How to Calculate Slip from Rotor Speed and Synchronous Speed

A four-pole induction motor is connected to a 50 Hz supply. Its rotating magnetic field travels at 1500 rpm, while its rotor turns at 1440 rpm.

The difference is 60 rpm. But what is the percentage slip?

It’s 4.0%, calculated by dividing the difference between the two speeds by the synchronous speed:

\[ s=\frac{1500-1440}{1500}\times100 =4.0\% \]

That’s the calculation you need to know. The important detail is the denominator: percentage slip is calculated relative to synchronous speed, not rotor speed.

Let’s work through why this is the correct method, how to find synchronous speed when it isn’t given, and what the slip value tells you about the motor.

01What do synchronous speed and rotor speed actually measure?

An induction motor has two different rotational speeds to keep track of.

• Synchronous speed (\(n_s\)) is the speed of the rotating magnetic field produced by the stator.
• Rotor speed (\(n_r\)) is the actual rotational speed of the motor’s rotor and shaft.

The stator itself does not rotate. Its magnetic field does.

This distinction matters because the rotor does not normally rotate at exactly the same speed as the magnetic field.

For an induction motor operating as a motor, the rotor turns slightly slower.

Why doesn’t the rotor catch up?

Remember how an induction motor produces torque.

The stator creates a rotating magnetic field. As this field moves relative to the conducting bars in the rotor, it induces an electromotive force (EMF) and current.

The induced current interacts with the magnetic field, producing forces that generate torque.

Here’s the important connection:

The rotor needs relative motion between itself and the rotating magnetic field to maintain electromagnetic induction.

Imagine walking along a moving travelator. If the travelator moves at 10 km/h and you move in the same direction at 9 km/h, the belt moves past your feet at a relative speed of 1 km/h.

An induction motor has a similar relationship. The magnetic field moves relative to the rotor, and that relative motion determines the rate at which magnetic flux changes through the rotor conductors.

The analogy isn’t exact: a magnetic field isn’t a physical belt. The actual induction comes from changing magnetic flux, as described by Faraday’s law.

If the rotor reached exactly the same speed as the rotating magnetic field, there would be no relative motion between them.

In the ideal induction-motor model, this means no changing magnetic flux through the rotor due to the fundamental rotating field, no induced rotor EMF, and therefore no induction torque.

The rotor must lag slightly behind the field to keep producing torque.

That difference in speed is called slip.

02How to calculate synchronous speed

Sometimes a question gives you both synchronous speed and rotor speed.

Other times, you need to calculate synchronous speed first.

The synchronous speed depends on two things:

• The AC supply frequency.
• The total number of magnetic poles in the stator.

The formula is:

\[ \boxed{n_s=\frac{120f}{P}} \]

Where:

The factor of 120 comes from converting the electrical frequency into mechanical revolutions per minute, accounting for the number of magnetic poles.

A useful point to remember is that the supply frequency alone does not determine the rotational speed. The number of poles matters too.

Synchronous speeds at 50 Hz

Australian AC supply has a frequency of 50 Hz. For a 50 Hz supply, the synchronous speeds are:

Notice the pattern: increasing the number of poles decreases the synchronous speed.

For example, a four-pole motor has twice the synchronous speed of an eight-pole motor when both operate at the same supply frequency.

03The formula for percentage slip

Once you know the synchronous speed and rotor speed, calculating slip is straightforward.

First, calculate the difference between the two speeds:

\[ \Delta n=n_s-n_r \]

Here, \(\Delta n\) is the slip speed, measured in rpm.

Then divide this difference by the synchronous speed:

\[ \boxed{s=\frac{n_s-n_r}{n_s}} \]

To express the answer as a percentage:

\[ \boxed{\%\text{ slip}= \frac{n_s-n_r}{n_s}\times100} \]

Where:

• \(n_s\) is synchronous speed in rpm.
• \(n_r\) is rotor speed in rpm.
• \(s\) is the dimensionless slip ratio.

Keep these two results separate:

For example, a speed difference of 60 rpm does not mean the motor has 60% slip.

You must divide by synchronous speed before converting to a percentage.

Why do we divide by synchronous speed?

The synchronous speed is the reference speed of the rotating magnetic field.

Dividing by it tells us how large the relative speed is compared with the field’s full rotational speed.

This also gives us two useful checks:

• If the rotor is stationary, \(n_r=0\), so slip is 100%.
• If the rotor reaches synchronous speed, \(n_r=n_s\), so slip is 0%.

Dividing by rotor speed instead would not give these correct reference points.

04Worked examples

05What happens to slip when the motor’s load changes?

