How to Analyse a Rotating Object Supported by Two Strings

A mass is rotating in a horizontal circle around a vertical pole. It is attached to the pole by two strings: one connected above the mass and one connected below it.

Both strings are pulling on the mass. But which one provides the centripetal force? Do their tensions add together, or do they oppose each other?

The answer is slightly more interesting than it first appears. The inward radial components of both tensions add together to provide the centripetal force. Their vertical components, however, act in opposite directions.

Getting these directions right is the key to solving the problem.

01Understand the forces acting on the object

Consider the standard two-string arrangement:

• The upper string connects the mass to an anchor above it.
• The lower string connects the mass to an anchor below it.
• Both anchors lie on the same vertical axis.
• The mass rotates in a horizontal circle of radius \(r\).

We will assume uniform circular motion, negligible air resistance, and massless, inextensible strings. Both strings are initially taut.

This is an extension of the standard one-string conical pendulum model. The underlying physics is the same: Newton’s second law, circular acceleration, and resolving forces into components.

What are the actual forces?

There are three real forces acting on the mass:

1. Upper tension, \(T_u\), directed upwards and towards the pole.
2. Lower tension, \(T_l\), directed downwards and towards the pole.
3. Gravitational force, \(mg\), directed vertically downwards.

Here, \(m\) is the mass in kilograms, and \(g\) is the gravitational acceleration in metres per second squared.

Notice something important: both tensions have an inward horizontal component.

The lower string pulls downwards, but it also pulls towards the pole. It does not pull the mass outwards.

Resolve the tensions into components

We will measure both string angles from the vertical.

• \(\theta_u\) is the angle between the upper string and the upward vertical.
• \(\theta_l\) is the angle between the lower string and the downward vertical.

This gives the following force components:

The horizontal components of both tensions point towards the centre of the circle.

Therefore, they add together.

The radial force equation

For uniform circular motion, the resultant inward force must equal the mass multiplied by centripetal acceleration:

\[ F_{\mathrm{radial}}=\frac{mv^2}{r} \]

Here:

• \(m\) is the object’s mass (kg).
• \(v\) is its speed (m s\(^{-1}\)).
• \(r\) is the radius of its circular path (m).

Resolving the two tensions horizontally:

\[ \boxed{ T_u\sin\theta_u+T_l\sin\theta_l =\frac{mv^2}{r} } \]

This is the most important equation in this problem.

The centripetal force is not a third force acting on the mass. It is the name given to the resultant inward radial force.

The vertical force equation

The object moves in a horizontal circle, so it has no vertical acceleration.

Its upward and downward forces must therefore balance:

\[ \boxed{ T_u\cos\theta_u-T_l\cos\theta_l-mg=0 } \]

Rearranging:

\[ T_u\cos\theta_u-T_l\cos\theta_l=mg \]

Notice the difference between the two equations:

• Radial components add because both point inwards.
• Vertical components oppose each other because one points upwards and the other downwards.

You need both equations to calculate the two tensions.

02When both strings make the same angle

Let’s start with the simpler arrangement.

Suppose the upper and lower strings make the same angle, \(\theta\), with the vertical.

This means:

\[ \theta_u=\theta_l=\theta \]

Our two equations become:

Radial direction:

\[ (T_u+T_l)\sin\theta=\frac{mv^2}{r} \]

Therefore:

\[ T_u+T_l=\frac{mv^2}{r\sin\theta} \]

Vertical direction:

\[ (T_u-T_l)\cos\theta=mg \]

Therefore:

\[ T_u-T_l=\frac{mg}{\cos\theta} \]

We now have two simultaneous equations.

Adding and subtracting them gives:

\[ \boxed{ T_u=\frac{1}{2} \left( \frac{mv^2}{r\sin\theta} +\frac{mg}{\cos\theta} \right) } \]

\[ \boxed{ T_l=\frac{1}{2} \left( \frac{mv^2}{r\sin\theta} -\frac{mg}{\cos\theta} \right) } \]

These expressions are useful when the two strings have equal angles.

However, don’t memorise them as your main method. The two original force equations are more useful because they also work when the angles differ.

03What if the strings make different angles?

This is where you need to be a little more careful.

Suppose the upper string makes an angle of \(35^\circ\) with the vertical, while the lower string makes an angle of \(50^\circ\).

You can no longer combine the tensions into \(T_u+T_l\) or \(T_u-T_l\) directly.

Why?

Because the two strings have different angles, so their horizontal and vertical components are different proportions of their respective tensions.

You must keep the angles separate.

