How to Find the Period of a Conical Pendulum from Its Geometry

A conical pendulum has a string of fixed length, but its bob travels in a horizontal circle. If you know the string length and the angle it makes with the vertical, you can calculate the time for one complete revolution without knowing the bob's mass or speed.

The key is to use the geometry to find the vertical distance from the pivot to the plane of circular motion.

The result is:

\[
\boxed{P=2\pi\sqrt{\frac{L\cos\theta}{g}}}
\]

where:
- \(P\) is the period, in seconds.
- \(L\) is the string length, in metres.
- \(\theta\) is the angle between the string and the vertical.
- \(g\) is gravitational field strength, approximately \(9.8\ \text{m s}^{-2}\).

If the question gives you the string length and the radius of the circular path instead, use:

\[
\boxed{P=2\pi\sqrt{\frac{\sqrt{L^2-r^2}}{g}}}
\]

Here, \(r\) is the horizontal radius of the bob's circular path.

Both equations describe the same result. The first uses the angle to find the vertical height. The second uses the radius and Pythagoras' theorem.

Let's build the reasoning so you know when, and why, to use each one.

## 1. Start with the geometry of the pendulum

Picture a small bob attached to a string hanging from a fixed pivot. When the bob rotates steadily, it traces a horizontal circle while the string sweeps out an imaginary cone.

Three lengths describe this arrangement:

- \(L\): the full length of the string.
- \(r\): the radius of the horizontal circle.
- \(h\): the vertical distance from the pivot to the plane of the circle.

The string is the hypotenuse of a right-angled triangle. Its horizontal projection is the radius, and its vertical projection is the height.

[IMAGE: Scientifically accurate conical pendulum illustration with a side view and small top view. Show a fixed pivot, the sloping string of length L, the bob moving in a horizontal circular path of radius r, and the vertical distance h from pivot to the plane of the circle. Mark theta between the string and the vertical. Include a clear right-angled geometry triangle showing L as the hypotenuse, r as the horizontal side, and h as the vertical side.]

Using trigonometry:

\[
\sin\theta=\frac{r}{L}
\]

Therefore,

\[
r=L\sin\theta
\]

For the vertical side:

\[
\cos\theta=\frac{h}{L}
\]

Therefore,

\[
\boxed{h=L\cos\theta}
\]

This vertical distance is the geometric quantity that determines the period.

Notice that the radius and the string length are not interchangeable. If you substitute \(L\) for \(r\), your circular-motion calculation will be wrong.

## 2. Identify the real radial force

Before calculating the period, we need to establish what makes the bob move in a circle.

There are two real forces acting on an ideal conical pendulum:

- Tension, \(T\), acting along the string towards the pivot.
- Weight, \(mg\), acting vertically downwards.

There is no separate physical force called centripetal force. The term describes the net inward force required for circular motion.

For this arrangement, the real radial force is the horizontal component of tension:

\[
F_{\text{radial}}=T\sin\theta
\]

Weight acts vertically, so it has no horizontal radial component.

The vertical component of tension balances the bob's weight because the bob has no vertical acceleration:

\[
T\cos\theta=mg
\]

The two Newton's second law equations are therefore:

\[
\begin{aligned}
T\cos\theta &= mg\\
T\sin\theta &= \frac{mv^2}{r}
\end{aligned}
\]

Here, \(v\) is the bob's tangential speed, in metres per second, and \(m\) is its mass, in kilograms.

[IMAGE: Free-body diagram of a conical pendulum bob viewed from the side. Show only the two real forces: tension T along the string towards the pivot and weight mg vertically downwards. Resolve tension into an upward component T cos theta and a horizontal inward radial component T sin theta. Clearly distinguish the actual tension vector from its components. Mark the inward direction towards the centre of the horizontal circle.]

If you need a fuller explanation of why tension, rather than the whole tension force, supplies the radial force, see [Conical Pendulums: Resolving Tension and Weight in HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-a-mass-moving-in-a-horizontal-circle-on-a-string/).

For this problem, we need one more step: connecting those forces to the period using the geometry.