Now consider what happens when an induction motor has to do more work.

Suppose a motor is driving a pump. If the pump requires more torque, the motor must produce more electromagnetic torque to keep the pump turning.

The rotor initially slows slightly.

But does the rotating magnetic field also slow down?

No. If the supply frequency and number of poles remain constant, synchronous speed stays constant.

The rotor speed changes, which changes the slip.

Within the motor’s normal operating range, increasing slip allows greater induced rotor current and electromagnetic torque to develop.

Consider the following illustrative data for a six-pole induction motor connected to a 50 Hz supply.

Its synchronous speed is 1000 rpm.

Illustrative data only:

There are three important observations.

1. Synchronous speed stays constant.

The stator’s rotating field continues at 1000 rpm because the supply frequency and pole count have not changed.

2. Rotor speed decreases as load increases.

The rotor needs to develop more torque to balance the increased mechanical load.

3. Slip increases.

The greater relative motion increases the rotor’s induced current frequency. The electromagnetic torque can then increase to meet the additional demand.

This is why slip is not necessarily a sign of an inefficient or faulty motor. A certain amount of slip is essential to normal induction-motor operation.

At no load, slip is generally small, but not exactly zero. The motor still needs torque to overcome friction, windage, and other losses.

06Does every motor need slip?

No. This is an important distinction between induction motors and synchronous motors.

Both types of motor can have a rotating magnetic field in the stator, but their rotors produce torque through different mechanisms.

Why can a synchronous motor operate with zero slip?

A synchronous motor has its own rotor magnetic field, produced by permanent magnets or a DC-excited winding.

This field interacts with the stator’s rotating magnetic field.

Once the rotor is synchronised, it maintains a fixed angular relationship with the rotating field. It can therefore produce torque while rotating at exactly synchronous speed.

When the mechanical load increases, the angular separation between the rotor and stator magnetic fields changes. This allows the motor to produce more torque without changing its steady-state rotational speed, provided it remains within its stable operating range.

An induction motor works differently. Its rotor relies on induced current, which requires relative motion between the rotor and the rotating field.

So don’t assume that every motor must have positive slip to produce torque. That requirement belongs to the normal induction-motor mechanism.

What if the rotor moves faster than synchronous speed?

There is another interesting case.

An induction machine can operate as a generator if an external mechanical source drives its rotor faster than the rotating magnetic field.

Consider a four-pole machine supplied at 50 Hz.

Its synchronous speed is:

\[ n_s=\frac{120(50)}{4} =1500\text{ rpm} \]

Suppose an external mechanical source drives the rotor at 1530 rpm.

The signed slip is:

\[ s=\frac{1500-1530}{1500} \]

\[ \boxed{s=-0.020=-2.0\%} \]

The slip is negative because the rotor is moving faster than the rotating magnetic field.

Under these conditions, the machine can generate electrical energy rather than consume electrical energy to drive a mechanical load.

The electromagnetic torque now opposes the rotor’s motion, acting as a braking torque. Mechanical energy supplied to the shaft is converted into electrical energy.

This is called generating operation.

The negative sign is meaningful. It tells you that the rotor is moving faster than the synchronous speed in the same direction.

For signed slip calculations, take the rotating field’s direction as positive. If the rotor rotates in the opposite direction, use a negative rotor speed rather than simply subtracting two positive speed magnitudes.

07Common mistakes when calculating slip

These are the mistakes most likely to cost you marks in a calculation or explanation question.

A useful calculation check is to ask yourself whether the result makes physical sense.

For a normal induction motor operating as a motor:

\[ 0<n_r<n_s \]

Therefore:

\[ 0<s<1 \]

And percentage slip must be between 0% and 100%.

If you calculate a slip of 150% for a normally operating induction motor, go back and check your speeds and denominator.

If you’re unsure about positive and negative motion quantities, the Speed and Velocity: HSC Physics Explained Clearly guide is a useful refresher.

08Questions and solutions

The following questions start with a standard slip calculation and build towards more demanding problems involving load changes, different pole numbers, measurement limitations, and generating operation.

Question 1

A six-pole induction motor operates on a 50 Hz supply. Its rotor rotates at 960 rpm.

Calculate:

(a) The synchronous speed.

(b) The slip speed.

(c) The percentage slip.

Assume the rotor and rotating magnetic field turn in the same direction.

Solution 1

The synchronous speed is 1000 rpm, the slip speed is 40 rpm, and the percentage slip is 4.0%.