The general equations remain:

Radial direction:

\[ T_u\sin\theta_u+T_l\sin\theta_l =\frac{mv^2}{r} \]

Vertical direction:

\[ T_u\cos\theta_u-T_l\cos\theta_l=mg \]

These equations work for both equal and unequal angles.

04The important catch: A string cannot push

There is one more condition you must check when analysing two strings.

A string can pull an object, but it cannot push it.

This means tension cannot be negative:

\[ T_u\geq0,\qquad T_l\geq0 \]

If your calculation produces a negative tension, don’t simply write down the negative number as your answer.

It means your assumed arrangement, with both strings taut, is not physically possible under those conditions.

The lower string will become slack.

What is the minimum speed needed to keep both strings taut?

Let’s return to the symmetric arrangement, where both strings make an angle \(\theta\) with the vertical.

From our earlier equation:

\[ T_l=\frac{1}{2} \left( \frac{mv^2}{r\sin\theta} -\frac{mg}{\cos\theta} \right) \]

For the lower string to remain taut:

\[ T_l\geq0 \]

Therefore:

\[ \frac{mv^2}{r\sin\theta} \geq\frac{mg}{\cos\theta} \]

Cancel the mass and rearrange:

\[ v^2\geq rg\tan\theta \]

So the minimum speed is:

\[ \boxed{v_{\min}=\sqrt{rg\tan\theta}} \]

This is an interesting result.

The minimum speed does not depend on the object’s mass.

A heavier object requires greater tension, but it does not require a higher minimum speed to keep the lower string taut.

At the minimum speed, the lower string has zero tension. The upper string alone supports the object.

A contrasting example: What happens when the speed is too low?

Consider a 0.50 kg object rotating at a radius of 0.80 m. Both strings make an angle of \(30^\circ\) with the vertical.

Suppose its speed is only 1.8 m s\(^{-1}\).

The minimum speed required to keep both strings taut is:

\[ v_{\min}=\sqrt{0.80(9.8)\tan30^\circ} \]

\[ v_{\min}=2.13\text{ m s}^{-1} \]

Our object’s speed is below this value.

If we calculate the lower-string tension using the two-taut-string equations, we obtain:

\[ T_l=-0.80\text{ N} \]

This is impossible.

The negative result tells us that the lower string cannot remain taut. It becomes slack, and the object cannot maintain the original circular path and geometry under the assumed two-string model.

This is a useful diagnostic whenever you solve a problem involving multiple strings.

Always check that every calculated tension is zero or positive.

Does the minimum speed change if the angles are different?

No. There is a useful extension of the same result.

For unequal angles, the two equations are:

\[ T_u\cos\theta_u-T_l\cos\theta_l=mg \]

\[ T_u\sin\theta_u+T_l\sin\theta_l =\frac{mv^2}{r} \]

At the instant the lower string just becomes slack:

\[ T_l=0 \]

The equations reduce to:

\[ T_u\cos\theta_u=mg \]

\[ T_u\sin\theta_u=\frac{mv^2}{r} \]

Dividing the radial equation by the vertical equation:

\[ \tan\theta_u=\frac{v^2}{rg} \]

Therefore:

\[ \boxed{v_{\min}=\sqrt{rg\tan\theta_u}} \]

Notice that the lower-string angle does not appear.

The upper string determines the minimum speed at which the lower string can remain taut. The lower-string angle affects the tension once both strings are taut, but not the minimum speed.

This is a useful result to understand rather than simply memorise.

05A reliable method for solving two-string circular motion problems

When you encounter this type of question, follow this sequence.

A few common mistakes are worth watching for:

One final limitation: these equations assume constant speed. If the object is speeding up or slowing down, it also has tangential acceleration. You would then need to analyse the tangential direction separately.

06Questions and solutions

The following questions progress from standard tension calculations to more challenging problems involving changing speeds, measured forces, and the conditions required for both strings to remain taut.

Use \(g=9.8\text{ m s}^{-2}\) unless stated otherwise.

Question 1

A 0.40 kg object rotates in a horizontal circle of radius 0.60 m. It is supported by two strings, each making an angle of \(40^\circ\) with the vertical.

The object moves at a constant speed of 3.5 m s\(^{-1}\).

Calculate the tension in the upper and lower strings.

Solution 1

The upper tension is \(8.91\text{ N}\), and the lower tension is \(3.79\text{ N}\). Both strings can remain taut.

