Topic 1: Newtonian Mechanics Energy & Momentummrslibretto.weebly.com/uploads/1/3/3/1/13312103/unit_15... · 2019-08-24 · Topic 1: Newtonian Mechanics ! In the previous example,

¨  Work (W)– the amount of energy transferred by a force acting through a distance. ¤ Scalar – but can be positive or negative

Units – N � m or Joules (J)

Topic 1: Newtonian Mechanics

Energy & Momentum

ΔE =W = F!d = Fdcosθ


Energy & Momentum

¨  Power (P) – the rate at which work is done and hence energy is transferred.

•  Units – J/s or Watts

Work, Energy & Power

P = ΔE

Δt = W

Δt= Fdcosθ

Δt= Fv cosθ


¨  Is work being performed on the satellite by the force of gravity?

Energy & Momentum

¨  Work may be done by individual forces, but there is no net work done on an object when . . .

. . . the net force is zero. The object is in equilibrium (at rest or constant velocity)


Energy & Momentum

¨  ratio of useful work done by a system to the total work done by the system

¨  ratio of useful energy output of a system to the total energy input to the system

¨  ratio of useful power output of a system to total power input to the system

e =

outputinput

x 100


Energy & Momentum


Energy & Momentum

K = ½ mv2

Ug = mgΔy

Us = ½ kx2 Ue = Pt = VIt

Q = mcΔT = mL


¨  Elastic Potential Energy Energy & Momentum

Displacement (m)

Force (N)

x

Fmax

Favg

!Fs = k

!x

Us =

12

kx2 = Fsx


¨  Gravitational Potential Energy Energy & Momentum

Path Independence: the change in Ug is the same for any path between the same two locations

ΔUg = mgΔy


¨  Conservative & Non-Conservative Forces Energy & Momentum



The total mechanical energy of an object remains constant if the net work done by external non-conservative forces is zero, Wnc = 0 J.

Eo = Ef



In the case of non-conservative forces being present, we can state:

Eo + Wnc = Ef


¨  One of the fastest roller coasters in the world is the Magnum XL-200 at Cedar Point Park in Sandusky, Ohio. The ride includes a vertical drop of 59.4 m. Assume that the coaster has a speed of nearly zero as it crests the top of the hill. Neglect friction and find the speed of the riders at the bottom of the hill.

Energy & Momentum


¨  One of the fastest roller coasters in the world is the Magnum XL-200 at Cedar Point Park in Sandusky, Ohio. The ride includes a vertical drop of 59.4 m. Assume that the coaster has a speed of nearly zero as it crests the top of the hill. Neglect friction and find the speed of the riders at the bottom of the hill.

Energy & Momentum

Eo = Ef

Ug = K

mgh = 1/2 mv2

v = 2gh = (2)(9.8 m/s2)(59.4 m) = 34.1 m/s


¨  In the previous example, we ignored non-conservative forces, such as friction. In reality, however, such forces are present when the roller coaster descends the hill. The actual speed of the riders at the bottom is 32.2 m/s. Assuming again that the coaster has a speed of nearly zero at the top of the hill, find the work done by non-conservative forces on a 55.0 kg rider during the descent down the hill.

Energy & Momentum


¨  In the previous example, we ignored non-conservative forces, such as friction. In reality, however, such forces are present when the roller coaster descends the hill. The actual speed of the riders at the bottom is 32.2 m/s. Assuming again that the coaster has a speed of nearly zero at the top of the hill, find the work done by non-conservative forces on a 55.0 kg rider during the descent down the hill.

Energy & Momentum

Eo + Wnc = Ef

Ug + Wnc = K

Wnc = K - Ug

Wnc = 1/2 mv2 - mgh

Wnc = 1/2 (55.0 kg)(32.2 m/s)2 - (55.0 kg)(9.8 m/s2)(59.4 m) = - 3536 J

Wnc = 3.5 x 103 J


¨  A pendulum is allowed to swing freely from rest. Derive the possible tension in the string as the pendulum bob passes through its lowest position.

Energy & Momentum



Energy & Momentum



Energy & Momentum

FT

Fg

Fc = ma

FT - mg = mv2

r

FT = mg + mv2

LWhat is v?