## 3. Derive the period from the geometry

The speed of an object moving in a circle is related to its angular speed by:

\[
v=\omega r
\]

Angular speed, \(\omega\), is measured in radians per second. It describes how quickly the bob sweeps through an angle around the circle.

Substitute \(v=\omega r\) into the radial force equation:

\[
T\sin\theta=\frac{m(\omega r)^2}{r}
\]

Simplifying:

\[
T\sin\theta=m\omega^2r
\]

Now use the geometric relationship \(r=L\sin\theta\):

\[
T\sin\theta=m\omega^2L\sin\theta
\]

For a genuine conical pendulum, the radius is non-zero, so we can cancel \(\sin\theta\):

\[
T=m\omega^2L
\]

Combine this with the vertical force equation:

\[
T\cos\theta=mg
\]

Substituting for tension:

\[
m\omega^2L\cos\theta=mg
\]

Cancel the mass:

\[
\omega^2L\cos\theta=g
\]

Rearranging:

\[
\omega^2=\frac{g}{L\cos\theta}
\]

The period is the time taken for one complete revolution. Since one revolution covers \(2\pi\) radians:

\[
P=\frac{2\pi}{\omega}
\]

Substituting the expression for angular speed gives:

\[
P=2\pi\sqrt{\frac{L\cos\theta}{g}}
\]

Using \(h=L\cos\theta\), we can write the result more simply:

\[
\boxed{P=2\pi\sqrt{\frac{h}{g}}}
\]

This is the central result: the period depends on the vertical distance from the pivot to the plane of circular motion.

The bob's mass has cancelled, and the circular radius is no longer needed once the vertical distance is known.

## 4. Worked examples: calculating the period

The method depends on which geometric measurements the question gives you. The underlying physics stays the same.

### Worked example: Find the period using the string length and angle

A \(0.30\ \text{kg}\) bob is attached to a \(1.20\ \text{m}\) string. It moves in a steady horizontal circle, with the string making an angle of \(30.0^\circ\) with the vertical.

Calculate the period of the motion. Use \(g=9.8\ \text{m s}^{-2}\).

**Step 1**: Find the vertical distance from the pivot to the circular path.

We know:
- \(L=1.20\ \text{m}\)
- \(\theta=30.0^\circ\)

Use the vertical component of the string length:

\[
h=L\cos\theta
\]

Substitute:

\[
h=1.20\cos30^\circ
\]

\[
h=1.039\ \text{m}
\]

**Step 2**: Calculate the period.

Use:

\[
P=2\pi\sqrt{\frac{h}{g}}
\]

Substitute:

\[
P=2\pi\sqrt{\frac{1.039}{9.8}}
\]

\[
P=2.05\ \text{s}
\]

The bob completes one revolution every \(2.05\) seconds.

Notice that the mass of \(0.30\ \text{kg}\) was not needed. It cancels when we derive the period equation.

### Worked example: Find the period using the string length and radius

A bob is attached to a \(1.50\ \text{m}\) string. It moves in a horizontal circle with a radius of \(0.90\ \text{m}\).

Calculate the period of the motion. Use \(g=9.8\ \text{m s}^{-2}\).

[IMAGE: Side-view geometry diagram of a conical pendulum with a 1.50 m sloping string and a 0.90 m horizontal radius. Show the unknown vertical distance h, the fixed pivot directly above the circle centre, and the bob at the end of the string. Do not show the calculated height or period.]

**Step 1**: Find the vertical distance using Pythagoras' theorem.

We know the string is the hypotenuse of the right-angled triangle.

Therefore:

\[
L^2=r^2+h^2
\]

Rearrange:

\[
h=\sqrt{L^2-r^2}
\]

Substitute:

\[
h=\sqrt{1.50^2-0.90^2}
\]

\[
h=\sqrt{2.25-0.81}
\]

\[
h=1.20\ \text{m}
\]

**Step 2**: Calculate the period.

\[
P=2\pi\sqrt{\frac{h}{g}}
\]

\[
P=2\pi\sqrt{\frac{1.20}{9.8}}
\]

\[
\boxed{P=2.20\ \text{s}}
\]

The bob completes one revolution every \(2.20\) seconds.

