Lecture 3

P3 Dynamics Mark Cannon

Hilary Term 2012

0-1

Lecture 3

Energy and Power

3-1

Review Newton’s second law: ⇐⇒

force = rate of change of momentum

F=

dV dm d mV = m +V dt dt dt

Impulse and momentum 2

Z ⇐⇒

impulse = change in momentum

F dt = m (V2 − V1 ) 1

During a frictionless collision: ? ? ? ?

no impulse acts parallel to the impact surface coefficient of restitution e determines the relative normal velocities if no external impulse acts, then momentum is conserved If e = 1 (perfectly elastic collision), then energy is conserved.

For continuous flow problems: force = rate of flux of momentum 3-2

Review

Incremental work done by a force F moving distance ds: dW = F · ds total work done in moving from position 1 to position 2: Z 2 W = F · ds . 1

Work done in moving an object of mass m between heights z1 and z2 : Z 2 W = mg dz = mg (z2 − z1 ) . 1

3-3

Energy

Energy can be: stored in and recovered from a mechanical system (e.g. flywheel, spring) dissipated through mechanical losses (e.g. friction, collisions) conserved if there are no losses

3-4

Kinetic Energy The kinetic energy of a particle is defined as T =

1 1 × mass × speed2 = mV 2 2 2 where V 2 = V · V

An alternative definition of kinetic energy is the work required to accelerate from rest to speed V hence the work required to change the kinetic energy from T1 to T2 is: W = T2 − T1

work and kinetic energy are both measured in joules (J)

3-5

Kinetic Energy Proof that Work = Change in Kinetic Energy m

Consider a variable force F acting on a particle of mass m at a variable position r

r

F

0

? Newton’s 2nd law gives F = m ¨r ? The work W done by F on m moving from position 1 to position 2 is Z 2 Z 2 ¨r · dr F · dr = m W = 1

but and

1 2

Z

dr dr = dt = r˙ dt dt d|˙r|2 r˙ · ¨r = 12 dt

=⇒ =⇒

¨r · r˙ dt

W =m W =

1 2

1Z

2

m 1

d|˙r|2 dt dt

so W =

1 m (|˙r2 |2 − |˙r1 |2 ) = T2 − T1 2

3-6

Kinetic Energy Example – Launching a boat 1 ms

force

-1

Find the distance travelled before stopping from the work done on the boat ? Work done is equal to change in kinetic energy: Fs =

1 m (V12 − V22 ) 2

? but m = 1500 kg, V1 = 1 ms2 , V2 = 0, and F = 70 × 9.81 × 0.3 so 70 × 9.81 × 0.3 × s =

1 × 1500 × (12 − 02 ) 2

giving s = 3.64 m

3-7

Potential Energy

The potential energy U is: the energy stored in a system due to its position e.g. in gravitational, electric or magnetic fields or because of elasticity

Any increase in potential energy is equal to the work done on the system In a conservative system this energy may be recovered as work

3-8

Potential Energy Example – hydroelectric energy storage

The pumped-storage plant at Raccoon Mountain, TN ? 80% efficiency (power recovered/power input) ? rapid start-up (e.g. 16 seconds to peak power output) 3-9

Potential Energy Examples Potential energy stored in a mass m raised from height z1 to height z2 . Work done:

2

Z

F dz = mg (z2 − z1 )

W = 1

hence U2 − U1 = mg (z2 − z1 ) Potential (or strain) energy stored in a stretched spring with stiffness k and unstretched length L1 Work done in stretching to length L2 : W =

1 k (L2 − L1 )2 2

but U1 = 0 when unstretched, giving U2 =

1 k (L2 − L1 )2 2

3 - 10

Potential Energy Example – Gravitational Potential Energy At the surface of the earth (radius R) the acceleration due to gravity is g . Elsewhere it is inversely proportional to the square of the radius r from the centre of the earth, so that the weight of a body of mass m at radius r is mg (R/r )2 . Find the potential energy U at radius r if U = 0 when r = ∞. Solution 2 R U(r ) = mg dr r Z 1 2 = mgR dr r2 mgR 2 +C =− r Z

But U → 0 as r → ∞ so C = 0. 3 - 11



Connection with Thermodynamics 1st law of Thermodynamics for a closed system (i.e. constant mass) gives heat input − work output = increase in (internal energy + mechanical energy) Increase in mechanical energy = ∆T + ∆U Increase in internal energy = energy change due to temperature increase In mechanics, heat output or increases in internal energy are typically caused by frictional losses Hence use ? work input = − work output ? heat output = − heat input to rearrange: work input

=

increase in mechanical energy + losses

losses

=

heat output + increase in internal energy

3 - 12

Work and Energy Examples Bending a longbow or crossbow, compressing the spring on an air gun, stretching the elastic on a catapult – work done is later released as kinetic energy.

Model of a self-propelling car designed by Leonardo da Vinci

3 - 13

Work and Energy Examples A flywheel on an internal combustion engine is used to compress the next charge of gas – kinetic energy is released as work.

BMW M5 engine 3 - 14

Work and Energy

Examples Driving a nail with a hammer – kinetic energy in the hammer provides momentum for a large impulse. Mechnical vibrations – kinetic energy and potential energy are repeatedly exchanged. A moving car stopped by its brakes – kinetic energy is lost as the brakes become hot.

3 - 15

Work and Energy Example – Mass on a slope z

m

s V

N mg

q

A mass m slides down a frictionless surface of slope θ. Find the speed V after the mass has descended through a vertical distance h.

