Tag: vertical loop

Summary

This is Lecture 9 for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link. Here we will cover the concept of Motion in a Vertical Circle, and the conditions to make a full vertical loop.

1. Introduction to Non-Uniform Circular Motion

Unlike motion in a horizontal circle at constant speed, motion in a vertical circle is inherently non-uniform. Because gravity acts vertically downwards, it accelerates the particle when it moves downwards and decelerates it when it moves upwards. Consequently, both the speed ($v$) and the magnitude of the tension (or normal reaction) change continuously from point to point.

2. General Equations of Motion at Any Angle

Consider a particle of mass $m$ tied to a light, inextensible string of length $L$ moving in a vertical circle of radius $L$. Let $\theta$ be the angular displacement of the string from the downward vertical line.

At any general position $\theta$:

1. Forces acting radially inwards: Tension $T$ and the radial component of gravity $mg \cos\theta$.
2. Radial equation of motion:
   $$\Sigma F_c = T – mg \cos\theta = \frac{mv^2}{L}$$
   $$T = mg \cos\theta + \frac{mv^2}{L}$$

From this equation, we can deduce two crucial facts:

• Maximum Tension occurs at the lowest point ($\theta = 0^\circ$, $\cos 0^\circ = 1$), where both speed and the gravity component assist tension:
  $$T_{max} = mg + \frac{mv_{bottom}^2}{L}$$
• Minimum Tension occurs at the highest point ($\theta = 180^\circ$, $\cos 180^\circ = -1$):
  $$T_{min} = \frac{mv_{top}^2}{L} – mg$$

3. Conditions for Completing the Loop (The String Case)

For a particle attached to a flexible string, the string must remain taut throughout the entire circular trajectory. If the string goes slack ($T < 0$), the particle ceases to move in a circle and falls into a projectile trajectory.

Condition at the Highest Point (Top):

To successfully cross the top point without the string going slack, the minimum allowable tension at the top is zero ($T_{top} \ge 0$).

Setting $T_{top} = 0$:

$$0 = \frac{mv_{top}^2}{L} – mg \implies v_{top} = \sqrt{gL}$$

The minimum critical velocity at the top of a vertical circle for a string is $\sqrt{gL}$.

Condition at the Lowest Point (Bottom):

We can use the Law of Conservation of Mechanical Energy to relate the minimum velocity at the bottom ($v_{bottom}$) to the minimum velocity at the top ($v_{top} = \sqrt{gL}$).

Taking the lowest point as our reference level ($U = 0$):

• $E_{bottom} = \frac{1}{2}mv_{bottom}^2 + 0$
• $E_{top} = \frac{1}{2}mv_{top}^2 + mg(2L)$

Equating $E_{bottom} = E_{top}$:

$$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}m(gL) + 2mgL$$

$$\frac{1}{2}mv_{bottom}^2 = \frac{5}{2}mgL$$

$$v_{bottom} = \sqrt{5gL}$$

The minimum critical velocity at the lowest point required to complete the vertical loop is $\sqrt{5gL}$.

4. Comparison Table: String vs. Rigid Rod (or Tube)

A common JEE trap involves distinguishing between a string and a rigid rod (or a particle constrained inside a smooth vertical circular tube).

Property	Flexible String / Thread	Rigid Rod / Circular Tube
Can support compression?	No ($T$ cannot be negative)	Yes (Rod can push or pull)
Condition at the top	$T \ge 0 \implies v_{top} \ge \sqrt{gL}$	Velocity can be zero ($v_{top} \ge 0$)
Minimum velocity at bottom to loop	$v_{bottom} \ge \sqrt{5gL}$	$v_{bottom} \ge \sqrt{4gL}$

(Note for Rod: If $v_{top} = 0$, energy conservation from bottom to top gives $\frac{1}{2}mv_{bottom}^2 = mg(2L) \implies v_{bottom} = \sqrt{4gL}$).

5. Summary of Motion Regimes (String Case)

Depending on the velocity imparted at the lowest point ($v_{bottom}$), the particle falls into one of three distinct mechanical categories:

1. $v_{bottom} \ge \sqrt{5gL}$: The particle completes the full vertical loop. The string remains taut throughout.
2. $\sqrt{2gL} < v_{bottom} < \sqrt{5gL}$: The particle leaves the circular path before reaching the top. Tension becomes zero ($T=0$) somewhere between $\theta = 90^\circ$ and $\theta = 180^\circ$, and the particle enters projectile motion.
3. $v_{bottom} \le \sqrt{2gL}$: The particle oscillates back and forth like a pendulum. It never reaches the horizontal diameter ($\theta = 90^\circ$).

