Visualizing Circular Motion in a Vertical Plane

Vertical circular motion refers to the movement of an object along a circular path in a vertical plane. Unlike horizontal circular motion, it involves varying forces due to gravity acting in different directions at different points. This type of motion is common in roller coasters, pendulums, and rotating rides. Understanding how speed, tension, and gravity interact is key to analyzing vertical circular motion in physics.

1.0Definition of Vertical Circular Motion

• Vertical circular motion is the movement of an object in a circular path within a vertical plane, where forces like gravity and tension vary throughout the motion. The object's speed changes—usually faster at the bottom and slower at the top—due to the influence of gravity.
• Suppose a particle of mass m is attached to a light inextensible string of length L. The particle is moving in a vertical circle of radius L about a fixed point O. It is imparted a velocity u in the horizontal direction at the lowest point A. Let v be its velocity at point P of the circle as shown in the figure.

2.0Velocity At A Point 

Velocity At A Point P:Then apply mechanical energy conservation between point A and P(considering potential energy zero at point A)

$$\begin{gathered} \text{Total (PE + KE) at A}=\text{Total (PE + KE) at P} \\ 0+\frac{1}{2}m{u}^2=mgh+\frac{1}{2}m{v}^2 \\ \frac{1}{2}m{u}^2=mg(L-L\cos \theta )+\frac{1}{2}m{v}^2(\text{as }h=L-L\cos \theta ) \\ \text{[where }L\text{ is the length of the string]} \\ v=\sqrt{u^2-2gL(1-\cos \theta )} \end{gathered}$$

3.0Tension At A Point

Tension At A Point P

$$\begin{gathered} \text{At point P required centripetal force }=\frac{m{v}^2}{L} \\ \text{Net force towards the centre}=T-mg\cos \theta \\ \text{This net force provides required centripetal force.} \\ T-mg\cos \theta =\frac{m{v}^2}{L} \\ T=mg\cos \theta +\frac{m{v}^2}{L} \end{gathered}$$

Special Case:

$$\begin{gathered} \text{At Point A: (Bottom Point)} \\ {T}_A=\frac{m{v}_A^2}{L}+mg \\ {T}_A=\frac{m{u}^2}{L}+mg(\text{Here }\theta =0^{\circ }) \\ \text{At Point B} \\ {v}_B=\sqrt{u^2-2gL} \\ {T}_B=\frac{m{v}_B^2}{L}\Rightarrow T_B=\frac{m{u}^2}{L}-2mg(\text{Here }\theta =90^{\circ }) \\ \text{At Point C: (Top point)} \\ {v}_C=\sqrt{u^2-4gL} \\ {T}_C=\frac{m{v}_C^2}{L}-mg\Rightarrow T_C=\frac{m{u}^2}{L}-5mg(\text{Here }\theta =180^{\circ }) \\ \text{From above equation,} \\ {T}_{\text{bottom}}-T_{\text{top}}=T_A-T_C=6mg, \end{gathered}$$

4.0Condition For Looping The Loop

What should be the minimum velocity of a particle connected with a string of length L at point A so that it just performs vertical circular motion about a fixed point.The point mass will complete the circle only and only if tension is never zero (except momentarily. If at all) if tension becomes zero at any point, string will go slack and subsequently, the only force acting on the body is gravity. Hence its subsequent motion will be similar to that of a projectile.

$\text{Tension at point }P,T-mg\cos \theta =\frac{m{v}^2}{L}$

• From the equation, it is evident that tension decreases with increase in θ because cos θ is a decreasing function by decreases with height. Hence tension is minimum at the top most point i.e. $T_{\min }=T_{\text{topmost}}$.
• However if tension is momentarily zero at the highest point the body would still be able to complete the circle. Hence condition for completing the circle (or looping the loop) is,

$$\begin{gathered} T_{\min }\ge 0\text{or}{T}_{\text{top}}\ge 0 \\ {T}_{\text{top}}+mg=\frac{m{v}_{\text{top}}^2}{L} \\ {T}_{\text{top}}=\frac{m{v}_{\text{top}}^2}{L}-mg \end{gathered}$$

For Looping the Loop,

$$\begin{gathered} T_{\text{top}}\ge 0 \\ \frac{m{v}_{\text{top}}^2}{L}-mg\ge 0 \\ \frac{m{v}_{\text{top}}^2}{L}\ge mg\Rightarrow v_{\text{top}}\ge \sqrt{gL}\dots \dots \dots .\text{(1)} \\ \text{If speed at the lowest point is u, then from conservation of mechanical energy between lowest point and top most point,} \\ \frac{1}{2}m{u}^2=\frac{1}{2}m{v}_{\text{top}}^2+mg\cdot 2L\dots \dots \dots \text{(2)} \\ \text{Using equations (1) and (2),} \\ u\ge \sqrt{5gL} \\ \text{For looping the loop, velocity at lowest point must be} \\ u\ge \sqrt{5gL} \end{gathered}$$
If velocity at lowest point is just enough for looping the loop, value of various quantities at point A, B, C and D are :

Condition For Looping For Massless Rod

• In case of light rod tension at the top most point can never be zero so velocity will become zero.
• For Completing the Loop $v_L\ge \sqrt{4gR}$

5.0Condition of Oscillation

Condition of Oscillation $0<u\le \sqrt{2gL}$

$$\begin{gathered} \text{The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero.} \\ \text{(In lower half circle (A to B))} \\ Here,T-mg\cos \theta =\frac{m{v}_A^2}{L} \\ T=\frac{m{v}_A^2}{L}+mg\cos \theta \\ \text{In the lower part of the circle, when velocity becomes zero and tension is non-zero, means when v = 0, but}T\ne 0.\text{So, to make the particle oscillate in the lower half cycle, maximum possible velocity at A can be given by,} \\ \frac{1}{2}m{v}_A^2+0=mgL+0 \\ {v}_A=\sqrt{2gL}\text{(1)} \\ \text{Thus for}0<u\le \sqrt{2gL} \\ \text{The particle oscillates in the lower half of the circle } \\ (0^{\circ }<\theta \le 90^{\circ }) \end{gathered}$$

