A particle of mass m is moving in a horizontal circle of radius r under centripetal force equal to \$-\cfrac{K}{{r}^{2}}\$, where K is a constant. The total energy of the particle is:

Centripetal force \$ = \frac{mv^{2}}{r} =\frac{k}{r^{2}}\$ ...(given)Kinetic Energy \$=\frac{1}{2}mv^{2} = \frac{1}{2} \frac{k}{r}\$Potential Energy \$ = -\int_{\infty }^{r}F dr \$the lower limit has been taken as\$ \infty \$ because potential energy is zero at infinty\$ = -\int_{\infty }^{r}\frac{k}{r^{2}}dr \$\$ =- k\int_{\infty }^{r}r^{-2} dr\$\$ = -k\left |\frac{r^{-1}}{-1} \right |_{\infty }^{r}\$\$ = \frac{-k}{r} \$Total energy \$ = \frac{k}{2r} - \frac{k}{r} = - \frac{k}{2r} \$

Class 11 Laws of Motion - Problems on Dynamics of Circular Motion

Q: A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?

Frequency \$f=\dfrac{14}{25}Hz\$Angular velocity \$\omega =2\pi f= \dfrac{2 \times 14\pi }{25}rad/s=3.52rad/s\$Acceleration is \$a=\omega ^{2}r=3.52^{2}\times 0.8=9.9m/s^{2}\$The direction is along the radius at every point towards the centre.

Q: A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 \$ms^{-1}\$. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is

Hence from the figure,We know that,\$T\sin\theta =\dfrac { { mv }^{ 2 } }{ r } \$ [centripetal] \$\longrightarrow \left( i \right) \$\$T\cos\theta =mg\quad \longrightarrow \left( ii \right) \$ (force)from \$(i)\$ and \$(ii))\$\$\dfrac { T\sin\theta }{ T\cos\theta } =\dfrac { { mv }^{ 2 } }{ r } \times \dfrac { 1 }{ mg } \$\$\tan\theta =\dfrac { { v }^{ 2 } }{ rg } \$Given that, radius \$=10\$ m velocity \$=10\$ m/s \$g=10m/{ s }^{ 2 }\$\$tan\theta =\dfrac { { 10 }^{ 2 } }{ 10\times 10 } =1\Rightarrow \boxed { \theta ={ 45 }^{ 0 } } \$

Q: A tube of length \$L\$ is filled completely with an incompressible liquid of mass \$M\$ and closed at both the ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity \$\omega\$. The force exerted by the liquid at the other end is

The force exerted by liquid at the other end is the centripetal force due to rotation. The case can be treated as rotation of the center of mass of the liquid at distance \$L/2\$ from the axis of rotation.Hence force exerted by liquid at the end \$=M\omega^2(\dfrac{L}{2})\$ \$=\dfrac{1}{2}M\omega^2L\$

Q: A mass of \$2\ kg\$ is whirled in a horizontal circle by means of a string at an initial speed of \$5\ rpm\$. Keeping the radius constant, the tension in the string is doubled. The new speed is nearly

\$\textbf{Step 1: Drawing Free body diagram} \$ \$\textbf{[Ref. Fig.]}\$Let \$r\$ be the radius of circle.\$\textbf{Step 2: Apply Newton's second law}\$Solving in ground frame:Applying Newton's second Law on mass (m) along centripetal direction (considering direction towards centre as Positive) \$\sum F_c=ma_c\$ \$T=m\,r\,w^2\$ \$\text{From initial condition FBD}\$ \$T=mrw^2\$ \$....(1)\$\$\text{From Final condition FBD}\$ \$2T=mrw^2_2\$ \$w^2_2=\dfrac{2T}{mr}\$ \$....(2)\$\$\textbf{Step 3: Equation solving}\$Substituting value of \$T\$ in equation \$(2)\$ \$w^2_2=\dfrac{2mr w^2}{mr}\$ \$\Rightarrow\$ \$w_2=\sqrt{2}w\$ \$\Rightarrow w_2=\sqrt{2}\times 5\,rpm\$ \$\approx 7\,rpm\$Hence, option \$C\$ is correct answer.