JEE-Mains/NEET Practice Questions

Question 1: Minimum Velocity Calculation

A stone of mass $1 \text{ kg}$ is tied to the end of a $2 \text{ m}$ long string and whirled in a vertical circle. What is the minimum speed required at the lowest point for the stone to successfully complete the circle? (Take $g = 10 \text{ m/s}^2$)

(a) $5 \text{ m/s}$

(b) $10 \text{ m/s}$

(c) $2\sqrt{5} \text{ m/s}$

(d) $\sqrt{10} \text{ m/s}$

Solution:

Using the critical velocity formula for the bottom of a vertical circle with a string:

$v_{min} = \sqrt{5gL}$

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

$v_{min} = \sqrt{5 \times 10 \times 2} = \sqrt{100} = 10 \text{ m/s}$.

Answer: (b)

Question 2: Tension Difference in a Vertical Circle

A particle of mass $m$ moves in a vertical circle of radius $R$ with a constant speed $v$. (Wait, speed is not constant in a vertical circle unless specified, but let’s read carefully: if speed is constant, $T = mg \cos\theta + \frac{mv^2}{R}$. Let’s use a standard variable speed problem instead).

Correction: A particle of mass $m$ is tied to a string of length $L$ and whirled in a vertical circle. If the tension in the string at the highest point is zero, what is the tension in the string when the particle is at the lowest point?

(a) $2mg$

(b) $4mg$

(c) $6mg$

(d) $3mg$

Solution:

1. At the highest point, $T_{top} = 0 \implies \frac{mv_{top}^2}{L} – mg = 0 \implies mv_{top}^2 = mgL$.
2. Using conservation of energy between top and bottom:
   $E_{top} = E_{bottom}$
   $\frac{1}{2}mv_{top}^2 + mg(2L) = \frac{1}{2}mv_{bottom}^2 + 0$
   $\frac{1}{2}(mgL) + 2mgL = \frac{1}{2}mv_{bottom}^2 \implies \frac{5}{2}mgL = \frac{1}{2}mv_{bottom}^2 \implies mv_{bottom}^2 = 5mgL$.
3. Tension at the lowest point:
   $T_{bottom} = mg + \frac{mv_{bottom}^2}{L} = mg + \frac{5mgL}{L} = mg + 5mg = 6mg$.
   Answer: (c)

Question 3: Projectile Departure from Circular Path

A small block slides inside a smooth vertical circular track of radius $R$. It is released from rest at the top of the track ($90^\circ$ from the bottom, i.e., horizontal level). At what angle $\theta$ (measured from the downward vertical) does the block lose contact with the track?

Solution:

1. Release height is $R$ above the center (or height $R$ above bottom if reference is bottom, wait: released from rest at the horizontal level, which is height $R$ above the bottom center? Let’s clarify: released from rest at the height of the center, i.e., $h = R$ above the lowest point).
2. Let it lose contact at angle $\theta$ from the top or vertical. Let’s measure $\theta$ from the downward vertical. At this point, the normal reaction $N = 0$.
3. Radial equation of motion when $N = 0$:
   $mg \cos\theta = \frac{mv^2}{R} \implies v^2 = gR \cos\theta$.
4. Apply energy conservation from the release point (height $R$ above bottom) to height $h = R \cos\theta$ above the bottom:
   $mgR = \frac{1}{2}mv^2 + mg(R \cos\theta)$
   Substitute $v^2 = gR \cos\theta$:
   $mgR = \frac{1}{2}m(gR \cos\theta) + mgR \cos\theta$
   $mgR = \frac{3}{2}mgR \cos\theta \implies \cos\theta = \frac{2}{3} \implies \theta = \cos^{-1}\left(\frac{2}{3}\right)$.

Question 4: Rod vs String Loop-the-Loop

A ball of mass $m$ is attached to a rigid light rod of length $L$ and rotated in a vertical circle. What is the minimum velocity that must be imparted to the ball at the lowest point so that it completes the vertical circle?

(a) $\sqrt{5gL}$

(b) $\sqrt{2gL}$

(c) $\sqrt{4gL}$

(d) $\sqrt{gL}$

Solution:

For a rigid rod, the ball can pass through the top point with zero velocity ($v_{top} = 0$) because the rod can support compression (unlike a string).

Using energy conservation between the lowest point and the highest point:

$E_{bottom} = E_{top}$

$\frac{1}{2}mv_{bottom}^2 + 0 = 0 + mg(2L)$

$\frac{1}{2}v_{bottom}^2 = 2gL \implies v_{bottom} = \sqrt{4gL}$.

Answer: (c)