Vertical Circular Motion

Example of Non-Uniform Circular Motion

A particle of mass m is attached to a light and inextensible string. The other end of the string is fixed at O and the particle moves in a vertical circle of radius r is equal to the length of the string as shown in the figure.

Tension at any Point on the Vertical Loop

Consider the particle when it is at the point P and the string makes an angle θ with vertical.

Forces acting on the particle are:

T = tension in the string along its length,

And, mg = weight of the particle vertically downward.

Hence, the net radial force on the particle is
$
\begin{aligned}
F_r & =T-m g \cos \theta \\
F_r & =\frac{m v^2}{r}
\end{aligned}
$
Where $r=$ length of the string

$
\text { So, } \frac{m v^2}{r}=T-m g \cos \theta
$
Or, Tension at any point on the vertical loop

$
T=\frac{m v^2}{r}+m g \cos \theta
$
Since the speed of the particle decreases with height, hence, tension is maximum at the bottom, where $\cos \theta=1($ as $\theta=0)$.

$
T_{\max }=\frac{m v_{\text {Bottom }}^2}{r}+m g
$
Similarly,

$
T_{\min }=\frac{m v_{T o p}^2}{r}-m g
$

Velocity at any Point on the vertical Loop

If $u$ is the initial velocity imparted to the body at the lowest point then, the velocity of the body at height $h$ is given by

$
v=\sqrt{u^2-2 g h}=\sqrt{u^2-2 g r(1-\cos \theta)}
$
Velocity at the lowest point (A) for the various conditions in Vertical circular motion.
Tension in the string will not be zero at any of the points and the body will continue the circular motion.

$
u_A>\sqrt{5 g r}
$
Tension at highest point C will be zero and the body will just complete the circle.

$
u_A=\sqrt{5 g r}
$
$
\sqrt{2 g r}<u_A<\sqrt{5 g r}
$
Both velocity and tension in the string become zero between $A$ and $B$ and the particle will oscillate along a semi-circular path.

$
u_A=\sqrt{2 g r}
$
The velocity of the particle becomes zero between A and B but the tension will not be zero and the particle will oscillate about the point A .

$
u_A<\sqrt{2 g r}
$

Critical Velocity

It is the minimum velocity given to the particle at the lowest point to complete the circle.

$u_A=\sqrt{5 g r}$

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Solved Examples Based on Vertical Circular Motion

Example 1:A small bob tied at one end of a string of the length 1m is describes a vertical circle so that the maximum and minimum tension in the string is in the ratio 5:1. The velocity of the bob at the highest position is __________m/s (take g= 10 m/s²)

1) 4

2) 5

3) 6

4) 7

Solution:

Let the speed of bob at the lowest position be $V_1$ and at the highest position be $V_2$ Maximum tension is at the lowest position and minimum tension is at the highest position.

Now, using, the conservation of mechanical energy,

$\begin{aligned} & \frac{1}{2} \mathrm{mv}_1^2=\frac{1}{2} \mathrm{mv}_2^2+\mathrm{mg} 2 l \\ & \Rightarrow \mathrm{v}_1^2=\mathrm{v}_2^2+4 g l \\ & \mathrm{Now}_{\mathrm{max}}-\mathrm{mg}=\frac{\mathrm{mv}_1^2}{l} \\ & \Rightarrow \mathrm{T}_{\mathrm{max}}=\mathrm{mg}+\frac{\mathrm{mv}_1^2}{l} \\ & \& \mathrm{~T}_{\min }+\mathrm{mg}=\frac{\mathrm{mv}_2^2}{l} \\ & \Rightarrow \mathrm{T}_{\min }=\frac{\mathrm{mv}_2^2}{l}-\mathrm{mg} \\ & \text { So } \\ & \frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{5}{1} \\ & \Rightarrow \mathrm{mg}+\frac{\mathrm{mv}_1^2}{l}=\left[\frac{\mathrm{mv}_2^2}{l}-\mathrm{mg}\right] 5 \\ & \Rightarrow \mathrm{mg}+\frac{\mathrm{m}}{l}\left[\mathrm{v}_2^2+4 \mathrm{gl}\right]=\frac{5 \mathrm{mv}_2^2}{\mathrm{l}}-5 \mathrm{mg} \\ & \Rightarrow \mathrm{mg}+\frac{\mathrm{mv}_2^2}{l}+4 \mathrm{mg}_2=\frac{5 \mathrm{mv}_2^2}{l}-5 \mathrm{mg} \\ & \Rightarrow 10 \mathrm{mg}=\frac{4 \mathrm{mv}_2^2}{l} \\ & \mathrm{v}_2^2=\frac{10 \times 10 \times 1}{4} \\ & \Rightarrow \mathrm{v}_2^2=25 \\ & \Rightarrow \mathrm{v}_2=5 \mathrm{~m} / \mathrm{s}\end{aligned}$

Thus, the velocity of the bob at the highest position is 5 m/s.

Hence, the answer is the option (2).

