Vertical Circular Motion

Prerequisites

Before studying this topic, review:

• Circular Motion - Centripetal force
• Conservation of Energy

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Introduction

Vertical circular motion is a classic problem combining energy conservation with circular motion dynamics. The key challenge: the speed keeps changing because gravity acts differently at various points of the circle.

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Types of Vertical Circular Motion

1. Object on a String (or Rod)

String/Rope: Can only pull, not push. Goes slack if tension becomes zero. Rod: Can push or pull. Motion continues even at low speeds.

2. Object on a Track

Inner track (like a roller coaster loop) — Normal force acts inward Outer track (ball in a bowl) — Normal force acts outward

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Analysis at Key Points

For an object of mass $m$ on a string of length $r$:

At the Lowest Point (Bottom)

Forces: Tension $T$ upward, Weight $mg$ downward

$$T - mg = \frac{mv_{bottom}^2}{r}$$ $$\boxed{T_{bottom} = mg + \frac{mv_{bottom}^2}{r}}$$

Tension is maximum here.

At the Highest Point (Top)

Forces: Tension $T$ downward, Weight $mg$ downward (both toward center)

$$T + mg = \frac{mv_{top}^2}{r}$$ $$\boxed{T_{top} = \frac{mv_{top}^2}{r} - mg}$$

Tension is minimum here.

At Any Angle θ from Vertical

$$T - mg\cos\theta = \frac{mv^2}{r}$$ $$T = mg\cos\theta + \frac{mv^2}{r}$$

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Minimum Speed Conditions

For the string to remain taut, tension must be non-negative.

Minimum Speed at Top

Setting $T_{top} = 0$:

$$0 = \frac{mv_{top}^2}{r} - mg$$ $$\boxed{v_{top}^{min} = \sqrt{gr}}$$

Minimum Speed at Bottom

Using energy conservation from bottom to top:

$$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2r)$$

Substituting $v_{top} = \sqrt{gr}$:

$$\frac{1}{2}v_{bottom}^2 = \frac{1}{2}(gr) + 2gr = \frac{5gr}{2}$$ $$\boxed{v_{bottom}^{min} = \sqrt{5gr}}$$

Interactive Demo: Visualize Vertical Circular Motion

See how speed and tension vary at different points in the vertical circle.

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Tension at Critical Speed

When launched from bottom at exactly $v = \sqrt{5gr}$:

At Bottom

$$T_{bottom} = mg + \frac{m(5gr)}{r} = mg + 5mg = \boxed{6mg}$$

At Top

$$T_{top} = \frac{m(gr)}{r} - mg = mg - mg = \boxed{0}$$

Tension Difference

$$T_{bottom} - T_{top} = 6mg$$

This is always true regardless of the actual speed (as long as the circle is completed):

$$\boxed{T_{bottom} - T_{top} = 6mg}$$

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Speed at Any Point

Using energy conservation from bottom:

$$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv^2 + mgh$$

where $h = r(1 - \cos\theta)$ (height above bottom)

$$v^2 = v_{bottom}^2 - 2gr(1 - \cos\theta)$$

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What Happens Below Critical Speed?

If $v_{bottom} < \sqrt{5gr}$:

Case 1: $v_{bottom} < \sqrt{2gr}$

Object oscillates back and forth (doesn’t reach horizontal level) - similar to SHM for small angles.

Case 2: $\sqrt{2gr} < v_{bottom} < \sqrt{5gr}$

Object rises past horizontal but string goes slack before reaching top. Object becomes a projectile.

The angle at which string goes slack:

$$\cos\theta = \frac{v_{bottom}^2 - 2gr}{3gr}$$

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Comparison: String vs Rod

Aspect	String	Rigid Rod
Can push?	No	Yes
$v_{top}^{min}$	$\sqrt{gr}$	0
$v_{bottom}^{min}$	$\sqrt{5gr}$	$\sqrt{4gr} = 2\sqrt{gr}$
Goes slack?	Yes	No

For a rigid rod, the object can be stationary at the top (rod pushes down):

$$v_{bottom}^{min} = \sqrt{2g(2r)} = 2\sqrt{gr}$$

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Worked Examples

Example 1: Critical Speed

Solution:

$$v_{bottom}^{min} = \sqrt{5gr} = \sqrt{5 \times 10 \times 1} = \sqrt{50} = \boxed{5\sqrt{2} \approx 7.07 \text{ m/s}}$$

Example 2: Tension Calculation

Solution:

At bottom:

$$T_{bottom} = mg + \frac{mv^2}{r} = 2(10) + \frac{2(64)}{0.5} = 20 + 256 = \boxed{276 \text{ N}}$$

Speed at top (energy conservation):

$$v_{top}^2 = v_{bottom}^2 - 4gr = 64 - 4(10)(0.5) = 64 - 20 = 44$$

At top:

$$T_{top} = \frac{mv_{top}^2}{r} - mg = \frac{2(44)}{0.5} - 20 = 176 - 20 = \boxed{156 \text{ N}}$$

Verify: $T_{bottom} - T_{top} = 276 - 156 = 120 = 6mg$ (correct!)

Example 3: Incomplete Circle

Solution:

Check: $\sqrt{2gr} = \sqrt{40} \approx 6.32$ m/s

Since $6 < 6.32$, ball won’t reach horizontal level. Let’s find where:

At height $h$, $v^2 = 36 - 2(10)h = 36 - 20h$

String goes slack when $T = 0$:

$$mg\cos\theta = \frac{mv^2}{r}$$

With $h = r(1-\cos\theta)$:

$$g\cos\theta = \frac{36 - 20(2)(1-\cos\theta)}{2}$$ $$20\cos\theta = 36 - 40 + 40\cos\theta$$ $$-20\cos\theta = -4$$ $$\cos\theta = 0.2$$ $$\theta = \boxed{78.5°}$$

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Practice Problems

1. A stone on a 0.8 m string is whirled in a vertical circle. Find the minimum speed at the highest point.

2. A 0.5 kg ball on a 1 m string moves with 10 m/s at the lowest point. Find tensions at bottom, top, and horizontal positions.

3. What minimum speed must a car have at the top of a circular bridge (radius 20 m) to maintain contact?

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