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Torque

Learn how to find the torque exerted by a force.

What is torque?

Torque is a measure of the force that can cause an object to rotate about an axis. Just as force is what causes an object to accelerate in linear kinematics, torque is what causes an object to acquire angular acceleration.
Torque is a vector quantity. The direction of the torque vector depends on the direction of the force on the axis.
Anyone who has ever opened a door has an intuitive understanding of torque. When a person opens a door, they push on the side of the door farthest from the hinges. Pushing on the side closest to the hinges requires considerably more force. Although the work done is the same in both cases (the larger force would be applied over a smaller distance) people generally prefer to apply less force, hence the usual location of the door handle.
Figure 1: Opening a door with maximum torque.
Torque can be either static or dynamic.
A static torque is one which does not produce an angular acceleration. Someone pushing on a closed door is applying a static torque to the door because the door is not rotating about its hinges, despite the force applied. Someone pedaling a bicycle at constant speed is also applying a static torque because they are not accelerating.
The drive shaft in a racing car accelerating from the start line is carrying a dynamic torque because it must be producing an angular acceleration of the wheels given that the car is accelerating along the track.
The terminology used when describing torque can be confusing. Engineers sometimes use the term moment, or moment of force interchangeably with torque. The radius at which the force acts is sometimes called the moment arm.

How is torque calculated?

The magnitude of the torque vector tau for a torque produced by a given force F is
tau, equals, F, dot, r, sine, left parenthesis, theta, right parenthesis
where r is the length of the moment arm and theta is the angle between the force vector and the moment arm. In the case of the door shown in Figure 1, the force is at right angles (90degrees) to the moment arm, so the sine term becomes 1 and
tau, equals, F, dot, r.
The direction of the torque vector is found by convention using the right hand grip rule. If a hand is curled around the axis of rotation with the fingers pointing in the direction of the force, then the torque vector points in the direction of the thumb as shown in Figure 2.
Figure 2: Direction of the torque vector found with the right-hand rule.

How is torque measured?

The SI unit for torque is the Newton-meter.
In imperial units, the Foot-pound is often used. This is confusing because colloquially the pound is sometimes used as a unit of mass and sometimes force. What is meant here is pound-force, the force due to earth gravity on a one-pound object. The magnitude of these units is often similar as 1, space, N, m, \simeq, 1, point, 74, space, f, t, dot, l, b, s.
Measuring a static torque in a non-rotating system is usually quite easy, and done by measuring a force. Given the length of the moment arm, the torque can be found directly. Measuring torque in a rotating system is considerably more difficult. One method works by measuring strain within the metal of a drive shaft which is transmitting torque and sending this information wirelessly.

What role does torque play in rotational kinematics?

In rotational kinematics, torque takes the place of force in linear kinematics. There is a direct equivalent to Newton’s 2ⁿᵈ law of motion (F, equals, m, a),
tau, equals, I, alpha.
Here, alpha is the angular acceleration. I is the rotational inertia, a property of a rotating system which depends on the mass distribution of the system. The larger I, the harder it is for an object to acquire angular acceleration. We derive this expression in our article on rotational inertia.

What is rotational equilibrium?

The concept of rotational equilibrium is an equivalent to Newton’s 1ˢᵗ law for a rotational system. An object which is not rotating remains not rotating unless acted on by an external torque. Similarly, an object rotating at constant angular velocity remains rotating unless acted on by an external torque.
The concept of rotational equilibrium is particularly useful in problems involving multiple torques acting on a rotatable object. In this case it is the net torque which is important. If the net torque on a rotatable object is zero then it will be in rotational equilibrium and not able to acquire angular acceleration.
Exercise 1:
Consider the wheel shown in Figure 3, acted on by two forces. What magnitude of the force F, start subscript, 2, end subscript will be required for the wheel to be in rotational equilibrium?
Figure 3: A wheel acted on by two torques.

How does torque relate to power and energy?