Q: A pendulum bob has a speed \$3\;m/s\$ while passing through its lowest position, length of the pendulum is \$0.5\;m\$ then its speed when it makes an angle of \$60^{\circ}\$ with the vertical is

\$\textbf{Step 1: Apply energy conservation}\$ In \$\Delta ABD\$, \$AB = AD \cos \theta\$ \$\Rightarrow AB = l \cos \dfrac{\pi}{3} = \dfrac{l}{2}\$ \$\Rightarrow h = l - \dfrac{l}{2} = \dfrac{l}{2}\$As there are no dissipative force energy will be conserved. \$\Rightarrow (E)_i = E_f\$ \$\Rightarrow K_1 + U_1 = K_2 + U_2\$Taking point \$C\$ is zero potential level \$\Rightarrow \dfrac{1}{2} mu^2 = \dfrac{1}{2} mv^2 + (mgh)\$\$\textbf{Step 2: Calculations}\$ \$\Rightarrow v = \sqrt{u^2 - 2gh}\$Putting given values \$v = \sqrt{(3)^2(m/s)^2 - 2 \times 9.8(m/s^2) \times \dfrac{0.5}{2}(m)} = 2 m/s\$Hence \$A\$ is the correct option.

Q: Three identical particles are joined together by a thread as shown in figure.All the three particles are moving in circular path in a horizontal plane. If the velocity of the outermost particle is \$v_0\$, then the ratio of tensions in the three sections of the string is

angular velocity \$\omega\$ is same for all\$T_c=m\omega^2\;(3\ell)\$\$T_B=T_c+m\omega^2\;(2\ell)=m\omega^2(5\ell)\$\$T_A=T_B+m\omega^2(\ell)=m\omega^2\;(6\ell)\$\$\therefore\;T_c\,\colon\,T_B\,\colon\,T_A\,\colon\,\colon\,3\,\colon\,5\,\colon\,6\$

Q: A body initially at rest and sliding along a frictionless track from a height h(as shown in the figure) just completes a vertical circle of diameter AB\$=\$D. The height h is equal to.

Let total energy is conserved,Total mechanical energy at start = Total mechanical energy at 'B'mgh = \$\dfrac{1}{2} m v_B ^2 \ + mgD\$ ...........(1)Minimum speed required to complete rotation is, \$v_b = \sqrt{rg} = \sqrt{Dg/2}\$Hence equation (1) becomes,mgh = \$\dfrac{1}{2} m \dfrac{D}{2} g\$ + mgDhence, \$h = \dfrac {5D}{4}\$

Q: Discuss the motion in vertical circle. Find an expression for the minimum velocities at the lowest point and top point. Also find tension at these points?

The motion of a Particle in Vertical Circle Whenever a body is released from a height, it travels vertically downward towards the surface of the earth. This is due to the force of gravitational attraction exerted on the body by the earth. The acceleration produced by this force is called acceleration due to gravity and is denoted by \$‘g’\$. Value of \$‘g’\$ on the surface of the earth is taken to the 9.8 \$m/s^2\$ and it is same for all the bodies. It means all bodies (whether an iron ball or a piece of paper), when dropped (u = 0) from same height should fall with the same rapidity and should take the same time to reach the earth, it is the minimum velocity given to the particle at the lowest point to complete the circle.The tendency of the string to become slack is maximum when the particle is at the topmost point of the circle.At the top, tension is given by T = \$\dfrac{mv_T^2}{R}\$ - mg, where \$v_T\$ = speed of the particle at the top. \$\dfrac{ mv_T^2}{R}\$ = \$T + mg\$For \$v_T\$ to be minimum, \$T = 0\$ \$v_T = \sqrt{gR}\$If \$V_B\$ be the critical velocity of the particle at the bottom, then from conservation of energy:\$Mg(2R)\$ + \$ \dfrac{1}{2} mv_T^2\$ = 0 + \$\dfrac{1}{2} mv_B^2\$As \$v_T = \sqrt{gR} \$ > \$2 mgR\$ + \$\dfrac{1}{2} mgR \$ =\$ \dfrac{1}{2 }mv_B^2\$so \$v_B = \sqrt{5 g R}\$Highest point \$H ( h = 2r )\$We have, \$v\$ = \$\sqrt{u^2 - 2gh}\$ \$v\$ = \$\sqrt{(\sqrt{5gr})^2 - 2g ( 2r )}\$ \$v\$ = \$\sqrt{5gr - 4gr} \$ = \$ \sqrt{gr}\$Tension at the lowest point: \$T - mg\$ = \$\dfrac{mv^2}{R}\$.........(since,v = \$\sqrt{5gr}\$)\$T = 6mg\$ Tension at highest point:\$T + mg\$ = \$\dfrac{mv^2}{R}\$.......(since, v = \$\sqrt{gr}\$)\$T = 0 \$

