# NEET Physics Two Dimensional Motion Conical Pendulum

Conical Pendulum

Category : NEET

Conical Pendulum This is the example of uniform circular motion in horizontal plane. A bob of mass m attached to a light and in-extensible string rotates in a horizontal circle of radius r with constant angular speed w about the vertical. The string makes angle $\theta$ with vertical and appears tracing the surface of a cone. So this arrangement is called conical pendulum. The force acting on the bob are tension and weight of the bob. From the figure   $T\sin \theta =\frac{m{{v}^{2}}}{r}$                          ?. (i) and $T\cos \theta =mg$                                     ?. (ii) Tension in the string: $T=mg\sqrt{1+{{\left( \frac{{{v}^{2}}}{rg} \right)}^{2}}}$ $T=\frac{mg}{\cos \theta }=\frac{mgl}{\sqrt{{{l}^{2}}-{{r}^{2}}}}$              [As $\cos \theta =\frac{h}{l}=\frac{\sqrt{{{l}^{2}}-{{r}^{2}}}}{l}$] (2) Angle of string from the vertical: $\tan \theta =\frac{{{v}^{2}}}{rg}$ (3) Linear velocity of the bob: $v=\sqrt{gr\tan \theta }$ (4) Angular velocity of the bob: $\omega =\sqrt{\frac{g}{r}\tan \theta }=\sqrt{\frac{g}{h}}=\sqrt{\frac{g}{l\cos \theta }}$ (5) Time period of revolution: ${{T}_{P}}=2\pi \sqrt{\frac{l\cos \theta }{g}}=2\pi \sqrt{\frac{h}{g}}=2\pi \sqrt{\frac{{{l}^{2}}-{{r}^{2}}}{g}}=2\pi \sqrt{\frac{r}{g\tan \theta }}$ Sample problems based on conical pendulum Problem 175. A point mass m is suspended from a light thread of length l, fixed at O, is whirled in a horizontal circle at constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass are [AMU (Med.) 2001] (a) (b) (c) (d) Solution: (c) Centrifugal force (F) works radially outward, Weight (w) works downward Tension (T) work along the string and towards the point of suspension Problem 176. A string of length L is fixed at one end and carries a mass M at the other end.  The string makes 2/p revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is [BHU 2002] (a) ML               (b) 2 ML            (c) 4 ML            (d) 16 ML Solution: (d) $T\sin \theta =M{{\omega }^{2}}R$           ..... (i) $T\,\sin \,\theta \,=\,M{{\omega }^{2}}L\sin \theta$                  ..... (ii) From (i) and (ii) $T=M{{\omega }^{2}}L$$=M4{{\pi }^{2}}{{n}^{2}}L$ $=M4{{\pi }^{2}}{{\left( \frac{2}{\pi } \right)}^{2}}L$$=16ML$ Problem 177.   A string of length $1m$ is fixed at one end and a mass of $100gm$ is attached at the other end. The string makes $2/\pi$ rev/sec around a vertical axis through the fixed point. The angle of inclination of the string with the vertical is ($g=10\ m/{{\sec }^{2}}$) (a) ${{\tan }^{-1}}\frac{5}{8}$            (b) ${{\tan }^{-1}}\frac{8}{5}$            (c) ${{\cos }^{-1}}\frac{8}{5}$            (d) ${{\cos }^{-1}}\frac{5}{8}$ Solution: (d) For the critical condition, in equilibrium $T\sin \theta =m\,{{\omega }^{2}}r$ and $T\cos \theta =mg$ $\therefore \,\,\tan \theta =\frac{{{\omega }^{2}}r}{g}$ $\Rightarrow \,\,\frac{4{{\pi }^{2}}{{n}^{2}}r}{g}=\frac{4{{\pi }^{2}}{{(2/\pi )}^{2}}\,.