*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

**, covered during time**

*S***, by the time**

*t***to get the speed:**

*t*

*V = S/t*In case of

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

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

*t*

*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

*is the rate, at which the*

**power***work*is performed.

If during the time

**the**

*t**work*performed is

**, we define the**

*W**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

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

**power**

*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

**that moves it forward must be equal in**

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

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

**must be constant as well.**

*F*The distance

**covered as a function of time**

*S***is**

*t*

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

**during the time**

*F***is**

*t*

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

*power*follows that the power this force

**exhorts is**

*F*

*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

**moves as a result of action of force**

*m***, where**

*F(t)***is time. The distance it covers is**

*t***.**

*S(t)*Then during an infinitesimal time interval from

**to**

*t*

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

**t***d*

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

**,**

*F(t) = m·a(t)*where

**is acceleration.**

*a(t)*Increment of distance is

*d*,

**S(t) = V(t)·**d**t**where

**is an object's speed.**

*V(t)*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= m·V·

**V(t)**Power exhausted by this force is, therefore

*P(t) =**d*

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

= m·V·d= m·V·

**V(t)/**d**t =**

= m·V(t)·a(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)=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

**and we lift it with constant linear speed**

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

*V***, where**

*ω=V/R***- radius of a well's wheel.**

*R*Since the speed is constant, the force

**that acts on a bucket equals to**

*F***, where**

*m·g***- acceleration of the free fall.**

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

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

*R**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

**in terms of unknown force**

*P***and linear speed of a bucket:**

*F*

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

**and a radius of its shaft:**

*F*

*τ = F·R*Substituting torque

**for**

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

*F·R*

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

**and formula for power in case of rotation with constant angular speed**

*P=F·V***. Instead of force**

*P=τ·ω***in case of straight line motion, we use torque**

*F***for rotation and, instead of linear speed**

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

*V***for rotation.**

*ω*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

= 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.

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