# NEET Physics Two Dimensional Motion Centrifugal Force

Centrifugal Force

Category : NEET

Centrifugal Force It is an imaginary force due to incorporated effects of inertia. When a body is rotating in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangential at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body with the same velocity), the body appears to be stationary before it is released. When the body is released, it appears to B, as if it has been thrown off along the radius away from the centre by some force. In reality no force is actually seen to act on the body. In absence of any real force the body tends to continue its motion in a straight line due to its inertia. The observer A easily relates this events to be due to inertia but since the inertia of both the observer B and the body is same, the observer B cannot relate the above happening to inertia. When the centripetal force ceases to act on the body, the body leaves its circular path and continues to moves in its straight-line motion but to observer B it appears that a real force has actually acted on the body and is responsible for throwing the body radially out-words. This imaginary force is given a name to explain the effects on inertia to the observer who is sharing the circular motion of the body. This inertial force is called centrifugal force. Thus centrifugal force is a fictitious force which has significance only in a rotating frame of reference. Sample problems based on centripetal and centrifugal force Problem 106.   A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string at an initial speed of 10 r.p.m.  Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is [MP PMT 2001] (a) 5 r.p.m.        (b) 10 r.p.m.      (c) 20 r.p.m.       (d) 14 r.p.m. Solution: (a) Tension in the string $T=m\,{{\omega }^{2}}r=m4{{\pi }^{2}}{{n}^{2}}r$ $T\propto {{n}^{2}}$ or $n\propto \sqrt{T}$     [As m and r are constant] $\therefore \,\,\frac{{{n}_{2}}}{{{n}_{1}}}=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}=\sqrt{\frac{T/4}{T}}\,\,\Rightarrow \,\,{{n}_{2}}=\frac{{{n}_{1}}}{2}=\frac{10}{2}=5\,rpm$ Problem 107. A cylindrical vessel partially filled with water is rotated about its vertical central axis.  It?s surface will [RPET 2000] (a) Rise equally   (b) Rise from the sides      (c) Rise from the middle   (d) Lowered equally Solution: (b) Due to the centrifugal force. Problem 108.   A proton of mass $1.6 \times 1{{0}^{27}}kg$ goes round in a circular orbit of radius 0.10 m under a centripetal force of $4 \times 1{{0}^{13}}N$. Then the frequency of revolution of the proton is about [Kerala PMT 2002] (a) $0.08 \times 1{{0}^{8}}cycles per sec$                    (b) $4 \times 1{{0}^{8}}cycles per sec$ (c) $8 \times 1{{0}^{8}}cycles per sec$                         (d) $12 \times 1{{0}^{8}}cycles per sec$ Solution: (a) $F=4\times {{10}^{-13}}\,N$;$m=1.6\times {{10}^{-27}}\,kg$; $r=0.1\,m$ Centripetal force $F=m4{{\pi }^{2}}{{n}^{2}}r$ $\therefore \,\,n=\sqrt{\frac{F}{4m\,{{\pi }^{2}}r}}$$=8\times {{10}^{6}}\,cycles/\sec$$=0.08\times {{10}^{8}}\,cycle/\sec$. Problem 109. Three identical particles are joined together by a thread as shown in figure.  All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v0, then the ratio of tensions in the three sections of the string is [UPSEAT 2003] (a) 3 : 5 : 7        (b) 3 : 4 : 5        (c) 7 : 11 : 6       (d) 3 : 5 : 6 Solution: (d) Let the angular speed of the thread is $\omega$  For particle $C\Rightarrow {{T}_{3}}=m{{\omega }^{2}}3l$ For particle ?B? ${{T}_{2}}-{{T}_{3}}=m{{\omega }^{2}}2l\Rightarrow {{T}_{2}}=m{{\omega }^{2}}5l$ For particle ?C?  ${{T}_{1}}-{{T}_{2}}=m{{\omega }^{2}}l$$\Rightarrow$${{T}_{1}}=m{{\omega }^{2}}6l$ $\therefore \,\,\,{{T}_{3}}:{{T}_{2}}:{{T}_{1}}=3:5:6$ Problem 110.   A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle with a uniform angular velocity of 2 rad/s. The tension of the string is (in N) [KCET 1998] (a) 2      (b)        $\frac{1}{3}$   (c)         4          (d)        $\frac{1}{4}$ Solution: (c) $T=m{{\omega }^{2}}r=1\times {{(2)}^{2}}\times (1)=4\,Newton$ Problem 111.   A cord can bear a maximum force of 100 N without breaking. A body of mass 1 kg tied to one end of a cord of length 1 m is revolved in a horizontal plane. What is the maximum linear speed of the body so that the cord does not break.  (a) 10 m/s          (b) 20 m/s          (c) 25 m/s          (d) 30 m/s Solution: (a) Tension in cord appears due to centrifugal force $T=\frac{m\,{{v}^{2}}}{r}$ and for critical condition this tension will be equal to breaking force (100 N) $\therefore \,\,\frac{m\,v_{\max }^{2}}{r}=100\Rightarrow v_{\max }^{2}=\frac{100\times 1}{1}\Rightarrow {{v}_{\max }}=10\,m/s$ Problem 112. A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity ${{\omega }_{0}}$. If the length of the string and angular velocity are doubled, the tension in the string which was initially ${{T}_{0}}$ is now [AIIMS 1985] (a) ${{T}_{0}}$                       (b) ${{T}_{0}}/2$         (c) $4{{T}_{0}}$                      (d) $8{{T}_{0}}$ Solution: (d) $T=m{{\omega }^{2}}l\,\,\,\therefore \,\,\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{\omega }_{2}}}{{{\omega }_{1}}} \right)}^{2}}\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)\Rightarrow \frac{{{T}_{2}}}{{{T}_{0}}}={{\left( \frac{2\omega }{\omega } \right)}^{2}}\left( \frac{2l}{l} \right)\Rightarrow {{T}_{2}}=8{{T}_{0}}$ Problem 113.   A stone is rotated steadily in a horizontal circle with a period T by a string of length l. If the tension in the string is kept constant and l increases by 1%, what is the percentage change in T.   (a) 1%               (b) 0.5%            (c) 2%               (d) 0.25% Solution: (b) Tension $=\frac{m\,4{{\pi }^{2}}l}{{{T}^{2}}}\,\,\,\therefore \,\,l\propto {{T}^{2}}\,\,or\,\,T\propto \sqrt{l}$       [Tension and mass are constant] Percentage change in Time period $=\frac{1}{2}$ (percentage change in length)         [If % change is very small]      $=\frac{1}{2}(1%)=0.5%$ Problem 114.   If mass speed and radius of rotation of a body moving in a circular path are all increased by 50%, the necessary force required to maintain the body moving in the circular path will have to be increased by (a) 225%           (b) 125%           (c) 150%            (d) 100% Solution: (b) Centripetal force $F=\frac{m\,{{v}^{2}}}{r}$ If m, v and r are increased by 50% then let new force $F'=\frac{\left( m+\frac{m}{2} \right)\,\,{{\left( v+\frac{v}{2} \right)}^{2}}}{\left( r+\frac{r}{2} \right)}=\frac{9}{4}\frac{m\,{{v}^{2}}}{r}=\frac{9}{4}F$ Percentage increase in force $\frac{\Delta F}{F}\times 100\,\,=\frac{F'\,-F}{F}\times 100%=\frac{500}{4}%\,=125%$ Problem 115.   Two masses $m$ and M are connected by a light string that passes through a smooth hole O at the centre of a table. Mass $m$ lies on the table and M hangs vertically. $m$ is moved round in a horizontal circle with O as the centre. If $l$ is the length of the string from O to $m$ then the frequency with which $m$ should revolve so that M remains stationary is (a) $\frac{1}{2\pi }\sqrt{\frac{Mg}{m\ l}}$          (b) $\frac{1}{\pi }\ \sqrt{\frac{Mg}{m\ l}}$         (c) $\frac{1}{2\pi }\sqrt{\frac{m\ l}{Mg}}$  (d) $\frac{1}{\pi }\ \sqrt{\frac{m\ l}{Mg}}$ Solution: (a) ?m? Mass performs uniform circular motion on the table. Let n is the frequency of revolution then centrifugal force $=m\,4{{\pi }^{2}}{{n}^{2}}l$ For equilibrium this force will be equal to weight Mg $m\,4{{\pi }^{2}}{{n}^{2}}l=Mg$ $\therefore \,\,n=\frac{1}{2\pi }\sqrt{\frac{Mg}{m\,l}}$ Problem 116. A particle of mass M moves with constant speed along a circular path of radius r under the action of a force F.  Its speed is [MP PMT 2002] (a) $\sqrt{\frac{r\,F}{m}}$         (b) $\sqrt{\frac{F}{r}}$              (c) $\sqrt{F\,m\,r}$         (d) $\sqrt{\frac{F}{m\,r}}$ Solution: (a) Centripetal force $F=\frac{m\,{{v}^{2}}}{r}\therefore \,\,v=\sqrt{\frac{r\,F}{m}}$           Problem 117.   In an atom for the electron to revolve around the nucleus, the necessary centripetal force is obtained from the following force exerted by the nucleus on the electron [MP PET 2002] (a) Nuclear force             (b) Gravitational force     (c) Magnetic force           (d) Electrostatic force Solution: (d)             Problem 118. A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be [AFMC 2002] (a) Double          (b) Half             (c) 4 times          (d) $\frac{1}{4}$ times Solution: (c) $F=\frac{m\,{{v}^{2}}}{r}$ $\therefore \,F\propto {{v}^{2}}$ or $\,\,\frac{{{F}_{2}}}{{{F}_{1}}}={{\left( \frac{{{v}_{2}}}{{{v}_{1}}} \right)}^{2}}\,={{\left( \frac{2v}{v} \right)}^{2}}=4\Rightarrow {{F}_{2}}=4{{F}_{1}}$ Problem 119. A bottle of soda water is grasped by the neck and swing briskly in a vertical circle. Near which portion of the bottle do the bubbles collect (a) Near the bottom                                 (b) In the middle of the bottle (c) Near the neck                         (d) Uniformly distributed in the bottle Solution: (c) Due to the lightness of the gas bubble they feel less centrifugal force so they get collect near the neck of the bottle. They collect near the centre of circular motion i.e. near the neck of the bottle. Problem 120. A body is performing circular motion. An observer ${{O}_{1}}$ is sitting at the centre of the circle and another observer ${{O}_{2}}$ is sitting on the body. The centrifugal force is experienced by the observer (a) ${{O}_{1}}\,\,only$ (b) ${{O}_{2}}\,only$   (c) $Both\text{ }by\,{{O}_{1}}\,and\,{{O}_{2}}$          (d) None of these Solution: (b) Centrifugal force is a pseudo force, which is experienced only by that observer who is attached with the body performing circular motion.

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