\,1}{10}=\frac{8}{5}$ Sample problems (Miscellaneous) Problem 178. If the frequency of the rotating platform is f and the distance of a boy from the centre is r, which is the area swept out per second by line connecting the boy to the centre (a) $\pi rf$                     (b) $2\pi rf$       (c) $\pi {{r}^{2}}f$                  (d) $2\pi {{r}^{2}}f$ Solution: (c) Area swept by line in complete revolution $=\pi {{r}^{2}}$ If frequency of rotating platform is f per second, then Area swept will be $\pi \,{{r}^{2}}f$ per second. Problem 179. Figure below shows a body of mass M moving with uniform speed v along a circle of radius R. What is the change in speed in going from ${{\operatorname{P}}_{1}}\,\,to\, {{P}_{2}}$. (a) Zero             (b) $\sqrt{2v}$  (c) $v/\sqrt{2}$  (d) $2\,v$ Solution: (a) In uniform circular motion speed remain constant.  $\therefore$ change in speed is zero. Problem 180.   In the above problem, what is change in velocity in going from ${{\operatorname{P}}_{1\,\,}}to {{P}_{2}}$ (a) Zero             (b) $\sqrt{2v}$  (c) $v/\sqrt{2}$  (d) 2 v Solution: (b) Change in velocity $=2v\sin (\theta /2)$$=2v\sin \left( \frac{90}{2} \right)$$=2v\sin 45$$=\frac{2v}{\sqrt{2}}$$=\sqrt{2}\,v$ Problem 181. In the above problem, what is the change in angular velocity in going from P1 to P2 (a) Zero             (b) $\sqrt{2}\,v/R$         (c) $v/\sqrt{2}\,R$          (d) $2\,v/R$ Solution: (a) Angular velocity remains constant, so change in angular velocity = Zero. Problem 182.   A particle of mass $m$ is fixed to one end of a light spring of force constant $k$ and unstretched length $l$. The system is rotated about the other end of the spring with an angular velocity $\omega$, in gravity free space.   The increase in length of the spring will be (a)        $\frac{m{{\omega }^{2}}l}{k}$ (b)        $\frac{m{{\omega }^{2}}l}{k-m{{\omega }^{2}}}$ (c)         $\frac{m{{\omega }^{2}}l}{k+m{{\omega }^{2}}}$   (d)        None of these Solution: (b) In the given condition elastic force will provides the required centripetal force $k\,x=m\,{{\omega }^{2}}r$ $k\,x=m\,{{\omega }^{2}}(l+x)\Rightarrow k\,x=m{{\omega }^{2}}l+m\,{{\omega }^{2}}x$ $\Rightarrow$ $x(k-m\,{{\omega }^{2}})=m\,{{\omega }^{2}}l$ $\therefore \,\,x=\frac{m\,{{\omega }^{2}}l}{k-m\,{{\omega }^{2}}}$ Problem 183. A uniform rod of mass $m$ and length $l$ rotates in a horizontal plane with an angular velocity $\omega$ about a vertical axis passing through one end. The tension in the rod at a distance $x$ from the axis is (a) $\frac{1}{2}\ m{{\omega }^{2}}x$ (b) $\frac{1}{2}\ m{{\omega }^{2}}\frac{{{x}^{2}}}{l}$                    (c) $\frac{1}{2}m{{\omega }^{2}}l\left( 1-\frac{x}{l} \right)$      (d) $\frac{1}{2}\frac{m{{\omega }^{2}}}{l}[{{l}^{2}}-{{x}^{2}}]$ Solution: (d) Let rod AB­ performs uniform circular motion about point A. We have to calculate the tension in the rod at a distance x from the axis of rotation. Let mass of the small segment at a distance x is dm So  $dT=dm\,{{\omega }^{2}}x$$=\left( \frac{m}{l} \right)\,dx\,.