Start by calculating the required radial force:

\[ F_c=\frac{mv^2}{r} \]

\[ F_c=\frac{0.40(3.5)^2}{0.60} \]

\[ F_c=8.17\text{ N} \]

Since both strings make the same angle:

\[ (T_u+T_l)\sin40^\circ=8.17 \]

Therefore:

\[ T_u+T_l=\frac{8.17}{\sin40^\circ} \]

\[ T_u+T_l=12.71 \]

Now consider the vertical forces:

\[ (T_u-T_l)\cos40^\circ=mg \]

\[ (T_u-T_l)\cos40^\circ=0.40(9.8) \]

\[ T_u-T_l=\frac{3.92}{\cos40^\circ} \]

\[ T_u-T_l=5.12 \]

We now have:

\[ T_u+T_l=12.71 \]

\[ T_u-T_l=5.12 \]

Adding:

\[ 2T_u=17.83 \]

\[ T_u=8.91\text{ N} \]

Subtracting:

\[ 2T_l=7.59 \]

\[ T_l=3.79\text{ N} \]

The upper string has greater tension because its vertical component must help balance the object’s weight.

Question 2

A 0.75 kg mass rotates in a horizontal circle of radius 0.90 m at a constant speed of 5.0 m s\(^{-1}\).

It is supported by two strings:

• The upper string makes an angle of \(30^\circ\) with the vertical.
• The lower string makes an angle of \(45^\circ\) with the vertical.

Calculate the tension in each string.

Solution 2

The upper-string tension is \(20.6\text{ N}\), and the lower-string tension is \(14.9\text{ N}\).

Because the angles differ, resolve each tension separately.

First, calculate the radial force:

\[ F_c=\frac{mv^2}{r} \]

\[ F_c=\frac{0.75(5.0)^2}{0.90} \]

\[ F_c=20.83\text{ N} \]

The radial equation is:

\[ T_u\sin30^\circ+T_l\sin45^\circ=20.83 \]

The vertical equation is:

\[ T_u\cos30^\circ-T_l\cos45^\circ =0.75(9.8) \]

Therefore:

\[ T_u\cos30^\circ-T_l\cos45^\circ=7.35 \]

Rearrange the vertical equation:

\[ T_u=\frac{7.35+T_l\cos45^\circ}{\cos30^\circ} \]

Substitute into the radial equation:

\[ \left( \frac{7.35+T_l\cos45^\circ}{\cos30^\circ} \right)\sin30^\circ +T_l\sin45^\circ=20.83 \]

Solving:

\[ T_l=14.87\text{ N} \]

Substitute this into the vertical equation:

\[ T_u= \frac{7.35+14.87\cos45^\circ} {\cos30^\circ} \]

\[ T_u=20.63\text{ N} \]

Both tensions are positive, confirming that both strings can remain taut.

The main trap is treating the two tensions as though they have equal angles. Their components must be calculated independently.

Question 3

A 0.30 kg object rotates in a horizontal circle of radius 0.50 m. It is supported by two strings that each make an angle of \(40^\circ\) with the vertical.

Initially, the object moves at 2.5 m s\(^{-1}\). Its speed is then doubled to 5.0 m s\(^{-1}\), while its radius and string angles remain unchanged.

Calculate the tensions in both strings at each speed.

Explain why doubling the speed does not simply double, or even quadruple, each individual tension.

Solution 3

At 2.5 m s\(^{-1}\), the upper and lower tensions are \(4.84\text{ N}\) and \(1.00\text{ N}\), respectively.

At 5.0 m s\(^{-1}\), they increase to \(13.6\text{ N}\) and \(9.75\text{ N}\).

The key is that doubling speed quadruples the required centripetal force, but the vertical force balance remains unchanged.

Initial speed: 2.5 m s\(^{-1}\)

Calculate the radial force:

\[ F_c=\frac{0.30(2.5)^2}{0.50} \]

\[ F_c=3.75\text{ N} \]

The radial equation gives:

\[ (T_u+T_l)\sin40^\circ=3.75 \]

\[ T_u+T_l=5.83 \]

The vertical equation gives:

\[ (T_u-T_l)\cos40^\circ=0.30(9.8) \]

\[ T_u-T_l=3.84 \]

Solving:

\[ T_u=\frac{5.83+3.84}{2} \]

\[ T_u=4.84\text{ N} \]

\[ T_l=\frac{5.83-3.84}{2} \]

\[ T_l=1.00\text{ N} \]

New speed: 5.0 m s\(^{-1}\)

The speed has doubled, so:

\[ F_c=\frac{0.30(5.0)^2}{0.50} \]

\[ F_c=15.0\text{ N} \]

The radial equation becomes:

\[ (T_u+T_l)\sin40^\circ=15.0 \]

\[ T_u+T_l=23.34 \]

The vertical equation has not changed:

\[ T_u-T_l=3.84 \]

Therefore:

\[ T_u=\frac{23.34+3.84}{2} \]

\[ T_u=13.59\text{ N} \]

\[ T_l=\frac{23.34-3.84}{2} \]

\[ T_l=9.75\text{ N} \]

Why don’t the individual tensions quadruple?