There is considerable confusion between torque, power and energy. For example, the torque of an engine is sometimes incorrectly described as its 'turning power'.
Torque and energy have the same dimensions (i.e. they can be written in the same fundamental units), but they are not a measure of the same thing. They differ in that torque is a vector quantity defined only for a rotatable system.
Power however, can be calculated from torque if the rotational speed is known. In fact, the horsepower of an engine is not typically measured directly, but calculated from measured torque and rotational speed. The relationship is:
\begin{aligned} P &= \frac{\mathrm{Force} \cdot \mathrm{Distance}}{\mathrm{Time}} \\ & = \frac{\mathrm{F} \cdot 2\pi r}{t} \\ &= 2\pi \tau \omega \qquad \mathrm{(\omega~in~ revolutions/sec)} \\ &= \tau \omega \qquad \mathrm{(\omega~in~radian/sec)}\end{aligned}
Along with horsepower, the peak torque produced by a vehicle engine is an important and commonly quoted specification. Practically speaking, peak torque is relevant for generally describing how quickly a vehicle will accelerate and its ability to pull a load. Horsepower (relative to weight) on the other hand is more relevant to the maximum speed of a vehicle.
It is important to recognize that while maximum torque and horsepower are useful general specifications, they are of limited use when making calculations involving the overall motion of a vehicle. This is because in practice both vary as a function of rotational speed. The general relationship can be non-linear and differs for different types of motor as shown in Figure 4.
Figure 4: Relationship of available torque to rotational speed for different sources.

How can we increase or decrease torque?

It is often necessary to increase or decrease the torque produced by a motor to suit different applications. Recall that the length of a lever can increase or decrease the force on an object at the expense of the distance through which the lever must be pushed. Similarly, the torque produced by a motor can be increased or decreased through the use of gearing. An increase in torque comes with a proportional decrease in rotational speed. The meshing of two gear teeth can be viewed as equivalent to the interaction of a pair of levers as shown in Figure 5.
Figure 5: The meshing of two gears viewed as the interaction of two levers.
The use of adjustable gearing is necessary to obtain good performance in vehicles powered by combustion engines. These engines produce maximum torque only for a narrow range of high rotational speeds. Adjustable gearing allows sufficient torque to be delivered to the wheels at any given rotational speed of the engine.
Bicycles require gearing because of the inability of humans to pedal with a cadence sufficient to achieve a useful speed when driving a wheel directly (unless one is cycling a penny-farthing).
Adjustable gearing is not typically required in vehicles powered by steam engines or electric motors. In both cases, high torque is available at low speeds and is relatively constant over a wide range of speeds.
Exercise 2a:
A gasoline engine producing 150, space, N, m of torque at a rotational speed of 300, space, r, a, d, slash, s is used to drive a winch and lift a weight as shown in figure 6. The winch drum has a radius of 0.25 m and is driven from the engine via a 1:50 speed reduction gear. What mass could be raised with this setup? (Assume the winch is in rotational equilibrium, i.e. the mass is traveling up at constant velocity).
Figure 6: Engine-driven winch used to lift a mass (exercise 2).
Exercise 2b:
At what speed would the weight be traveling upward?

Data sources

Cyclist : Hansen, E.A, Smith G. Factors affecting cadence choice during submaximal cycling and cadence influence on performance. International Journal of Sports Physiology and Performance. March 2009; 4(1):3-17.
Diesel engine: Mercedes 250 CDI
Otto cycle engine: Mercedes E250
Electric motor: Tesla Model S 85
Steam locomotive: 2-8-0 "Consolidation" Locomotive at 70% boiler capacity
Penny-farthing : Wikimedia Commons

Want to join the conversation?

• please explain how the pulley works
• A pulley is a device that can change the direction of tension. For example, with a pulley, you can pull down on a rope, but the force of tension will be converted so it pulls to the right of an object.
(1 vote)
• Isnt a Newton-meter a joule?
• The units for torque (Newton-meter) and energy (Joules) are dimensionally the same, but torque and energy are not equivalent. The difference is that in the unit for torque there is also radians which we treat as dimensionless (ie. radians have no units) but that unit is still in the expression for torque which is not in the expression for energy.