The important step was finding \(h\) before using the period equation. Substituting the radius directly would give the wrong result.

## 5. Choosing the correct geometric method

You can now solve the period problem using whichever measurements the question provides.

| Information provided | First calculation | Period equation |
|---|---|---|
| String length \(L\) and angle \(\theta\) from vertical | \(h=L\cos\theta\) | \(P=2\pi\sqrt{h/g}\) |
| String length \(L\) and radius \(r\) | \(h=\sqrt{L^2-r^2}\) | \(P=2\pi\sqrt{h/g}\) |
| Vertical distance \(h\) | No extra geometry needed | \(P=2\pi\sqrt{h/g}\) |
| String length \(L\) and angle \(\alpha\) from horizontal | \(h=L\sin\alpha\) | \(P=2\pi\sqrt{h/g}\) |

A useful way to organise your working is to calculate the vertical distance first, regardless of how the question presents its measurements.

Then calculate the period.

### Three mistakes worth avoiding

**Mistake 1: Using the full string length instead of its vertical component.**

The expression \(2\pi\sqrt{L/g}\) is not the general conical-pendulum period. For a non-zero angle, the vertical distance is smaller than the full string length.

**Mistake 2: Treating the radius as the string length.**

The radius is the horizontal projection of the string. You need \(r=L\sin\theta\), not \(r=L\).

**Mistake 3: Treating the entire tension as the radial force.**

Only \(T\sin\theta\) acts horizontally towards the centre. The vertical component balances the bob's weight.

A useful diagnostic is to check your answer against the geometry. The vertical distance must satisfy \(0<h\leq L\), and the radius must satisfy \(0\leq r<L\) for a genuine conical orbit.

## 6. A contrasting case: not all circular motion has the same radial force

Consider a car travelling around a flat circular bend at constant speed.

The car needs a net inward force to keep changing direction. However, there is no string pulling it towards the centre.

Instead, static friction between the tyres and the road provides the inward force.

[IMAGE: An educational illustration of a car travelling around a flat circular bend, viewed from slightly above. Show the curved road, the car, the centre of the circular path, and an inward friction force arrow acting on the car. Include weight and normal force vertically only if the force diagram is clear.]

Compare the two situations:

| Situation | Real inward radial force |
|---|---|
| Conical pendulum | Horizontal component of tension, \(T\sin\theta\) |
| Car on a flat circular road | Static friction between tyres and road |

Both situations obey the same circular-motion requirement:

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

But the actual physical force supplying that inward resultant is different.

This is why you should identify the real forces before writing a circular-motion equation. Centripetal force describes the net radial force; it is not an extra force to add to your free-body diagram.

### Extension: Comparing a conical pendulum with an ordinary pendulum

An ordinary simple pendulum swings backwards and forwards through a vertical arc. A conical pendulum rotates continuously in a horizontal circle.

For small-angle oscillations, the approximate period of an ordinary simple pendulum is:

\[
P_{\text{simple}}\approx2\pi\sqrt{\frac{L}{g}}
\]

For a conical pendulum:

\[
P_{\text{conical}}=2\pi\sqrt{\frac{L\cos\theta}{g}}
\]

These equations look similar, but the physical motion is different.

The ordinary pendulum formula uses the full string length. The conical pendulum formula uses its vertical projection.

Do not transfer the ordinary pendulum formula to a conical pendulum just because both systems have a bob hanging from a string.

## 7. Testing the geometric prediction experimentally

The period equation also gives us a way to test whether the ideal conical-pendulum model describes an actual apparatus.

The question is straightforward: if we measure the string length and circular radius, does the calculated period agree with the time recorded for complete revolutions?

### Setting up the investigation

Imagine a clamp stand with a fixed pivot at the top. A light string hangs from the pivot, with a small dense bob attached at its lower end.

The bob is set rotating until it follows a steady horizontal circular path.

We need two geometric measurements:

1. Use a ruler to measure the string length \(L\), from the pivot to the centre of the bob.
2. Measure the circular radius \(r\), from the vertical projection of the pivot to the bob's path. A calibrated overhead video or a suitable reference at the height of the circular path can help.