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Problems on Dynamics of Circular Motion

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Circular Motion - L1

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Problems on Circular Motion IV

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Important Questions

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?

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A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 $m s^{- 1}$ . A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is

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A tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity $ω$ . The force exerted by the liquid at the other end is

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A mass of $2 k g$ is whirled in a horizontal circle by means of a string at an initial speed of $5 r p m$ . Keeping the radius constant, the tension in the string is doubled. The new speed is nearly

Medium

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A pendulum bob has a speed $3 m / s$ while passing through its lowest position, length of the pendulum is $0.5 m$ then its speed when it makes an angle of $6 0^{\circ}$ with the vertical is

Medium

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A particle of mass m is moving in a horizontal circle of radius r under centripetal force equal to $- \frac{K}{r ^{2}}$ , where K is a constant. The total energy of the particle is:

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Three identical particles are joined together by a thread as shown in figure.
All the three particles are moving in circular path in a horizontal plane. If the velocity of the outermost particle is $v_{0}$ , then the ratio of tensions in the three sections of the string is

Medium

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A body initially at rest and sliding along a frictionless track from a height h(as shown in the figure) just completes a vertical circle of diameter AB $=$ D. The height h is equal to.

Medium

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Discuss the motion in vertical circle. Find an expression for the minimum velocities at the lowest point and top point. Also find tension at these points?

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A gramophone record is revolving with angular velocity . A coin is placed at a distance r from the centre of the record. The static coefficient of friction is $μ$ . The coin will revolve with the record if

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Circular Motion - L1

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?

A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 ms−1. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is

A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is

A mass of 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 rpm. Keeping the radius constant, the tension in the string is doubled. The new speed is nearly

A pendulum bob has a speed 3m/s while passing through its lowest position, length of the pendulum is 0.5m then its speed when it makes an angle of 60∘ with the vertical is

A particle of mass m is moving in a horizontal circle of radius r under centripetal force equal to −r2K​, where K is a constant. The total energy of the particle is:

Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in circular path 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

A body initially at rest and sliding along a frictionless track from a height h(as shown in the figure) just completes a vertical circle of diameter AB=D. The height h is equal to.

Discuss the motion in vertical circle. Find an expression for the minimum velocities at the lowest point and top point. Also find tension at these points?

A gramophone record is revolving with angular velocity . A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ . The coin will revolve with the record if

A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 $m s^{- 1}$ . A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is

A tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity $ω$ . The force exerted by the liquid at the other end is

A mass of $2 k g$ is whirled in a horizontal circle by means of a string at an initial speed of $5 r p m$ . Keeping the radius constant, the tension in the string is doubled. The new speed is nearly

A pendulum bob has a speed $3 m / s$ while passing through its lowest position, length of the pendulum is $0.5 m$ then its speed when it makes an angle of $6 0^{\circ}$ with the vertical is

A particle of mass m is moving in a horizontal circle of radius r under centripetal force equal to $- \frac{K}{r ^{2}}$ , where K is a constant. The total energy of the particle is:

Three identical particles are joined together by a thread as shown in figure.
All the three particles are moving in circular path in a horizontal plane. If the velocity of the outermost particle is $v_{0}$ , then the ratio of tensions in the three sections of the string is

A body initially at rest and sliding along a frictionless track from a height h(as shown in the figure) just completes a vertical circle of diameter AB $=$ D. The height h is equal to.

A gramophone record is revolving with angular velocity . A coin is placed at a distance r from the centre of the record. The static coefficient of friction is $μ$ . The coin will revolve with the record if