Example 2: A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N . If the maximum speed with which the stone can revolve is $\frac{\mathrm{K}}{\pi} \mathrm{rev} . / \mathrm{min}$

The value of $K$ is : (Assume the string is massless and unstretchable)

1) 400

2) 300

3) 600

4) 800

Solution

Here tension (T) provides centripetal force

$\begin{aligned} & T=\frac{\mathrm{mV}}{\mathrm{R}}=\mathrm{MR} \omega^2 \\ & \omega=\frac{\mathrm{K}}{\pi}\left(\frac{\mathrm{rev}}{\min }\right)=\frac{\mathrm{K} \times 2 \pi \mathrm{rad}}{\pi \times 60 \mathrm{~s}} \\ & T_{\max }=80=\mathrm{mR} \omega_{\max }^2 \\ & \quad 80=\left(10^{-1}\right) \times(2) \times \frac{\mathrm{K}^2}{900} \\ & \mathrm{~K}^2=400 \times 900 \\ & \mathrm{~K}=20 \times 30=600\end{aligned}$

Hence, the answer is the option (3).
Example 3: A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

1) The same throughout the motion.

2) The minimum at the highest position of the circular path.

3) The minimum at the lowest position of the circular path.

4) The minimum when the rope is in the horizontal position.

Solution:

Tension at any general point P
$
\begin{aligned}
& T_P=T=\frac{m v^2}{R}+m g \cos \theta \\
& \text { at } \theta=0 \\
& \text { (Lowest poition) } \\
& T_{\max }=\frac{m v^2}{R}+m g \Rightarrow \text { maximum } \\
& \theta=180^{\circ}, \cos \theta=-1 \\
& \mathrm{~T}_{\min }=\frac{\mathrm{mv}^2}{\mathrm{R}}-\mathrm{mg}
\end{aligned}
$

Hence, the answer is the option (2).

Example 4: A ball is released from rest from point P of a smooth semi-spherical vessel as shown in the figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while the angular position of point Q is $\alpha$ with respect to point P. Which of the following graphs represents the correct relation between A and $\alpha$ when the ball goes from Q to R?

1)

2)

3)

4)

Solution:

Let the speed of the ball at pt. Q be V
By energy conservation,

$
\begin{aligned}
& \mathrm{TE}_{\mathrm{p}}=\mathrm{TE}_{\mathrm{Q}} \\
& \mathrm{mgR}+0=\frac{1}{2} \mathrm{mv}^2+\mathrm{mg}(\mathrm{R}-\mathrm{R} \sin \alpha) \\
& \frac{1}{2} \mathrm{mv}^2=\mathrm{mgR} \sin \alpha \\
& \mathrm{V}=\sqrt{2 g \mathrm{~g}(\sin \alpha)} \rightarrow(1)
\end{aligned}
$

$
\begin{aligned}
& \mathrm{N}-\mathrm{mg} \sin \alpha=\left(\frac{\mathrm{mv}^2}{\mathrm{R}}\right)=\mathrm{F}_{\mathrm{c}} \\
& \mathrm{N}-\mathrm{mg} \sin \alpha=2 \mathrm{mg} \sin \alpha(\text { from eqn } 1) \\
& \mathrm{N}=3 \mathrm{mg} \sin \alpha \rightarrow(2) \\
& \frac{\mathrm{F}_{\mathrm{c}}}{\mathrm{N}}=\mathrm{A}=\frac{2 \mathrm{mg} \sin \alpha}{3 \mathrm{mg} \sin \alpha}=\frac{2}{3} \rightarrow(3)
\end{aligned}
$
A is independent of $\alpha$

Hence, the answer is the option (3).

Example 5: A pendulum of length 2 m consists of a wooden bob of mass 50 g . A bullet of mass 75 g is fired towards the stationary Bob with a speed v . The bullet emerges out of the Bob with a speed $\overline{3}$ and the Bob just completes the vertical circle. The value of v is $\mathrm{ms}^{-1}$.

$
\text { (if } \left.g=10 \mathrm{~m} / \mathrm{s}^2\right) \text {. }
$

1) 10

2) 12

3) 14

4) 16

Solution

Linear momentum conservation
$
\begin{aligned}
& 75 \times \mathrm{V}=50 \times \mathrm{V}^{\prime}+75 \times \frac{\mathrm{V}}{3} \\
& \mathrm{~V}^{\prime} \times 50=2 \times \frac{75}{3} \times \mathrm{V} \\
& \mathrm{V}^{\prime}=\frac{2 \times 75}{50 \times 3} \times \mathrm{V} \\
& \quad=\frac{2 \times 3}{2 \times 3} \mathrm{~V}=\mathrm{V}
\end{aligned}
$
The bob just completes the vertical circular motion

$
\begin{aligned}
& \therefore \mathrm{V}=\sqrt{5 \mathrm{gl}} \\
& \mathrm{V}=\sqrt{5 \times 10 \times 2} \\
& \mathrm{~V}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$

Hence, the answer is the option (1).

Summary

Vertical circular motion involves an object moving along a circular path in a vertical plane, influenced by forces like gravity and tension. In such motion, the tension in the string and the velocity of the object vary at different points, being maximum at the lowest point and minimum at the highest. Understanding the critical velocity and the conditions under which the object completes the circle or deviates from the path is essential in analyzing vertical circular motion. This concept is applied in various real-life scenarios, such as pendulums, roller coasters, and spinning objects.