*Notes to a video lecture on http://www.unizor.com*
__Definition of Mechanical Power__
To analyze the motion, we often use a concept of

*speed*.

Let's assume that an object moves in some inertial reference frame, and

the distance covered by it from its initial position along its

trajectory is a function of time

*S(t)*.

Recall the definition of

*speed* of an object as the rate, at which this object covers the

*distance* along its motion along a trajectory.

In case of a

*uniform* motion we can simply divide the distance

*S*, covered during time

*t*, by the time

*t* to get the speed:

*V = S/t*
In case of

*non-uniform* motion the speed changes and at any particular moment of time

*t* an instantaneous speed can be calculated using differentials:

*V(t) = **d***S(t)/**d**t**
To analyze the mechanical work performed to achieve certain results, we often use a concept of

*power*.

Let's assume that something or someone performs certain work and, as the time goes by, the work performed is a function of time

*W(t)*.

The

**power** is the rate, at which the

*work* is performed.

If during the time

*t* the

*work* performed is

*W*, we define the

*average power* of the whoever or whatever performs the work as

*P = W/t*
Most likely, at equal in length but different time intervals the amount

of work performed will be different. For example, when a car starts, its

engine should give a car an acceleration, which requires more work per

unit of time than to maintain a constant speed on a smooth straight

road.

In cases like this we can talk about an instantaneous

**power** as a function of time that can be calculated using differentials:

*P(t) = **d***W(t)/**d**t**
Consider an example of an object in uniform motion against the force of friction with a constant speed

*V*.

The force

*F* that moves it forward must be equal in

magnitude and opposite in direction to the force of friction to maintain

the constant speed. Since the friction is constant, the force

*F* must be constant as well.

The distance

*S* covered as a function of time

*t* is

*S(t) = V·t*
Therefore, the work performed by the force

*F* during the time

*t* is

*W(t) = F·S(t) = F·V·t*
From the definition of

*power* follows that the power this force

*F* exhorts is

*P(t) = **d***W(t)/**d**t = F·V**
As an example, the car engine exhorts the same power and consumes the

same amount of gas per unit of time, if the car uniformly moves along a

straight road. This power is used to generate a force sufficient to

overcome the friction of wheels and air resistance.

Consider a more general case, when the motion is not uniform.

Assume, an object of mass

*m* moves as a result of action of force

*F(t)*, where

*t* is time. The distance it covers is

*S(t)*.

Then during an infinitesimal time interval from

*t* to

*t+**d***t** the work performed by this force will be

*d***W(t) = F(t)·**d**S(t)**
Considering the Newton's second law,

*F(t) = m·a(t)*,

where

*a(t)* is acceleration.

Increment of distance is

*d***S(t) = V(t)·**d**t**,

where

*V(t)* is an object's speed.

Also, by definition of acceleration,

*a(t) = **d***V(t)/**d**t**
Therefore,

*d***W(t) = (m·**d**V(t)/**d**t)·V(t)·**d**t =**

= m·V·d**V(t)**
Power exhausted by this force is, therefore

*P(t) = **d***W(t)/**d**t =**

= m·V·d**V(t)/**d**t = **

= m·V(t)·a(t)
From the definition of

*power* as amount of work per unit of time or, more precisely, the first derivative of work by time

*P(t) = **d***W(t)/**d**t**
follows that the unit of measurement of power is

*joule/sec* called

*watt*.

Expanding the definition of

*joule* as

*newton·meter*,

**1 watt = 1J/sec = 1N·m/sec**
Obvious extensions of unit of power

*watt* are

*kilowatt = 1,000 watt* and

*megawatt = 1,000,000 watt*.

There is an old unit of power called

*horsepower*.

*Metric horsepower*, derived from lifting up against a force of gravity on Earth a weight of mass 75

*kg* with a constant speed of 1

*m/sec*, is related to

*watt* unit as

*1(metric HP) =*

=75(kg)·9.8(m/sec²)·1(m/sec)≅

≅ 735.5(W)
For historical reasons there is also a

*mechanical horsepower*, defined as 33,000

*pound-feet per minute*, related to

*watt* unit as

*1(mechanical HP) ≅ 745.7(W)*
So, a car engine of 200

*mechanic horsepower* has the power of about 149,140

*watt*.

*Watt*, as a unit of measurement, was called in honor of James

Watt, an 18th century Scottish scientist who was one of the first to

research a concept of power, developed steam engines and measured the

power of a horse.

Now let's address the concept of

*power* in a case of rotation with constant angular speed. An example is lifting a bucket of water from a well.

Assume, a bucket of water has a mass

*m* and we lift it with constant linear speed

*V* with an angular speed of the well's wheel

*ω=V/R*, where

*R* - radius of a well's wheel.

Since the speed is constant, the force

*F* that acts on a bucket equals to

*m·g*, where

*g* - acceleration of the free fall.

At the same time, if the wheel is turned by some motor and

*R* is the radius of its shaft, the motor manufacturer provides technical characteristic not only of the

*power*, but also of a

*torque* of a motor.

Remember that the

*torque* equals to

*τ = F·R*
So, on one hand, we have expressed the power of a motor

*P* in terms of unknown force

*F* and linear speed of a bucket:

*P = F·V = F·R·ω*
On another hand, we expressed the torque of this motor in terms of the same unknown force

*F* and a radius of its shaft:

*τ = F·R*
Substituting torque

*τ* for

*F·R* in the formula for power, we can find the relationship between the power of a motor and is torque:

*P = τ·ω*
Notice the similarity between the formula for power in case of uniform motion along a straight line

*P=F·V* and formula for power in case of rotation with constant angular speed

*P=τ·ω*. Instead of force

*F* in case of straight line motion, we use torque

*τ* for rotation and, instead of linear speed

*V* for straight line motion, we use angular speed

*ω* for rotation.

Let's check this with real data about a particular engine.

Below is a graph representing the power and torque of Ford Motor Company 6.7L Power Stroke diesel V-8.

As you see, the power and torque grow relatively monotonically until

some engine limitations start playing significant role. While in the

area of monotonic growth, we can take a particular angular speed, say,

1400 RPM (revolutions per minute) and see that the power of an engine

equals approximately 174 HP (mechanical horsepower) and the torque is

about 650 Lb-Ft (pound-feet).

Let's check if the relationship between power and torque derived above is held in this case.

First of all, we transform all units into standard physical measures defined in SI:

*1 RPM (revolutions per minutes) =*

= 2π radian per minute =

= 2π/60 radians/sec =

= 0.1048 rad/sec
*1 HP = 745.7 W = 745.7 J/sec*
*1 Lb-Ft = 1.35582 J*
Substituting all the above, we see that

angular speed is equal to

*1400·0.1048 = 146.6 rad/sec*
power is

*174·745.7 = 129,751.8 J/sec*
torque equals to

*650·1.35582 = 881.3 J*
Now we can check the relationship between the power, torque and angular speed

*P = τ·ω*
Indeed,

*881.3·146.6 = 129,198.6*
As we see, the difference between power and a product of torque by

angular speed is minimal, attributable to imprecise measurement of the

parameters and graph reading.