Energy & Momentum

FT

Fg

Fc = ma

FT - mg = mv2

r

FT = mg + mv2

L

Eo = EfUg = K

mgh = 1/2 mv2

gL = 1/2 v2

v2 = 2gL



Energy & Momentum

FT

Fg

Fc = ma

FT - mg = mv2

r

FT = mg + mv2

L

Eo = EfUg = K

mgh = 1/2 mv2

gL = 1/2 v2

v2 = 2gL

FT = mg + m(2gL)

LFT = mg + 2mg

FT = 3mg


¨  A rollercoaster has a “loop-de-loop” section as shown. Derive the minimum height the car must start from to successfully make it through the loop?

Energy & Momentum



Energy & Momentum

Eo = Ef

mgho = mghf + 1/2 mv2

gho = g(2r) + 1/2 v2

Loop height = 2R

What is v?



Energy & Momentum

∑F = ma

FN + Fg = mv2

r

0 N + mg = mv2

rv2 = gr

Eo = Ef


gho = g(2r) + 1/2 v2



Energy & Momentum

Eo = Ef


gho = g(2r) + 1/2 v2

gho = g (2r) + 1/2 ( g r)

ho = 2.5 r

∑F = ma

FN + Fg = mv2

r

0 N + mg = mv2

rv2 = gr


Energy & Momentum


¨  Linear Motion - the quantity of motion of a moving body, measured as a product of its mass and velocity.

Energy & Momentum

�Units – kg � m/s

Which object possesses more momentum?

!p = m

!v


¨  Linear Motion - the quantity of motion of a moving body, measured as a product of its mass and velocity.

Energy & Momentum

�Units – kg � m/s

When could granny possess more momentum than the truck?

!p = m

!v


Energy & Momentum ¨  Impulse (J) – the change in momentum of a

system ¤ Vector


¨  Derivation of Impulse Energy & Momentum

J = Δ!p =!FΔt = m

!v


¨  Calculate the impulse experienced by the 1500 kg car as it bounce off the wall.

Energy & Momentum


¨  Calculate the impulse experienced by the 1500 kg car as it bounce off the wall.

Energy & Momentum

J = Δp = mΔv = m (v - vo)

J = Δp = 1500kg +2.60m/s - (-15.0m/s)( )J = Δp = 2.64x104 kg im/s


¨  Compare the forces, impact time, impulse, acceleration, and change in momentum for each vehicle in the impending collision.

Energy & Momentum



Energy & Momentum

Newton’s 3rd Law FT = FC



Energy & Momentum

FT = FC Newton’s 2nd Law

ma = ma So

aT < aC



Energy & Momentum

FT = FC &

ΔtT = ΔtC So

JT = JC ΔpT = ΔpC



Energy & Momentum

ΔpT = ΔpC

mΔv = mΔv So

ΔvT < ΔvC


Energy & Momentum Conservation of Momentum – the total momentum of an isolated system remains constant.

Isolated System – the next external forces acting on the system is zero.

po = pf


¨  Elastic collision - a collision in which the total kinetic energy is conserved ¤ po = pf, Eto = Etf, (mechanical) Eo = Ef, KEo = KEF

¨  Inelastic collision - a collision in which the total kinetic energy is not conserved ¤ po = pf, Eto = Etf, (mechanical) Eo ≠ Ef, KEo ≠ KEF

Energy & Momentum


¨  A ballistic pendulum consists of a block of wood (m2) suspended by a wire of negligible mass. A bullet (m1) is fired with a speed v01. Just after the bullet collides with it, the block (with the bullet in it) has a speed vf and then swings to a maximum height (h) above the initial position (see part b of the

drawing). Find the speed v01 of the bullet, assuming that air resistance is negligible.

Energy & Momentum

Can we say that the KE of bullet in (a) is equal to the Ug of bullet + block in (b)? Why?


¨  A ballistic pendulum consists of a block of wood (m2) suspended by a wire of negligible mass. A bullet (m1) is fired with a speed v01. Just after the bullet collides with it, the block (with the bullet in it) has a speed vf and then swings to a maximum height (h) above the initial position (see part b of the

drawing). Find the speed v01 of the bullet, assuming that air resistance is negligible.

Energy & Momentum

What about using these two points?