3 - 16

Work and Energy Example – Mass on a slope z

m

s V

N mg

q

Solution: First, try solving by using equations of motion. Newton’s second law gives mg sin θ = ma = mV where

dV ds

dz sin θ Eliminate ds, and separate variables V and z to give ds = −

−g dz = V dV Integrate from z = 0 to z = −h to give V 2 = 2gh 3 - 16

Work and Energy Example – Mass on a slope z

m

s V

N mg

q

Solution contd: We can get a much simpler solution using energy. Since there are no losses, the sum of the kinetic and the potential energy remains constant. Hence 1 mV 2 = mgh 2 and so as before V 2 = 2gh

3 - 16

Work and Energy Example – Riding a bicycle up a hill D d

q

A cyclist of mass M rides a bicycle of mass m up a hill. The gearing is such that the bike moves a distance D along the road while the pedals rotate 180o . The diameter of the pedal crank is d and the cyclist rests his full weight on the down-going pedal throughout the rotation from the highest to the lowest position. Find the steepest slope that he can climb at a steady speed.

3 - 17

Work and Energy Example – Riding a bicycle up a hill D d

q

Solution: During a downward pedal stroke: ? the total energy remains constant, ? kinetic energy is unchanged (since constant speed), hence potential energy does not change. ? Rider, mass M falls d − D sin θ as the bicycle, mass m rises D sin θ, hence Mg (d − D sin θ) = mgD sin θ giving sin θ =

Md . (M + m)D

3 - 17

Work and Energy Example – Riding a bicycle up a hill D d

q

Comment The solution approaches sin θ = d/D if m  M then the rider is at the same height at the end of the pedal stroke as at the beginning work is done only as the rider steps up a distance d on to the next pedal. Question: why ride a bicycle up a hill when you could walk up the steps?

3 - 17

Power Power is the rate of doing work, with units of J s−1 = Watts (W) For linear motion: the work dW done in moving a force F a distance ds is dW = F · ds hence the rate of doing work is dW ds =F· dt dt or P =F·V in words power = force · velocity

3 - 18

Power Power is the rate of doing work, with units of J s−1 = Watts (W) For a rotating shaft: the work dW done turning a torque M through angle dθ is dW = M · dθ so the rate of doing work is dW dθ =M· =M·ω dt dt or P =M·ω in words power = torque · angular velocity

3 - 18

Power

Example – Acceleration of a car The engine of a car develops 100 bhp (brake horsepower) at 5000 rpm (revolutions per minute) and the power varies in direct proportion with the speed. The engine drives the road wheels through a gearbox which reduces the angular velocity by a factor of 7. The road wheels have radius 0.3 m. If there are no losses due to friction or air resistance, find the time required for the car to accelerate from 30 mph to 50 mph. Check that the engine can do this without exceeding 5000 rpm.

3 - 19

Power Solution: Conversion factors: 1 bhp = 746 W, 1 mile = 5280 ft, 1 ft = 0.3048 m First find the road thrust F Maximum engine speed: ωmax =

5000 × 2π = 523.60 rad s−1 60

Power at engine speed ω P = 100 × 746 ×

ω = 142.48 ω W 523.60

Engine torque Me =

P = 142.48 N m ω

Wheel torque Mw = 142.48 × 7 = 997.33 N m Road thrust F = 3 - 20

997.33 Mw = = 3324.43 N rw 0.3

Power Solution contd: Next check the engine speed: Road speed V30 = 30 mph =

30 × 5280 × 0.3048 = 3600

13.41 m s−1 22.35 m s−1

V50 = 50 mph = Engine speed at 50 mph ω50 =

60 22.35 ×7× = 4980.4 rpm < 5000 0.3 2π

=⇒ ok

Finally find the time taken: Acceleration

F 3324.43 = = 2.216 m s−2 m 1500 hence the required time is a=

t=

V50 − V30 22.35 − 13.41 = = 4.03 s a 2.216

3 - 21

Forces and moments A force F at right angles to a moment arm R produces a moment M: M

R

F

M = RF

If F is inclined at an angle α then M is reduced to M

R

a F M = RF cos α

M is right-handed about an axis facing out of the paper The moment can be represented as a vector M with magnitude M and direction along the right-handed axis direction A moment about the axis of spin of a shaft is called a torque

3 - 22

Right-Hand Axes Sign Convention Stretch the thumb and first two fingers of your right hand in perpendicular directions. Remember the axis orientation via

z +ve

axis 1 (or x) = first finger

thumb

+ve

y

axis 2 (or y ) = second finger

second finger

axis 3 (or z) = thumb first finger

Positive rotations are clockwise to an observer facing away from the origin.

+ve x

Hence if you point your right-hand thumb in the direction of any axis, then positive rotation about that axis is in the direction that your fingers curl.

3 - 23

Moment vector definition The components of moment acting about O are     Mx −Rz Fy + Ry Fz My  =  Rz Fx − Rx Fz  Mz −Ry Fx + Rx Fy The RHS can be replaced by an equivalent matrix multiplication: 0      Mx 0 −Rz Ry Fx My  =  Rz   0 −Rx Fy  Mz −Ry Rx 0 Fz

Fz

z

Fy Mz

Fx y Rx

Rz

Ry Mx x

The RHS is the matrix representation of a vector cross product (HLT p.19) so we can define the moment of F about O by M=R×F

3 - 24

My

Summary Kinetic energy of a particle: T = 12 mV 2 Potential energy in a system: U = energy stored due to its position in a field work input: W = ∆T + ∆U + losses Power input required to move a force F at velocity V: P =F·V Representation of moments as vectors M=R×F Power input required to rotate a moment M at angular velocity ω: P =M·ω

3 - 25