Circular Motion: Analysing a Mass on a String

Expert reviewed 22 November 2024 8 minute read


Review of the Basic Principles of Circular Motion

Before diving into specific scenarios, let’s review some key concepts:

  • Uniform Circular Motion: This occurs when an object moves in a circular path with constant speed.
  • Centripetal Force: The net force directed toward the centre of the circular path, causing the object to continually change direction.
  • Tension: The force exerted by a string or rope when stretched.

In the case of a mass on a string, the tension in the string provides the centripetal force necessary for circular motion.

Horizontal Circular Motion

Consider a mass attached to a string, swinging in a horizontal circle.

Force Analysis

In this scenario, two main forces act on the mass:

  • Tension (T): Directed inward along the string.
  • Weight (mg): Directed downward due to gravity.

For purely horizontal motion, the tension must be significantly greater than the weight. The centripetal force is provided entirely by the tension:

Fc=T=mv2rF_c = T = \frac{mv^2}{r}

Where:

  • FcF_c is the centripetal force
  • mm is the mass of the object
  • vv is the velocity
  • rr is the radius of the circular path

What are the key Relationships in Circular Motion?

  • Tension and Mass: TmT\propto m (directly proportional)
  • Tension and Velocity: Tv2T\propto v^2 (directly proportional to squared velocity)
  • Tension and Radius: T1rT \propto \frac{1}{r} (inversely proportional)

Practice Question 1

A 0.5 kg mass is swung in a horizontal circle with a radius of 1 m at a speed of 4 m/s. Calculate the tension in the string.

T=mv2r=0.5(4)21=8 NT = \frac{mv^2}{r} \\= \frac{0.5 \cdot (4 )^2}{1 } \\= 8 \text{ N}

Conical Pendulum: String at an Angle

When the string makes an angle with the vertical, we have a conical pendulum.

placeholder

Force Analysis

In this case, we need to consider the components of the tension:

  • Horizontal component: Tx=TsinθT_x = Tsin θ (provides centripetal force)
  • Vertical component: Ty=TcosθT_y = Tcos θ (balances weight)

Key Equations

  • Centripetal force:
mv2r=Tsinθ\frac{mv^2}{r} = T\sin\theta
  • Vertical force balance:
mg=Tcosθmg = Tcos θ
  • Combining these:
v2rg=tanθ\frac{v^2}{rg} = \tan\theta

From this, we can derive an expression for velocity:

v=rgtanθv = \sqrt{rg\tan\theta}

Practice Question 2

A 0.2 kg mass is attached to a 0.8 m string and swung so that it makes a 30° angle with the vertical. Calculate the velocity of the mass.

v=rgtanθ=0.8×9.8×tan30°2.13 m/sv = \sqrt{rg\tan\theta} \\= \sqrt{0.8 \times 9.8 \times \tan 30°} \\\approx 2.13 \text{ m/s}

Vertical Circular Motion

When a mass on a string is swung in a vertical circle, the analysis becomes more complex due to the changing direction of forces.

Force Analysis

The tension and net force vary depending on the position:

  • At the top:
Tt=mv2rmgT_t = \frac{mv^2}{r} - mg
  • At the bottom:
Tb=mv2r+mgT_b = \frac{mv^2}{r} + mg

Minimum Velocity

For the mass to complete a vertical circle, there’s a minimum velocity required at the top:

vmin=grv_{min} = \sqrt{gr}

If the velocity is less than this, the mass will fall before completing the circle.

Key Takeaways

  • In circular motion with a mass on a string, tension provides the centripetal force.
  • For horizontal motion, tension must be greater than weight.
  • In conical pendulum motion, the angle affects the required velocity.
  • Vertical circular motion involves varying tension, with a minimum velocity requirement.
  • Understanding these principles is crucial for analysing more complex rotational systems.

Return to Module 5: Advanced Mechanics