Energy & Momentum

Kbot = ΔUgtop

1/2mv2 = mgh

v = 2gh

KE Ug

po = p

m1v1 + m2v2 = (m1 + m2)v

m1v1 = (m1 + m2) 2gh

v1 = (m1 + m2) 2gh

m1


¨  Conservation of Momentum in Two Dimensions ¤ Momentum is a vector quantity, and is conserved in an

isolated system. ¤ However, in two dimensions, the x and y components of

the total momentum are conserved separately.

Energy & Momentum

Rotational Motion

¨  Translational Motion – all points on a body travel on parallel path (not necessarily straight lines)

¨  Rotational Motion – all points on a body travel in a circular path with a common axis. ¤  can be combined with translational motion

Rotational Motion

¨  Axis of Rotation – an imaginary line with which an object rotates around

¨  Angular Displacement (Δθ) – the angle through which a rigid object rotates about a fixed axis

¤ Units – radians (rad) or degrees (˚) or revolution (rev)

axis of rotation

Δθ = θ - θo

θ =

Sr

Rotational Motion

¨  For a full circle the arc length (S) or circumference, is equivalent to 2πr. Therefore:

1 revolution = 360° = 2πr and

1 radian =

360°2π

= 180°π

= 57.3°

Rotational Motion

¨  Angular Velocity (ω) – the rate of change of angular displacement

¤ Units – radians per second (rad/s)

¨  Angular Acceleration (α) – the time rate change of angular velocity

¤ Units – radians per second2 (rad/s2)

ω =

ΔθΔt

= 2π f

α =

Δωt

Rotational Motion

¤  If an object changes its angular velocity at a constant rate, the average angular velocity is midway between the initial and final values:

ω =

ω i + ω f

2

Rotational Motion

ω = ωo + αt

θ = θo + ωot +1/2 αt2

ω2 = ωo2 + 2α (θ – θo)

¨  A bicycle wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the initial angular speed of the wheel is 2.00 rad/s: ¤  through what angle does the wheel

rotate in 2.00 seconds? ¤  what is the angular speed of the wheel

after 2.00 s?

Rotational Motion

¨  A bicycle wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the initial angular speed of the wheel is 2.00 rad/s: ¤  through what angle does the wheel

rotate in 2.00 seconds? ¤  what is the angular speed of the wheel

after 2.00 s?

θ = θo + ωot + 1/2α t2

θ = (2.00 rad/s)(2.00 s) + 1/2(3.50 rad/s2)(2.00 s)2

θ = 4 rad + 7 radθ = 11 radians

11 radians 57.3°1 rad

⎛⎝⎜

⎞⎠⎟

= 630.3°

ω = ωo + α t

ω = (2.00 rad /s) + (3.50 rad/s2)(2.00 s)ω = 9.00 rad/s

630.3° 1 rev360°

⎛⎝⎜

⎞⎠⎟

= 1.75 revolutions

11 rad 1 rev2πrad

⎛⎝⎜

⎞⎠⎟

= 1.75 revolutions

Rotational Motion

¨  In the familiar ice-skating stunt known as “crack the whip”, a number of skaters attempt to maintain a straight line as they skate around a central person who remains in place. Compare the angular velocity and the tangential velocity of the skaters as shown.

Rotational Motion

¨  Uniform Circular Motion – constant speed and constant radius ¤ Acceleration present – only centripetal acceleration

¨  Non-Uniform Circular Motion – non-constant speed ¤ Acceleration present –

1.  Centripetal Acceleration – rate of change of direction of linear (tangential) velocity

2.  Tangential Acceleration - rate of change of magnitude of linear (tangential) velocity

3.  Angular Acceleration – rate of change of angular velocity

Rotational Motion

Rotational Motion

¨  When the tires roll, there is a relationship between the angular speed at which the tires rotate and the linear speed (assumed constant) at which the car moves forward.

Similar reasoning reveals that:

Moment of Inertia

¨  Moment of Inertia (I) – “rotational inertia” - a quantity expressing a body’s tendency to resist angular acceleration. ¤ depends on mass and distribution of mass

¤ Units – kg m2

I = ∑m r2

Moment of Inertia

Moment of Inertia

¨  A solid circular disk has a mass of 1.2 kg and radius of 0.16 m. Each of three identical thin rods has a mass of 0.15 kg. The rods are attached perpendicularly to the plane of the disk at its outer edge to for a three-legged stool (see the drawing). Find the moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center.