The radial force increases by a factor of four because:

\[ F_c\propto v^2 \]

However, the vertical force balance remains:

\[ (T_u-T_l)\cos40^\circ=mg \]

Since the mass, gravity, and angle are unchanged, the difference between the two tensions must remain constant.

The sum of the tensions quadruples, but their difference stays the same.

This is why the upper tension increases by a different factor from the lower tension.

Question 4

A 0.50 kg object rotates in a horizontal circle of radius 0.75 m.

The upper string makes an angle of \(40^\circ\) with the vertical, and the lower string makes an angle of \(55^\circ\) with the vertical.

Two force sensors record the following tension values:

• Upper string: 12.4 N.
• Lower string: 8.0 N.

Determine whether these measurements are consistent with uniform horizontal circular motion.

If they are consistent, calculate the object’s speed.

Solution 4

The measurements are consistent with uniform horizontal circular motion to the precision of the recorded values. The object’s speed is approximately \(4.67\text{ m s}^{-1}\).

The important first step is to check the vertical force balance before calculating the speed.

Step 1: Check the vertical forces.

For horizontal circular motion, vertical acceleration is zero:

\[ T_u\cos40^\circ-T_l\cos55^\circ=mg \]

Substitute the measurements:

\[ 12.4\cos40^\circ-8.0\cos55^\circ \]

\[ =9.50-4.59 \]

\[ =4.91\text{ N} \]

The object’s weight is:

\[ mg=0.50(9.8)=4.90\text{ N} \]

These values agree to rounding.

Therefore, the vertical force balance is satisfied.

Step 2: Calculate the inward radial force.

Both horizontal tension components point towards the centre:

\[ F_c=T_u\sin40^\circ+T_l\sin55^\circ \]

Substitute:

\[ F_c=12.4\sin40^\circ+8.0\sin55^\circ \]

\[ F_c=7.97+6.55 \]

\[ F_c=14.52\text{ N} \]

Step 3: Calculate the speed.

Use:

\[ F_c=\frac{mv^2}{r} \]

Rearrange:

\[ v=\sqrt{\frac{F_cr}{m}} \]

Substitute:

\[ v=\sqrt{\frac{14.52(0.75)}{0.50}} \]

\[ \boxed{v=4.67\text{ m s}^{-1}} \]

This question tests something beyond simply selecting the right equation.

The measured tensions must satisfy both Newton’s second law equations. A calculation in the radial direction alone cannot establish that the object is moving in a horizontal circle.

Question 5

Extension challenge

A 0.40 kg object rotates in a horizontal circle of radius 0.80 m around a vertical axle.

Its upper string makes an angle of \(35^\circ\) with the vertical.

The lower anchor can be repositioned, allowing the lower string to make either a \(25^\circ\) or a \(60^\circ\) angle with the vertical. The upper anchor and the object’s position are unchanged when the lower anchor is moved.

A motor initially maintains the object’s speed at 2.0 m s\(^{-1}\).

A student argues:

“The lower string at \(25^\circ\) should be more effective at supporting the object because its tension has a larger vertical component. Therefore, changing the lower angle might allow both strings to remain taut, even if the speed is too low.”

Investigate this claim.

(a) Determine the minimum speed required for both strings to remain taut, and explain whether changing the lower-string angle affects this minimum.

(b) At 2.0 m s\(^{-1}\), determine whether either lower-string arrangement can remain taut. Support your conclusion with calculations.

(c) The motor increases the speed to 2.8 m s\(^{-1}\). Calculate the tensions in both strings for each lower-string angle.

(d) Suppose the object’s mass is doubled without changing the speed, radius, or string angles. Explain what happens to the two tensions and the minimum speed.

Solution 5

Neither lower-string arrangement can remain taut at 2.0 m s\(^{-1}\). The minimum speed is approximately \(2.34\text{ m s}^{-1}\), regardless of the lower-string angle.

At 2.8 m s\(^{-1}\), both arrangements can maintain two taut strings, although their individual tensions differ.

Doubling the mass doubles both tensions but does not change the minimum speed.

This question is testing an important distinction: changing a force’s direction changes its components, but it does not necessarily change the condition that determines whether a string can remain taut.