\,{{\omega }^{2}}x$$=\frac{m\,{{\omega }^{2}}}{l}$ [x d x] Integrating both sides $\int\limits_{x}^{l}{dT}=\frac{m\,{{\omega }^{2}}}{l}\int\limits_{x}^{l}{x\,dx}$$\Rightarrow$$T=\frac{m{{\omega }^{2}}}{l}\left[ \frac{{{x}^{2}}}{2} \right]_{x}^{l}$ $\therefore \,\,\,T=\frac{m\,{{\omega }^{2}}}{2l}\left[ \,{{l}^{2}}-{{x}^{2}} \right]$ Problem 184. A long horizontal rod has a bead which can slide along its length, and initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular motion about A with constant angular acceleration $\alpha$. If the coefficient of friction between the rod and the bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is [IIT-JEE (Screening) 2000] (a) $\sqrt{\frac{\mu }{\alpha }}$             (b) $\frac{\mu }{\sqrt{\alpha }}$             (c) $\frac{1}{\sqrt{\mu \alpha }}$       (d) Infinitesimal Solution: (a) Let the bead starts slipping after time t For critical condition Frictional force provides the centripetal force $m{{\omega }^{2}}L=\mu \,R$$=\mu \,m\times {{a}_{t}}=\mu mL\alpha$ m (at)2L = mmLa Þ $t=\sqrt{\frac{\mu }{\alpha }}$          (As w = at)           Problem 185. A smooth table is placed horizontally and an ideal spring of spring constant $k=1000\ N/m$ and unextended length of $0.5m$ has one end fixed to its centre. The other end is attached to a mass of $5kg$ which is moving in a circle with constant speed $20m/s$. Then the tension in the spring and the extension of this spring beyond its normal length are (a) $500\ N,\ 0.5\ m$      (b) $600\ N,\ 0.6\ m$      (c) $700\ N,\ 0.7\ m$      (d) $800\ N,\ 0.8\ m$ Solution: (a) $k=1000,$ $m=5\,kg,$ $l=0.5\,m,$ $v=20\,m/s$ (given) Restoring force $= kx=\frac{m\,{{v}^{2}}}{r}=\frac{m\,{{v}^{2}}}{l+x}\Rightarrow 1000\,x=\frac{5\,{{(20)}^{2}}}{0.5+x}\Rightarrow x=0.5\,\,m$ and Tension in the spring $=kx=1000\times \frac{1}{2}=500\,N$ Problem 186.   A particle describes a horizontal circle at the mouth of a funnel type vessel as shown in figure. The surface of the funnel is frictionless. The velocity $v$ of the particle in terms of $r$ and $\theta$ will be (a) $v=\sqrt{rg/\tan \ \theta }$    (b) $v=\sqrt{rg\ \tan \theta }$     (c) $v=\sqrt{rg\ \cot \theta }$      (d) $v=\sqrt{rg}/\cot \theta$ Solution: (c) For uniform circular motion of a particle $\frac{m\,{{v}^{2}}}{r}=R\cos \theta$ ?. (i) and $mg=R\sin \theta$                            ?. (ii) Dividing (i) by (ii)                        $\frac{{{v}^{2}}}{rg}=\cot \theta$$\Rightarrow \,\,v=\sqrt{rg\cot \theta }$ Problem 187. Figure shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring constant $k$ fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P [see. Fig], where the radius of the track is horizontal (a) $\sqrt{\frac{mgR}{3k}}$       (b) $\sqrt{\frac{3gR}{mk}}$       (c) $\sqrt{\frac{3mgR}{k}}$       (d) $\sqrt{\frac{3mg}{kR}}$ Solution: (c) For the given condition, centrifugal force at P should be equal to mg i.e. $\frac{mv_{P}^{2}}{R}=mg$$\therefore \,\,{{v}_{P}}=\sqrt{Rg}$ From this we can easily calculate the required velocity at the lowest point of circular track. $v_{p}^{2}=v_{L}^{2}-2gR$            (by using formula : ${{v}^{2}}={{u}^{2}}-2gh$) ${{v}_{L}}=\sqrt{v_{P}^{2}+2gR}=\sqrt{Rg+2gR}=\sqrt{3gR}$ It means the block should possess kinetic energy $=\frac{1}{2}\,mv_{L}^{2}$$=\frac{1}{2}\,m\,\times 3gR$ And by the law of conservation of energy $\frac{1}{2}k{{x}^{2}}=\frac{1}{2}\,3m\times \,g\,R\Rightarrow \,x=\sqrt{\frac{3m\,g\,R}{k}}$.

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