So really energy unit is [N*m] or [J] and torque unit is [N*m*rads], but again radians are unitless since they are defined as the ratio of circumference/radius which is [m/m].
• In the middle of the article, it says P = (F * 2pi* r)/t.
F*r is torque. So P = ((torque)2pi)/t. Then I am stuck.

Why is P = 2pi
(torque)*(angular velocity)?
The unit of angular velocity is rad/sec. I don't know why the angular velocity pops up.
• My first reply was just wrong, and here is the correction:

You are correct that F*r = torque, and that angular velocity is rad/sec. To see why P = 2pi(torque)*(angular velocity) plug in the units rad/revolution next to 2pi, and like the article says plug in rev/sec next to angular velocity. The revolutions cancel and rad/sec is leftover. Since P = torque * angular velocity for the final equation the article shows in that part of the article, angular velocity is just 2pi rad/sec. I hope this is far more helpful than my first reply. This link https://en.wikipedia.org/wiki/Torque, in the "Conversion to other units" section may also be helpful.
• I didn't understand how we solved exercise 2b, can someone please explain it in a better way?
• The formula for finding the (tangential) velocity is v=rω
This comes from taking the derivative of both sides of s=rθ (where s is arc length) with respect to time.

We also know that the engine has an angular velocity of 300 but then is reduced by a factor of 50.

Engine        Reduction      Winch
speed               ratio            speed

Plugging this back into our equation gives us:
v = r ω
= 1.5 m/sec

(Does this help?)
• In 2a, can anyone explain how tau(w)=tau(m)/r? R is only the factor by which speed reduces right? How is that going to affect our torque?
• Power= τ*ω
P of winch drum= τ(of winch drum)*ω(of winch drum); or in other words:
P_w=τ_w *ω_w where that w stands for winch drum
The power remains the same, therefore, P of winch drum= P of machine; I will replace P of machine with P_m
P_w= P_m; therefore, τ_w *ω_w=τ_m *ω_m
The exercise says "is driven from the engine via a speed reduction gear"; therefore, ω_w=ω_m*R (R stands for ratio).
P_w=τ_w *ω_m *R P_w has to equal P_m; τ_w *ω_m *R= τ_m *ω_m;
τ_w *R= τ_m; or τ_w=τ_m/R
• Can we solve this problem using law of conservation of momentum?
• No, I suppose momentum will not remain conserved because a constant force is acting on the system. which will lead to acceleration and thus velocity will not remain constant. which will in turn vary momentum
• so how to tackle a question with the wheel size of a car and torque ? what happens when increased and decreased the radius of tire ? can u explain in detail ?
• Let us take the force at the edge of tyre that is applied on the road (and then road applies equal frictional force to push the vehicle forward) as F . The radius of tyre as r and then calculate the relation between the torque available at the wheel shaft and the force applied on the road.
F*r=T (T=torque at wheel) (angle is 90 degrees so we eliminate that)
F=T/r
Therefore, when we increase the radius of the wheel, the force available at the road decreases and it will be difficult accelerating the vehicle.

There are also other practical effects of increasing wheel size such as increased friction, raised centre of gravity among others.
• I don't get the answer on 2b.
Shouldn't the angular speed of the winch drum be 2π*6/s? (delta theta over delta time)
I get 3π for answer (12π*0.25)
What am I missing?
• We aren't solving for angular speed, rather tangential (or linear) velocity.

The formula for finding the tangential velocity is v=rω
This comes from taking the derivative of both sides of s=rθ (where s is arc length) with respect to time.

We also know that the engine has an angular velocity of 300 but then is reduced by a factor of 50.

Engine        Reduction      Winch
speed               ratio            speed

Plugging this back into our equation gives us:
v = r ω
= 1.5 m/sec

(remember that radians are a dimensionless unit)
• What is the relationship between torque and velocity? In Example 2 B, I wasn't sure where the formula came from to find the velocity...
• The formula for finding the (tangential) velocity is v=rω
This comes from taking the derivative of both sides of s=rθ (where s is arc length) with respect to time.

We also know that the engine has an angular velocity of 300 but then is reduced by a factor of 50.

Engine        Reduction      Winch
speed               ratio            speed

Plugging this back into our equation gives us:
v = r ω