We also need to measure the time taken for a known number of complete revolutions.

[IMAGE: Experimental conical pendulum apparatus showing a clamp stand and fixed pivot, a string with a rotating bob, and a horizontal reference scale at the bob's orbital height. Include an overhead camera or viewing position for measuring the circular radius. Show the spatial arrangement clearly without suggesting that the apparatus is a photograph of a real experiment.]

### What should the experiment show?

The geometric model predicts:

\[
P=2\pi\sqrt{\frac{\sqrt{L^2-r^2}}{g}}
\]

If the bob follows steady conical motion and our measurements are accurate, this predicted period should agree with the measured period.

A first attempt might be to time just one revolution using a stopwatch. However, starting and stopping the stopwatch introduces reaction-time errors.

A better method is to time 20 or 25 complete revolutions, then divide the total time by the number of revolutions:

\[
P=\frac{t}{N}
\]

Here:
- \(t\) is the total recorded time, in seconds.
- \(N\) is the number of complete revolutions.
- \(P\) is the average period, in seconds.

Timing several revolutions makes the relative effect of stopwatch reaction time smaller.

### Example experimental results

The following are illustrative data, not measurements from a real experiment.

| Measurement | Value |
|---|---:|
| String length, \(L\) | \(1.00\ \text{m}\) |
| Circular radius, \(r\) | \(0.77\ \text{m}\) |
| Number of revolutions, \(N\) | 25 |
| Total recorded time, \(t\) | \(40.0\ \text{s}\) |

First, calculate the vertical distance:

\[
h=\sqrt{L^2-r^2}
\]

\[
h=\sqrt{1.00^2-0.77^2}
\]

\[
h=0.638\ \text{m}
\]

Now calculate the predicted period:

\[
P=2\pi\sqrt{\frac{0.638}{9.8}}
\]

\[
P_{\text{predicted}}=1.60\ \text{s}
\]

Next, calculate the measured period:

\[
P_{\text{measured}}=\frac{40.0}{25}
\]

\[
P_{\text{measured}}=1.60\ \text{s}
\]

The predicted and measured periods agree to the precision of the supplied data.

This supports the geometric model for the illustrative measurements.

It does not prove that every real conical pendulum will behave perfectly. Air resistance, string stretch, measurement errors, and variations in the bob's speed can all affect the result.

There is also an important measurement limitation: when \(r\) is close to \(L\), the vertical distance is relatively small. Small errors in the radius measurement can then produce noticeable errors in the calculated height and period.

## 8. Questions and solutions

The following questions move from direct geometric calculations to more demanding applications of the period relationship.

Use \(g=9.8\ \text{m s}^{-2}\) throughout unless stated otherwise. Assume ideal conical motion, a fixed pivot, a light inextensible string, and negligible air resistance.

### Question 1

A \(0.25\ \text{kg}\) bob is attached to a \(0.80\ \text{m}\) string. It rotates in a horizontal circle, with the string making an angle of \(35^\circ\) with the vertical.

Calculate the period of the motion.

[IMAGE: Side-view conical pendulum setup with a fixed pivot, 0.80 m string, bob, and angle of 35 degrees measured from the vertical. Do not show the vertical projection or the calculated period.]

### Solution 1

The period is \(1.62\ \text{s}\). We first calculate the vertical projection of the string, then substitute it into the period equation.

**Step 1**: Calculate the vertical distance.

\[
h=L\cos\theta
\]

\[
h=0.80\cos35^\circ
\]

\[
h=0.655\ \text{m}
\]

**Step 2**: Calculate the period.

\[
P=2\pi\sqrt{\frac{h}{g}}
\]

\[
P=2\pi\sqrt{\frac{0.655}{9.8}}
\]

\[
\boxed{P=1.62\ \text{s}}
\]

The bob completes one revolution every \(1.62\) seconds.

The mass was included in the question, but it was not needed. Mass cancels from the derivation of the period equation.

### Question 2

A conical pendulum has a string length of \(1.30\ \text{m}\) and a horizontal circular radius of \(0.50\ \text{m}\).

