Edexcel IAL - Mechanics 3- 4.4 Vertical Circular Motion- Study notes  - New syllabus

Motion of a Particle in a Vertical Circle

When a particle moves in a vertical circle, its motion is influenced by both the constraint force (such as tension or reaction) and gravity. Unlike horizontal circular motion, the speed is generally not constant because the particle gains and loses gravitational potential energy.

Problems of this type are solved using a combination of:

• Radial (centripetal) acceleration
• Newton’s second law in the radial direction
• Conservation of mechanical energy

Radial Acceleration

At any point on the vertical circle, the radial acceleration is directed towards the centre and has magnitude

\( a = \dfrac{v^2}{r} \)

Forces Acting on the Particle

Depending on the system, the forces may include:

• Weight \( mg \), acting vertically downward
• Tension in a string, or normal reaction from a track, acting towards the centre

Newton’s second law is applied radially, taking the direction towards the centre as positive.

Key Positions on the Vertical Circle

At different points, the radial force balance changes:

• Lowest point: weight acts away from the centre
• Highest point: weight acts towards the centre

Energy Considerations

If air resistance is neglected, mechanical energy is conserved:

$\text{Kinetic energy + gravitational potential energy = constant}$

This allows the speed at different points of the circle to be related.

Condition for Complete Vertical Motion

For a particle attached to a light string, the string must remain taut throughout the motion.

At the top of the circle, the minimum condition is

\( T \geq 0 \)

The limiting case occurs when the tension is zero at the highest point.

Additional Conditions in Motion in a Vertical Circle

In problems involving a particle attached to a light string moving in a vertical circle, two important results are frequently required:

• The minimum speed at the lowest point for complete circular motion
• The difference between the maximum and minimum tension in the string

Minimum Speed at the Lowest Point

For the particle to complete a full vertical circle, the string must remain taut at all times.

At the highest point, the limiting condition occurs when the tension is zero:

\( mg = \dfrac{mv^2}{r} \)

Hence the minimum speed at the highest point is

\( v_{\text{top}} = \sqrt{gr} \)

Using conservation of energy between the lowest and highest points:

\( \dfrac{1}{2}mu^2 = \dfrac{1}{2}mv_{\text{top}}^2 + 2mgr \)

Substituting \( v_{\text{top}}^2 = gr \):

\( u^2 = 5gr \)

Therefore, the minimum speed at the lowest point is

\( \boxed{u = \sqrt{5gr}} \)

Tension at the Highest and Lowest Points

Let \( T_{\max} \) be the tension at the lowest point and \( T_{\min} \) the tension at the highest point.

At the lowest point

\( T_{\max} – mg = \dfrac{mv_{\text{bottom}}^2}{r} \)

At the highest point

\( T_{\min} + mg = \dfrac{mv_{\text{top}}^2}{r} \)

Using energy conservation:

\( v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr \)

Difference between Maximum and Minimum Tension

Subtracting the two tension equations gives

\( T_{\max} – T_{\min} = 2mg + \dfrac{m(v_{\text{bottom}}^2 – v_{\text{top}}^2)}{r} \)

Substituting \( v_{\text{bottom}}^2 – v_{\text{top}}^2 = 4gr \):

\( T_{\max} – T_{\min} = 2mg + 4mg \)

\( \boxed{T_{\max} – T_{\min} = 6mg} \)