(a) Find the minimum speed.

At the instant the lower string becomes slack:

\[ T_l=0 \]

The upper string must therefore provide both the vertical support and the inward radial force:

\[ T_u\cos\theta_u=mg \]

\[ T_u\sin\theta_u=\frac{mv^2}{r} \]

Dividing:

\[ \tan\theta_u=\frac{v^2}{rg} \]

Therefore:

\[ v_{\min}=\sqrt{rg\tan\theta_u} \]

Substitute:

\[ v_{\min}=\sqrt{0.80(9.8)\tan35^\circ} \]

\[ \boxed{v_{\min}=2.34\text{ m s}^{-1}} \]

The lower-string angle does not appear in this equation.

At the threshold, the lower string has zero tension. Only the upper string acts, so changing the lower string’s angle cannot alter the minimum speed.

(b) Check the two arrangements at 2.0 m s\(^{-1}\).

We need to determine whether the lower tension is positive.

For unequal angles, the force equations are:

\[ T_u\cos\theta_u-T_l\cos\theta_l=mg \]

\[ T_u\sin\theta_u+T_l\sin\theta_l =\frac{mv^2}{r} \]

Eliminating \(T_u\) gives a useful expression for the lower tension:

\[ T_l= \frac{m\cos\theta_u}{\sin(\theta_u+\theta_l)} \left( \frac{v^2}{r}-g\tan\theta_u \right) \]

For the given speed:

\[ \frac{v^2}{r}=\frac{2.0^2}{0.80}=5.0 \]

Also:

\[ g\tan35^\circ=6.86 \]

The quantity in brackets is therefore:

\[ 5.0-6.86=-1.86 \]

For a lower angle of \(60^\circ\):

\[ T_l\approx-0.61\text{ N} \]

For a lower angle of \(25^\circ\):

\[ T_l\approx-0.70\text{ N} \]

Both results are negative.

Neither arrangement can maintain the specified horizontal circular motion with both strings taut.

The student’s argument is tempting because changing the lower angle does change its vertical component for a given tension.

However, the minimum-speed condition occurs when the lower tension is zero. At that exact point, the lower string contributes nothing, regardless of its angle.

(c) Calculate the tensions at 2.8 m s\(^{-1}\).

Now:

\[ \frac{v^2}{r}=\frac{2.8^2}{0.80} \]

\[ \frac{v^2}{r}=9.8 \]

The lower tension is:

\[ T_l= \frac{m\cos35^\circ}{\sin(35^\circ+\theta_l)} (9.8-9.8\tan35^\circ) \]

For the \(60^\circ\) lower-string arrangement:

\[ T_l= \frac{0.40\cos35^\circ}{\sin95^\circ} (9.8-9.8\tan35^\circ) \]

\[ T_l=0.97\text{ N} \]

Substitute into the vertical equation:

\[ T_u\cos35^\circ-0.97\cos60^\circ=3.92 \]

\[ T_u=5.38\text{ N} \]

For the \(25^\circ\) lower-string arrangement:

\[ T_l= \frac{0.40\cos35^\circ}{\sin60^\circ} (9.8-9.8\tan35^\circ) \]

\[ T_l=1.11\text{ N} \]

Substitute into the vertical equation:

\[ T_u\cos35^\circ-1.11\cos25^\circ=3.92 \]

\[ T_u=6.02\text{ N} \]

The results are:

Both arrangements now have positive tensions.

Changing the lower angle changes how the inward and vertical force components are distributed between the two strings. However, it does not change the minimum speed.

(d) Double the mass.

The mass appears as a common factor in both Newton’s second law equations.

If the mass doubles while the geometry and speed remain unchanged, both calculated tensions double.

For example, in the \(60^\circ\) arrangement:

\[ T_u=2(5.38)=10.76\text{ N} \]

\[ T_l=2(0.97)=1.94\text{ N} \]

The minimum speed remains:

\[ v_{\min}=2.34\text{ m s}^{-1} \]

This is because the mass cancels when the vertical and radial equations are divided.

The heavier object needs greater forces to maintain the same motion, but the speed at which the lower string first becomes taut is unchanged.

The central idea is that the minimum-speed condition is controlled by the upper string’s angle, the orbit radius, and gravity, not by the mass or the lower-string angle.

Once you can resolve the forces in a two-string arrangement, the next useful step is understanding what happens when one of those forces disappears. In particular, explore When a String Breaks: Which Way Does the Object Move? to connect force analysis with the object’s instantaneous velocity and subsequent path.