Calculate:

a. The vertical distance from the pivot to the plane of circular motion.

b. The angle between the string and the vertical.

c. The period of the motion.

[IMAGE: Side-view right-angled geometry of a conical pendulum. Label the sloping string 1.30 m and the horizontal radius 0.50 m. Leave the vertical height and angle unlabelled as unknowns. Show the pivot directly above the centre of the circular path.]

### Solution 2

The vertical distance is \(1.20\ \text{m}\), the angle is \(22.6^\circ\), and the period is \(2.20\ \text{s}\).

**Step 1**: Calculate the vertical distance using Pythagoras' theorem.

\[
h=\sqrt{L^2-r^2}
\]

\[
h=\sqrt{1.30^2-0.50^2}
\]

\[
h=\sqrt{1.69-0.25}
\]

\[
\boxed{h=1.20\ \text{m}}
\]

**Step 2**: Calculate the angle.

The horizontal radius is opposite the angle, and the string length is the hypotenuse.

\[
\sin\theta=\frac{r}{L}
\]

\[
\sin\theta=\frac{0.50}{1.30}
\]

\[
\theta=\sin^{-1}(0.3846)
\]

\[
\boxed{\theta=22.6^\circ}
\]

**Step 3**: Calculate the period.

\[
P=2\pi\sqrt{\frac{h}{g}}
\]

\[
P=2\pi\sqrt{\frac{1.20}{9.8}}
\]

\[
\boxed{P=2.20\ \text{s}}
\]

The string length is the hypotenuse, not the circular radius. Using the correct right-angled geometry is what makes the calculation work.

### Question 3

Two conical pendulums use strings of identical length, \(1.20\ \text{m}\).

Their circular radii are different:

- Pendulum A has a radius of \(0.36\ \text{m}\).
- Pendulum B has a radius of \(0.96\ \text{m}\).

Calculate the period of each pendulum.

A student predicts that pendulum B must take longer to complete a revolution because it travels around a larger circle.

Evaluate this prediction using your calculations.

[IMAGE: Two side-by-side side-view conical pendulum arrangements, both with 1.20 m strings. Show the first horizontal circular radius as 0.36 m and the second as 0.96 m. Make the different string angles and circular radii visually clear. Do not show force arrows or calculated periods.]

### Solution 3

Pendulum A has a period of \(2.15\ \text{s}\), while pendulum B has a shorter period of \(1.70\ \text{s}\). The student is incorrect: the larger circular path is completed more quickly because the bob has a greater speed and angular speed.

**Step 1**: Calculate the vertical distance for pendulum A.

\[
h_A=\sqrt{L^2-r_A^2}
\]

\[
h_A=\sqrt{1.20^2-0.36^2}
\]

\[
h_A=\sqrt{1.44-0.1296}
\]

\[
h_A=1.145\ \text{m}
\]

**Step 2**: Calculate its period.

\[
P_A=2\pi\sqrt{\frac{1.145}{9.8}}
\]

\[
\boxed{P_A=2.15\ \text{s}}
\]

**Step 3**: Calculate the vertical distance for pendulum B.

\[
h_B=\sqrt{L^2-r_B^2}
\]

\[
h_B=\sqrt{1.20^2-0.96^2}
\]

\[
h_B=\sqrt{1.44-0.9216}
\]

\[
h_B=0.720\ \text{m}
\]

**Step 4**: Calculate its period.

\[
P_B=2\pi\sqrt{\frac{0.720}{9.8}}
\]

\[
\boxed{P_B=1.70\ \text{s}}
\]

**Step 5**: Explain the difference.

Pendulum B travels around a larger circle.

Its circumference is:

\[
C_B=2\pi(0.96)=6.03\ \text{m}
\]

For pendulum A:

\[
C_A=2\pi(0.36)=2.26\ \text{m}
\]

Pendulum B travels approximately 2.67 times further in one revolution.

However, its period is shorter. This means its tangential speed must be considerably greater.

Using \(v=2\pi r/P\):

\[
v_A=\frac{2\pi(0.36)}{2.15}
=1.05\ \text{m s}^{-1}
\]

\[
v_B=\frac{2\pi(0.96)}{1.70}
=3.55\ \text{m s}^{-1}
\]

The larger orbit is accompanied by a much greater speed.

