Vertical Circular Motion: A Work-Energy Approach

The Physics of Vertical Circular Motion

When an object moves in a vertical circle, such as a ball on a string, its speed is not constant. Gravity acts on the object, slowing it down on the way up and speeding it up on the way down. This means there is a tangential component of acceleration in addition to the radial (centripetal) component. This is the defining feature of non-uniform circular motion.

To analyze this motion, we must resolve forces into two perpendicular directions:

1. Radial direction: Along the radius, pointing towards the center of the circle. The net force in this direction provides the centripetal force, $F_c=m{v}^2/r$.
2. Tangential direction: Tangent to the circular path. The net force in this direction provides the tangential acceleration, $F_t=m{a}_t$, which changes the object's speed.

The following video provides an excellent conceptual breakdown of these components and why the motion must be non-uniform.

The Core Technique: Combining Force and Energy Equations

As established, to solve any problem in vertical circular motion, we need two key relationships:

1. Newton's Second Law (Radial): At any point in the path, described by an angle $\theta$ from the lowest point, we can write an equation for the net radial force. For a mass $m$ on a string of length $r$, with tension $T$ and speed $v$:

   $$T-mg\cos \theta =\frac{m{v}^2}{r}⟹T=\frac{m{v}^2}{r}+mg\cos \theta$$

   This equation links the tension $T$ to the speed $v$ at that specific point.

2. Conservation of Mechanical Energy: Since tension is always perpendicular to the velocity, it does no work. If we neglect air resistance, the only force doing work is gravity, which is a conservative force. Therefore, mechanical energy is conserved. If the particle starts at the bottom with speed $u$ and reaches a point with speed $v$ at a height $h=r(1-\cos \theta )$ above the bottom:

   $$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+mgh⟹v^2=u^2-2gr(1-\cos \theta )$$

   This equation links the speed $v$ at any point to the initial speed $u$.

By combining these two fundamental equations, we can solve for any unknown variable in the system.

The Three Regimes of Motion

Depending on the initial speed $u$ given at the lowest point, a particle attached to a string will exhibit one of three distinct behaviors. Understanding these conditions is a frequent requirement in JEE Advanced problems.

Let's summarize these crucial conditions for a particle on a string of radius $r$:

• Condition 1: Looping the Loop ($u\ge \sqrt{5gr}$)
  The particle completes the full circle. The critical requirement is that the string remains taut at the highest point, which means $T_{top}\ge 0$. This leads to a minimum required speed at the top, $v_{top}\ge \sqrt{gr}$. Using energy conservation, this translates to a minimum initial speed at the bottom of $u_{min}=\sqrt{5gr}$.

• Condition 2: Leaving the Circle ($\sqrt{2gr}<u<\sqrt{5gr}$)
  The particle moves past the horizontal position ($\theta >90^{\circ }$), but the string becomes slack ($T=0$) before it reaches the top. At the moment the string goes slack, the particle has a non-zero velocity and becomes a projectile, following a parabolic path.

• Condition 3: Oscillation ($u\le \sqrt{2gr}$)
  The particle's velocity becomes zero before the tension does ($v=0$ for $\theta \le 90^{\circ }$). It fails to reach the horizontal level and simply swings back and forth in the lower arc of the circle.

Important Variations for JEE Advanced

The standard string-and-mass problem has several common variations that you must be prepared for. The underlying physics (combining force and energy equations) remains the same, but the constraints change.

1. Particle on a Light Rigid Rod

If the mass is attached to a light rigid rod instead of a string, the rod can sustain both tension (pulling) and compression (pushing). The condition for completing the loop is simply that the particle reaches the highest point. The speed at the very top can be zero.

• Minimum speed at top: $v_{top}=0$
• Minimum speed at bottom: Using energy conservation, $\frac{1}{2}m{u}_{min}^2=mg(2r)⟹u_{min}=\sqrt{4gr}$.
  This is a lower threshold than for a string.

2. Particle Sliding on a Smooth Spherical Surface

When a particle slides on the outside of a smooth sphere, the normal force $N$ replaces tension $T$. Since the surface can only push outwards, $N$ must be non-negative. The particle leaves the surface when $N=0$. This is dynamically identical to a string going slack.

When a particle moves inside a smooth spherical shell or a loop-the-loop track, the track can provide a normal force from either side (like a bead in a hollow tube). This case is dynamically identical to the rigid rod, and the minimum speed at the bottom to complete the loop is $\sqrt{4gr}$.

The following resource sections detail these important variations.

Advanced Problem Archetypes

Let's now apply this framework to some classic JEE Advanced problems.

Example 1: Combining with Pseudo-Forces

This problem connects directly to our previous lesson.

In this problem, all the standard results apply, but you must replace $g$ with $(g+a)$. The work done by the pseudo force is accounted for in this effective gravity term.

Example 2: Finding Conditions Mid-Path

A common question asks to find the speed or angle when tension has a specific value.

Example 3: Projectile Motion After Leaving the Circle

The most advanced problems involve tracking the motion after the circular path condition is violated.

Conclusion

Today, we've dissected non-uniform circular motion in a vertical plane, a topic rich with conceptual nuances and diverse problem types. You've learned that the key to unlocking these problems is not one, but a combination of two powerful tools: Newton's Second Law applied radially, and the principle of Conservation of Mechanical Energy.

Key Takeaways:

• Vertical circular motion is non-uniform because gravity creates a tangential acceleration, causing the speed to vary.
• The analysis always involves two main equations:
  1. Force (Radial): $\sum F_r=m{v}^2/r$ relates instantaneous force (Tension or Normal) to instantaneous speed.
  2. Energy Conservation: $\Delta K=-\Delta U_g$ relates speed at two different points in the path.
• The initial speed $u$ at the bottom determines the motion's regime:
  • Oscillation: $u\le \sqrt{2gr}$
  • Leaving the Circle: $\sqrt{2gr}<u<\sqrt{5gr}$
  • Looping: $u\ge \sqrt{5gr}$ (for a string or sliding off a sphere)
• For a rigid rod or a bead inside a track, the minimum looping speed is lower: $u_{min}=\sqrt{4gr}$.

This lesson serves as a perfect bridge to our next module, "Work, Energy, Momentum & Collisions." We will begin by generalizing the application of the work-energy theorem to various position-dependent forces using integration and exploring the powerful concept of potential energy curves to analyze motion.