Step 1: Calculate synchronous speed.

\[ n_s=\frac{120f}{P} \]

\[ n_s=\frac{120(50)}{6} \]

\[ n_s=1000\text{ rpm} \]

Step 2: Calculate slip speed.

\[ \Delta n=n_s-n_r \]

\[ \Delta n=1000-960 \]

\[ \boxed{\Delta n=40\text{ rpm}} \]

Step 3: Calculate percentage slip.

\[ s=\frac{1000-960}{1000} \]

\[ s=0.04 \]

\[ \boxed{\text{Slip}=4.0\%} \]

The rotor turns 40 rpm slower than the rotating magnetic field. This relative motion allows current to be induced in the rotor.

Question 2

A six-pole induction motor operates on a constant 50 Hz supply.

Two steady operating conditions are recorded:

• Condition A: rotor speed of 970 rpm.
• Condition B: rotor speed of 940 rpm.

Calculate the percentage slip and rotor current frequency for both conditions.

Explain why the synchronous speed remains unchanged, even though the rotor speed changes.

Solution 2

Condition A has 3.0% slip and a rotor current frequency of 1.5 Hz. Condition B has 6.0% slip and a rotor current frequency of 3.0 Hz.

Step 1: Calculate the synchronous speed.

Both conditions have the same supply frequency and pole count.

\[ n_s=\frac{120(50)}{6} \]

\[ n_s=1000\text{ rpm} \]

Step 2: Calculate slip for Condition A.

\[ s_A=\frac{1000-970}{1000} \]

\[ s_A=0.03 \]

Therefore:

\[ \boxed{s_A=3.0\%} \]

Calculate rotor current frequency:

\[ f_{r,A}=sf \]

\[ f_{r,A}=0.03(50) \]

\[ \boxed{f_{r,A}=1.5\text{ Hz}} \]

Step 3: Calculate slip for Condition B.

\[ s_B=\frac{1000-940}{1000} \]

\[ s_B=0.06 \]

Therefore:

\[ \boxed{s_B=6.0\%} \]

Calculate rotor current frequency:

\[ f_{r,B}=0.06(50) \]

\[ \boxed{f_{r,B}=3.0\text{ Hz}} \]

Step 4: Explain the change.

The supply frequency remains 50 Hz, and the motor still has six poles. Therefore, the synchronous speed remains 1000 rpm.

The lower rotor speed in Condition B indicates a greater speed difference between the rotor and the magnetic field.

This produces a greater slip and a higher induced rotor current frequency.

The slip has increased from 3% to 6%, which is an increase of 3 percentage points, or a 100% increase relative to its original value.

The key is that increasing mechanical load changes rotor speed, not synchronous speed.

Question 3

Two induction motors operate on separate 50 Hz AC supplies.

Motor A has six magnetic poles and a rotor speed of 970 rpm.

Motor B has four magnetic poles and a rotor speed of 1455 rpm.

A student argues that Motor B has greater slip because its speed difference is larger.

Determine:

(a) The synchronous speed of each motor.

(b) The slip speed of each motor.

(c) The percentage slip of each motor.

(d) The rotor current frequency of each motor.

Is the student’s argument correct? Explain.

Solution 3

Both motors have the same percentage slip of 3.0% and the same rotor current frequency of 1.5 Hz. The student’s argument is incorrect.

Step 1: Calculate Motor A’s synchronous speed.

Motor A has six poles:

\[ n_{s,A}=\frac{120(50)}{6} \]

\[ \boxed{n_{s,A}=1000\text{ rpm}} \]

Its slip speed is:

\[ \Delta n_A=1000-970 \]

\[ \boxed{\Delta n_A=30\text{ rpm}} \]

Step 2: Calculate Motor B’s synchronous speed.

Motor B has four poles:

\[ n_{s,B}=\frac{120(50)}{4} \]

\[ \boxed{n_{s,B}=1500\text{ rpm}} \]

Its slip speed is:

\[ \Delta n_B=1500-1455 \]

\[ \boxed{\Delta n_B=45\text{ rpm}} \]

Step 3: Calculate the percentage slip of each motor.

For Motor A:

\[ s_A=\frac{30}{1000} \]

\[ \boxed{s_A=3.0\%} \]

For Motor B:

\[ s_B=\frac{45}{1500} \]

\[ \boxed{s_B=3.0\%} \]

Step 4: Calculate the rotor current frequencies.

For Motor A:

\[ f_{r,A}=0.03(50) \]

\[ \boxed{f_{r,A}=1.5\text{ Hz}} \]

For Motor B:

\[ f_{r,B}=0.03(50) \]

\[ \boxed{f_{r,B}=1.5\text{ Hz}} \]

Step 5: Evaluate the student’s argument.