6.0Condition of Leaving The Circle

Condition of Leaving The Circle $\left(\sqrt{2gL}<u<\sqrt{5gL}\right)$

In upper half cycle (B to C)

$$\begin{gathered} Here,T+mg\cos \theta =\frac{m{v}^2}{L} \\ T=\left(\frac{m{v}^2}{L}-mg\cos \theta \right)\dots \dots \dots ..(2) \\ \text{In this part of circle tension force can be zero without having zero velocity, i.e. when}T=0,v\ne 0 \\ \text{From equation (2), it is clear that tension decreases if velocity decreases.} \\ \text{So, to complete the loop, tension force should not be zero in between B to C. Tension will be minimum at }C,i.e.T_C\ge 0\text{is the required condition.} \end{gathered}$$

$$\begin{gathered} \text{At top,}{T}_C+mg=\frac{m{v}_C^2}{L} \\ If{T}_C=0, \\ mg=\frac{m{v}_C^2}{L} \\ {v}_C^2=gL\Rightarrow v_C=\sqrt{gL} \\ \text{By COME (Between A and C),} \\ \frac{1}{2}m{v}_A^2+0 \\ \frac{1}{2}m{v}_C^2+mg(2L) \\ {v}_A^2=v_C^2+4gL\Rightarrow v_A^2=5gL\Rightarrow v_A=\sqrt{5gL} \\ \text{Therefore, if}\sqrt{2gL}<u<\sqrt{5gL},\text{the particle leaves the circle.} \\ \text{Note:}\text{After leaving the circle, the particle will follow a parabolic path.} \end{gathered}$$

7.0Solved Questions On Visualizing Circular Motion in a Vertical Plane

Q-1.A ball of mass 4 kg is attached to a 1-meter-long cord and swings in a vertical circle. If the maximum tension the cord can withstand is 183.2 N, what is the maximum speed the ball can have during the motion without breaking the cord?

Solution:

$$\begin{gathered} \text{Maximum tension}T=\frac{m{v}^2}{r}+mg \\ \text{(Tension will be maximum at lowest point)} \\ \frac{m{v}^2}{r}-T=mg\Rightarrow \frac{4v^2}{1}=183.2-4\times 9.8\Rightarrow v=6\text{m/s} \end{gathered}$$

Q-2.A pendulum bob of mass m is held such that the string is perfectly horizontal, and then it is released from rest. What will be the tension in the string when the bob reaches the lowest point in its swing?

Solution:

The situation is shown in fig. Let v be the velocity of the bob at the lowest position. In this position the P.E. of bob is converted into K.E. hence

$$\begin{gathered} mgL=\frac{1}{2}m{v}^2\Rightarrow v^2=2gL(1) \\ \text{If T be the tension in the string, then} \\ T-mg=\frac{m{v}^2}{L}\dots \dots \dots (2) \\ \text{From equation (1) and (2),} \\ T=3mg \end{gathered}$$

Q-3. A body weighing 0.4 kg is whirled in a vertical circle with a string making 2 revolutions per second. If the radius of the circle is 1.2 m. Find the tension

(a) at the top of the circle,

 (b) at the bottom of the circle.

$\text{Given:}g=10{\text{m/s}}^{2}and\pi =3.14$

$$\begin{gathered} \text{Solution:} \\ m=0.4\text{kg},T=\frac{1}{2}\text{s},r=1.2\text{m} \\ \omega =\frac{2\pi }{1/2}=4\pi \text{rad/s}=12.56\text{rad/s} \\ \text{(a) At the top of the circle}T=\frac{m{v}^2}{r}-mg=mr{\omega }^2-mg=m(r{\omega }^2-g) \\ T=0.4(1.2\times 12.56\times 12.56-9.8)\text{N}=72\text{N} \\ \text{(b) At the bottom of the circle}T=m(r{\omega }^2+g) \\ T=80\text{N} \end{gathered}$$

Q-4.Two-point masses, each mass  m is connected to a light rod of length 2l and it is free to rotate in the vertical plane as shown. Calculate the minimum horizontal velocity given to lower mass so that it completes the circular motion in the vertical plane.

Solution:

For the rod to just complete the loop velocity at the top most point is zero.Particles have the same angular velocity but different linear velocity.

$$\begin{gathered} \text{Loss of KE}=\text{Gain of PE} \\ \frac{1}{2}m{v}^2+\frac{1}{2}m{\left(\frac{v}{2}\right)}^2=mg(2l)+mg(4l) \\ v=4\sqrt{\frac{3gl}{5}} \end{gathered}$$

Q-5.Find minimum speed at A so that the ball can reach at point B as shown in figure. Also discuss the motion of particle when T = 0,; v = 0 simultaneously at $=90°.\theta =90^{\circ }$.            

Solution:

$$\begin{gathered} \text{From Energy Conservation ,} \\ \frac{1}{2}m{v}_A^2+0=0+mgL \\ \text{(For minimum speed at A, }(v_B=0)) \\ {v}_{\min }=\sqrt{2gL} \\ \text{At the position B, (v = 0) and (T = 0)} \\ T-mg\cos \theta =\frac{m{v}_B^2}{L}\text{……….(1)} \\ \text{(Putting }{v}_B=0\text{ and }\theta =90^{\circ }\text{ in equation)}\text{……….(2)} \\ \text{The ball will return back, motion is oscillatory.} \end{gathered}$$