIT-JEE 1994] (a) $2\pi m{{k}^{2}}{{r}^{2}}t$        (b) $m{{k}^{2}}{{r}^{2}}t$  (c) $\frac{m{{k}^{4}}{{r}^{2}}{{t}^{5}}}{3}$ (d) Zero Solution: (b) ${{a}_{c}}={{k}^{2}}r\,{{t}^{2}}\Rightarrow \frac{{{v}^{2}}}{r}={{k}^{2}}r\,{{t}^{2}}\Rightarrow {{v}^{2}}={{k}^{2}}\,{{r}^{2}}\,{{t}^{2}}\Rightarrow \,\,v=k\,r\,t$ Tangential acceleration ${{a}_{t}}=\frac{dv}{dt}=k\,r$ As centripetal force does not work in circular motion. So power delivered by tangential force $P={{F}_{t}}v=m\,{{a}_{t}}v=m\left( kr \right)\text{ }krt=m{{k}^{2}}{{r}^{2}}t$ Problem 144.   A simple pendulum is oscillating without damping.  When the displacement of the bob is less than maximum, its acceleration vector $\vec{a}$ is correctly shown in [IIT-JEE Screening 2002] (a) (b) (c) (d) Solution: (c) ${{a}_{c}}=\text{centripetal acceleration}$, ${{a}_{t}}=\text{tangential acceleration}$, ${{a}_{N}}=\text{ net acceleration}=\text{Resultant of}\,\,{{a}_{c}}\,\text{and}\,{{a}_{t}}$ ${{\operatorname{a}}_{N}}=\sqrt{a_{c}^{2}+a_{t}^{2}}$ Problem 145. The speed of a particle moving in a circle of radius $0.1\,m$ is $v=1.0\,\,t$ where $t$ is time in second. The resultant acceleration of the particle at $t=5s$ will be (a) $10\ m/{{s}^{2}}$ (b) $100\ m/{{s}^{2}}$            (c) $250\ m/{{s}^{2}}$                        (d) $500\ m/{{s}^{2}}$ Solution: (c) $v=1.0\,t\Rightarrow {{a}_{t}}=\frac{dv}{dt}=1\,m/{{s}^{2}}$ and ${{a}_{c}}=\frac{{{v}^{2}}}{r}=\frac{{{(5)}^{2}}}{0.1}=250\,m/{{s}^{2}}$  [At $t=5\,\sec ,\,\,v=5\,m/s$] $\therefore \,\,\,{{a}_{N}}=\sqrt{a_{c}^{2}+a_{t}^{2}}=\sqrt{{{(250)}^{2}}+{{1}^{2}}}\Rightarrow {{a}_{N}}=250\,m/{{s}^{2}}$ (approx.) Problem 146. A particle moving along the circular path with a speed v and its speed increases by ?g? in one second. If the radius of the circular path be r, then the net acceleration of the particle is (a) $\frac{{{v}^{2}}}{r}+g$    (b) $\frac{{{v}^{2}}}{{{r}^{2}}}+{{g}^{2}}$         (c) ${{\left[ \frac{{{v}^{4}}}{{{r}^{2}}}+{{g}^{2}} \right]}^{\frac{1}{2}}}$               (d) ${{\left[ \frac{{{v}^{2}}}{r}+g \right]}^{\frac{1}{2}}}$ Solution: (c) ${{a}_{t}}=g$ (given) and ${{a}_{c}}=\frac{{{v}^{2}}}{r}$  and ${{a}_{N}}=\sqrt{a_{t}^{2}+a_{c}^{2}}=\sqrt{{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}+{{g}^{2}}}=\sqrt{\frac{{{v}^{4}}}{{{r}^{2}}}+{{g}^{2}}}$ Problem 147. A car is moving with speed 30 m/sec on a circular path of radius 500 m. Its speed is increasing at the rate of $2m/se{{c}^{2}}$. What is the acceleration of the car [Roorkee 1982; RPET 1996; MH CET 2002; MP PMT 2003] (a) $2 m/{{s}^{2}}$    (b) $2.7 m/{{s}^{2}}$ (c) $1.8 m/{{s}^{2}}$  (d) $9.8 m/{{s}^{2}}$ Solution: (b) ${{\operatorname{a}}_{t}}= 2\,\,m/{{s}^{2}}$ and ${{a}_{c}}=\frac{{{v}^{2}}}{r}=\frac{30\times 30}{500}=1.8\,m/{{s}^{2}}\,\,\,\therefore \,\,a=\sqrt{a_{t}^{2}+a_{c}^{2}}=\sqrt{{{2}^{2}}+{{(1.8)}^{2}}}=2.7m/{{s}^{2}}$ Problem 148. For a particle in circular motion the centripetal acceleration is [CPMT 1998] (a) Less than its tangential acceleration      (b) Equal to its tangential acceleration (c) More than its tangential acceleration     (d) May be more or less than its tangential acceleration Solution: (d) Problem 149. A particle is moving along a circular path of radius 3 meter in such a way that the distance travelled measured along the circumference is given by $S=\frac{{{t}^{2}}}{2}+\frac{{{t}^{3}}}{3}$. The acceleration of particle when $t=2\ \sec$ is (a) $1.3 m/{{s}^{2}}$ (b) $13 m/{{s}^{2}}$  (c) $3 m/{{s}^{2}}$    (d) $10 m/{{s}^{2}}$ Solution: (b) $s=\frac{{{t}^{2}}}{2}+\frac{{{t}^{3}}}{3}\Rightarrow v=\frac{ds}{dt}=t+{{t}^{2}}$ and ${{a}_{t}}=\frac{dv}{dt}=\frac{d}{dt}(t+{{t}^{2}})=1+2t$ At t = 2 sec, v = 6 m/s and ${{a}_{t}}=5\,m/{{s}^{2}}$, ${{a}_{c}}=\frac{{{v}^{2}}}{r}=\frac{36}{3}=12\,m/{{s}^{2}}$ ${{a}_{N}}=\sqrt{a_{c}^{2}+a_{t}^{2}}=\sqrt{{{(12)}^{2}}+{{(5)}^{2}}}=13\,m/{{s}^{2}}$

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