The geometric reason is that pendulum B has a smaller vertical distance \(h\). Its angular speed is therefore greater, giving it a shorter period.

This is a useful example of why you cannot predict a period just by looking at the circumference.

### Question 4

A student investigates a conical pendulum using a string of length \(1.00\ \text{m}\).

They measure the horizontal radius of the bob's circular path as \(0.77\ \text{m}\). They then record the time taken for 25 complete revolutions as \(40.0\ \text{s}\).

a. Calculate the period predicted by the geometric model.

b. Calculate the period measured experimentally.

c. Assess whether the results support the model.

d. Explain why timing 25 revolutions is preferable to timing a single revolution.

[IMAGE: Conical pendulum experimental arrangement showing a clamp stand, fixed pivot, rotating bob, 1.00 m string, and a horizontal measuring reference at the bob's orbital height. Include a stopwatch or video timing setup. Do not include calculated values.]

### Solution 4

The geometric model predicts a period of \(1.60\ \text{s}\), and the experimental timing gives \(1.60\ \text{s}\). The results agree to the precision of the supplied data.

**Step 1**: Calculate the vertical distance using the geometry.

\[
h=\sqrt{L^2-r^2}
\]

\[
h=\sqrt{1.00^2-0.77^2}
\]

\[
h=\sqrt{0.4071}
\]

\[
h=0.638\ \text{m}
\]

**Step 2**: Calculate the predicted period.

\[
P_{\text{predicted}}
=2\pi\sqrt{\frac{0.638}{9.8}}
\]

\[
\boxed{P_{\text{predicted}}=1.60\ \text{s}}
\]

**Step 3**: Calculate the experimental period.

The period is the total time divided by the number of revolutions.

\[
P_{\text{measured}}=\frac{t}{N}
\]

\[
P_{\text{measured}}=\frac{40.0}{25}
\]

\[
\boxed{P_{\text{measured}}=1.60\ \text{s}}
\]

**Step 4**: Assess the agreement.

The two results agree to three significant figures.

This supports the model for the measured arrangement. However, agreement does not eliminate possible experimental errors.

**Step 5**: Explain the improved timing method.

Timing one revolution means that stopwatch reaction time can make up a noticeable fraction of the recorded time.

Timing 25 revolutions makes the total interval much longer. The same start-and-stop reaction-time error then has a smaller relative effect.

Dividing the total time by 25 gives a more reliable estimate of the average period.

### Question 5

A bob of mass \(0.20\ \text{kg}\) is attached to a \(1.00\ \text{m}\) string and moves in a steady conical pendulum motion.

The string has a maximum safe tension of \(5.0\ \text{N}\).

An engineer wants the bob to complete one revolution every \(1.00\ \text{s}\).

a. Calculate the vertical distance and horizontal radius required for this period.

b. Determine the tension required for the \(0.20\ \text{kg}\) bob. Is the proposed motion physically possible without exceeding the safe tension?

c. The bob is replaced with one of mass \(0.10\ \text{kg}\). Can this lighter bob achieve the same period? Explain whether its speed changes.

d. For the original \(0.20\ \text{kg}\) bob, calculate the shortest possible period without exceeding the maximum safe tension.

Assume the string remains ideal, the pivot is fixed, and the motion is uniform whenever the tension constraint is satisfied.

[IMAGE: A single conical pendulum arrangement with a fixed pivot, 1.00 m string and bob. Show the bob's horizontal circular path and label the bob's initial mass as 0.20 kg and the maximum safe string tension as 5.0 N. Include the target period of 1.00 s as supplied information. Do not show force components, calculated angles, or derived dimensions.]

### Solution 5

The \(0.20\ \text{kg}\) bob cannot achieve the target period safely because it would require a tension of \(7.90\ \text{N}\). The lighter \(0.10\ \text{kg}\) bob can achieve the same period, and the shortest safe period for the original bob is \(1.26\ \text{s}\).