The student has correctly identified that Motor B has a larger absolute speed difference.

However, slip is not simply the difference between the two speeds. It is the difference divided by synchronous speed.

Motor B has a larger synchronous speed, so its 45 rpm difference represents the same fraction of synchronous speed as Motor A’s 30 rpm difference.

Both motors therefore have identical percentage slip and rotor current frequency.

This is why comparing slip speeds alone can be misleading when motors have different pole counts.

Question 4

A laboratory compares a four-pole induction motor and a four-pole synchronous motor. Both operate on a 50 Hz supply.

Each motor has a tachometer that rounds its display to the nearest 10 rpm.

At light load, both tachometers display 1500 rpm.

A mechanical load is then added. The synchronous motor continues to display 1500 rpm, while the induction motor displays 1470 rpm.

Assume the induction motor remains in normal motoring operation, has nonzero mechanical losses, and the synchronous motor remains within its stable operating range.

Answer the following:

(a) Calculate the approximate percentage slip of the loaded induction motor.

(b) Does the initial reading of 1500 rpm prove that the induction motor had zero slip? Determine the range of possible percentage slip values consistent with its initial display.

(c) Explain why the synchronous motor can maintain exactly synchronous speed under load, while the induction motor requires nonzero slip to produce induction torque.

Solution 4

The loaded induction motor has approximately 2.0% slip. Its initial 1500 rpm display does not prove zero slip; its actual slip must be greater than zero and no more than approximately 0.33%. The synchronous motor can maintain zero slip because it uses a different torque-producing mechanism.

Step 1: Calculate the synchronous speed.

Both motors have four poles and operate at 50 Hz.

\[ n_s=\frac{120(50)}{4} \]

\[ \boxed{n_s=1500\text{ rpm}} \]

Step 2: Calculate the loaded induction motor’s slip.

Using the displayed rotor speed of 1470 rpm:

\[ s=\frac{1500-1470}{1500} \]

\[ s=0.02 \]

Therefore:

\[ \boxed{\text{Slip}=2.0\%} \]

This is the nominal slip calculated from the displayed speed.

Step 3: Examine the initial tachometer reading.

A reading rounded to the nearest 10 rpm of 1500 rpm means the actual speed lies between 1495 rpm and 1505 rpm.

However, the induction motor is operating as a motor, so its rotor must be slower than the 1500 rpm rotating field.

Therefore:

\[ 1495\leq n_r<1500 \]

The smallest possible slip is greater than zero, while the largest possible slip occurs at 1495 rpm:

\[ s_{\max}=\frac{1500-1495}{1500} \]

\[ s_{\max}=0.00333 \]

Converting to a percentage:

\[ s_{\max}=0.333\% \]

Therefore:

\[ \boxed{0\%<s\leq0.33\%} \]

The tachometer cannot resolve the small difference between the rotor and rotating field.

This is a measurement limitation, not evidence that the induction motor has zero slip.

Step 4: Compare the two motor mechanisms.

The induction motor requires relative motion between the rotating field and rotor to induce current. Its rotor must therefore run slightly slower than synchronous speed during normal motoring.

Even under light load, it needs some electromagnetic torque to overcome friction, windage, and other losses.

The synchronous motor has a magnetised rotor. Its magnetic field interacts directly with the rotating stator field.

The rotor maintains a fixed angular relationship with that field. When load increases, this angular relationship changes, allowing the motor to produce more torque without changing its steady-state rotational speed.

Therefore, the synchronous motor can maintain exactly 1500 rpm, while the induction motor requires nonzero slip.

Question 5

A four-pole induction machine is connected to a 50 Hz electrical supply.

It is tested in two different steady operating conditions. In both conditions, the rotor turns in the same direction as the stator’s rotating magnetic field.

The following information is recorded:

The frequency instrument reports frequency magnitude only; it does not indicate phase sequence.

An engineer claims:

“Both conditions must have the same rotor speed because the rotor current frequencies are identical. The machine must also be operating in the same mode.”

Evaluate this claim.

(a) Calculate the synchronous speed at 50 Hz.

(b) Calculate the signed slip for both conditions.

(c) Determine the rotor speed in each condition.

(d) Explain why both conditions have the same rotor current frequency despite having different rotor speeds.

The supply frequency is then reduced to 40 Hz while the counterweight continues to hold the rotor at its original Condition B speed. Assume the electrical transient has settled, the rotating field remains in the same direction, and the supply can accept returned electrical power.

(e) Calculate the new synchronous speed, signed slip, and rotor current frequency.

(f) Explain whether the machine is still operating as a generator and describe the direction of its electromagnetic torque.