This question exposes an important distinction: mass cancels from the period equation, but it does not cancel from the tension equation. A period may be mathematically consistent with the geometry but physically impossible for a string with a limited strength.

**Step 1**: Calculate the vertical distance required for a period of \(1.00\ \text{s}\).

Start with:

\[
P=2\pi\sqrt{\frac{h}{g}}
\]

Rearrange:

\[
h=\frac{gP^2}{4\pi^2}
\]

Substitute:

\[
h=\frac{(9.8)(1.00)^2}{4\pi^2}
\]

\[
\boxed{h=0.248\ \text{m}}
\]

**Step 2**: Calculate the required circular radius.

Use Pythagoras' theorem:

\[
r=\sqrt{L^2-h^2}
\]

\[
r=\sqrt{1.00^2-0.248^2}
\]

\[
\boxed{r=0.969\ \text{m}}
\]

The bob must travel in a relatively wide circle.

**Step 3**: Calculate the tension required for the original bob.

The vertical component of tension balances weight:

\[
T\cos\theta=mg
\]

From the geometry:

\[
\cos\theta=\frac{h}{L}
\]

Therefore:

\[
T=\frac{mgL}{h}
\]

Substitute the original mass:

\[
T=\frac{(0.20)(9.8)(1.00)}{0.248}
\]

\[
\boxed{T=7.90\ \text{N}}
\]

The maximum safe tension is only \(5.0\ \text{N}\).

Therefore:

\[
7.90>5.0
\]

The proposed motion is not physically safe for the original bob.

Although the period equation gives a valid geometric arrangement, the string cannot provide the tension required to maintain it.

**Step 4**: Repeat the tension calculation for the lighter bob.

The mass is now:

\[
m=0.10\ \text{kg}
\]

The target period and string length have not changed. Because mass cancels from the period equation, the required geometry is also unchanged.

The tension becomes:

\[
T=\frac{mgL}{h}
\]

\[
T=\frac{(0.10)(9.8)(1.00)}{0.248}
\]

\[
\boxed{T=3.95\ \text{N}}
\]

This is below the maximum safe tension of \(5.0\ \text{N}\).

The lighter bob can therefore achieve the target period safely under the stated assumptions.

Its tangential speed is also unchanged because its radius and period are unchanged:

\[
v=\frac{2\pi r}{P}
\]

\[
v=\frac{2\pi(0.969)}{1.00}
\]

\[
\boxed{v=6.09\ \text{m s}^{-1}}
\]

The lighter bob travels at the same speed, but requires less tension to maintain the motion.

**Step 5**: Find the shortest safe period for the original bob.

The shortest safe period occurs when the tension reaches its maximum allowed value:

\[
T_{\max}=\frac{mgL}{h_{\min}}
\]

Rearrange:

\[
h_{\min}=\frac{mgL}{T_{\max}}
\]

Substitute:

\[
h_{\min}=
\frac{(0.20)(9.8)(1.00)}{5.0}
\]

\[
h_{\min}=0.392\ \text{m}
\]

Now calculate the minimum safe period:

\[
P_{\min}=2\pi\sqrt{\frac{h_{\min}}{g}}
\]

\[
P_{\min}=2\pi\sqrt{\frac{0.392}{9.8}}
\]

\[
P_{\min}=2\pi\sqrt{0.040}
\]

\[
\boxed{P_{\min}=1.26\ \text{s}}
\]

The original bob cannot safely complete a revolution in less than \(1.26\) seconds with this string.

The tempting conclusion is that mass does not matter because it cancels from the period equation. The missing constraint is the string's maximum tension.

Mass does not determine the period of an ideal conical pendulum with a specified geometry, but it does affect whether the string can physically sustain that geometry.

## 9. What to calculate next

Once you can find the period from the string's geometry, you can reverse the method to find an unknown angle or radius from a measured period.

For the angle calculation, see [How to Find the Angle of a Conical Pendulum](https://tutorgum.com/learn/resources/hsc-physics-how-to-find-the-angle-of-a-conical-pendulum/).

The same approach to resolving forces and identifying the real radial force also applies to rotating objects supported by multiple strings. The next step is learning how to handle systems where more than one tension contributes to the inward force.