Solution 5

The engineer’s claim is incorrect. Condition A has a rotor speed of 1470 rpm and positive slip of 2%. Condition B has a rotor speed of 1530 rpm and negative slip of 2%. Both have a rotor current frequency magnitude of 1.0 Hz. At 40 Hz, Condition B has a slip of -27.5% and a rotor current frequency of 11 Hz, and it remains in generating operation.

This question tests an important distinction: rotor current frequency tells you the magnitude of the relative motion, but not whether the rotor is slower or faster than the rotating magnetic field.

Step 1: Calculate the synchronous speed at 50 Hz.

The machine has four poles:

\[ n_s=\frac{120(50)}{4} \]

\[ \boxed{n_s=1500\text{ rpm}} \]

Step 2: Calculate the magnitude of slip from the rotor current frequency.

For an induction machine:

\[ f_r=|s|f \]

Rearrange:

\[ |s|=\frac{f_r}{f} \]

Substitute:

\[ |s|=\frac{1.0}{50} \]

\[ \boxed{|s|=0.02} \]

Therefore, both conditions have a slip magnitude of 2%.

But this does not tell us the sign of slip.

We need to use the direction of energy transfer.

Step 3: Determine the signed slip for Condition A.

Condition A takes electrical power from the supply and uses it to drive a pump.

The machine is operating as a motor, so its rotor must turn slower than the rotating magnetic field.

Therefore, slip is positive:

\[ \boxed{s_A=+0.02} \]

Step 4: Determine the signed slip for Condition B.

Condition B receives mechanical power from the descending counterweight and returns electrical power to the supply.

The machine is operating as a generator.

For an induction generator, the rotor must turn faster than the rotating magnetic field.

Therefore, slip is negative:

\[ \boxed{s_B=-0.02} \]

Step 5: Calculate the rotor speeds.

Rearrange the slip equation:

\[ s=\frac{n_s-n_r}{n_s} \]

\[ n_r=n_s(1-s) \]

For Condition A:

\[ n_{r,A}=1500(1-0.02) \]

\[ \boxed{n_{r,A}=1470\text{ rpm}} \]

For Condition B:

\[ n_{r,B}=1500(1-(-0.02)) \]

\[ \boxed{n_{r,B}=1530\text{ rpm}} \]

The two rotor speeds are different, even though the rotor current frequency magnitudes are identical.

Step 6: Explain why the frequency readings match.

Rotor current frequency depends on the magnitude of slip:

\[ f_r=|s|f \]

Both conditions have a slip magnitude of 0.02 and a supply frequency of 50 Hz.

Therefore:

\[ f_r=0.02(50) \]

\[ \boxed{f_r=1.0\text{ Hz}} \]

The frequency instrument reports how rapidly the rotor currents alternate, but it does not reveal the sign of slip.

The relative motion is equally large in magnitude in both cases, but its direction is reversed.

This is the critical difference between the two operating modes.

Step 7: Recalculate synchronous speed when the supply changes to 40 Hz.

The pole count remains four.

\[ n_s=\frac{120(40)}{4} \]

\[ \boxed{n_s=1200\text{ rpm}} \]

The counterweight holds the rotor at 1530 rpm.

The rotor is now travelling considerably faster than the rotating magnetic field.

Step 8: Calculate the new signed slip.

\[ s=\frac{n_s-n_r}{n_s} \]

\[ s=\frac{1200-1530}{1200} \]

\[ s=-0.275 \]

Therefore:

\[ \boxed{s=-27.5\%} \]

Step 9: Calculate the new rotor current frequency.

Use the magnitude of slip:

\[ f_r=|s|f \]

\[ f_r=0.275(40) \]

\[ \boxed{f_r=11\text{ Hz}} \]

Step 10: Explain the direction of torque and energy transfer.

The rotor is still rotating faster than the stator’s magnetic field.

The machine therefore remains in generating operation.

Its electromagnetic torque acts opposite to the rotor’s direction of motion. This is a braking torque, so the descending counterweight must continue supplying mechanical energy to maintain the rotor’s speed.

That mechanical energy is converted into electrical energy and returned to the supply.

The key result is that the magnitude of slip determines the magnitude of rotor current frequency, but the sign of slip distinguishes motoring from generating operation.

09Where to go next

Calculating slip tells you how quickly the rotor moves relative to the rotating magnetic field. The next step is understanding how that relative motion produces induced rotor currents, electromagnetic torque, and changes in motor performance.

For the complete physical explanation, continue with AC Induction Motors: Rotating Fields, Rotor Currents and Slip.