Moment of Inertia

¨  A solid circular disk has a mass of 1.2 kg and radius of 0.16 m. Each of three identical thin rods has a mass of 0.15 kg. The rods are attached perpendicularly to the plane of the disk at its outer edge to for a three-legged stool (see the drawing). Find the moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center.

Istool = Idisk +3Irod

Istool =12 MdiskR

2 +3MrodR2

Istool =12 1.2 kg( ) 0.16 m( )2

+3 0.15 kg( ) 0.16 m( )2

Istool = 0.027 kg ⋅m2

Torque of a Rotating Body

τ = r⊥ F = r F sinθ

A bicycle wheel can be accelerated by either pulling on the chain that is on the gear or by pulling on a string wrapped around the tire. The wheel’s radius is 0.38 meter, while the radius of the gear is 0.14 meter. If you obtained the needed acceleration with a force of 15-Newtons on the chain, what force would you need to exert on the strap?


τ = r⊥ F = r F sinθ

A bicycle wheel can be accelerated by either pulling on the chain that is on the gear or by pulling on a string wrapped around the tire. The wheel’s radius is 0.38 meter, while the radius of the gear is 0.14 meter. If you obtained the needed acceleration with a force of 15-Newtons on the chain, what force would you need to exert on the strap?

τ chain = τwheel

Fchainrgear = Fstringrwheel

Fstring = Fchainrgear

rwheel

Fstring = (15 N)(0.14 m)

0.38 mFstring = 5.5 N


¨  Newton’s 2nd Law can be rewritten to represent rotational motion as:

¨  The advantage of using τ = I α is that it can be applied to any rigid body rotating about a fixed axis, and not just to a particle.


¨  A merry-go-round made of a solid steel wheel has a mass of 150 kilograms and a diameter of 4.4 meters. A person pushes the handlebar of the stationary merry-go-round for 15.0 seconds to make it rotate to a final speed of 8.0 rev/s. Determine: ¤  the torque applied to the merry-go-round.

¤  how much force was exerted on the handlebar.


¨  A merry-go-round made of a solid steel wheel has a mass of 150 kilograms and a diameter of 4.4 meters. A person pushes the handlebar of the stationary merry-go-round for 15.0 seconds to make it rotate to a final speed of 8.0 rev/s. Determine: ¤  the torque applied to the merry-go-round.

¤  how much force was exerted on the handlebar.

τ = I α

τ = (1/2 mr2) (Δω

t)

τ = (1/2 mr2) (2π f

t)

τ = 1/2 (150 kg)(2.2 m)2 2π (8.0 rev/s)

15.0 sτ = 1216.42 N im = 1.2 x 103 N im

F =

τr

= 1.2 x 103 N im

2.2 m = 550 N

Rotational Work

¨  A rope is wrapped around a wheel and is under a constant tension F. If the rope is pulled out a distance s, the wheel rotates through an angle of:

¨  The rotational work can be determined since:

θ = sr

W = F dW = F s

W = F r θ

WR = τ θ

Rotational Kinetic Energy

¨  The rotational work done by a net external torque causes the rotational kinetic energy to change. The kinetic energy of the entire rotating body is the sum of the kinetic energies of the particles.

K = ∑ 12

m vT 2

K = ∑ 12

m r2 ω 2

K = (∑ 12

I) ω 2

KR = 12

I ω 2

Total Mechanical Energy

ET

TotalMechanical

Energy

! = 12

mv2

Translationalkinetic energy

"#$ +

12

Iω 2

Rotationalkinetic energy

! + mgh

Gravitationalpotential energy

!

Which cylinder will reach the bottom of the incline first?

The solid cylinder wins!!!

Eo = Ef

12

mvo2 +

12

Iωo2 + mgho =

12

mvf2 +

12

Iω f2 + mghf

mgho = 12

mvf2 +

12

Iω f2

mgho = 12

mvf2 +

12

I vf

r⎛⎝⎜

⎞⎠⎟

2

vf = 2 mgho

m + Ir2

Hoop

vf = 2 mgho

m + mr2

r2

vf = gho

Solid Cylinder

vf = 2 mgho

m + 1 / 2 mr2

r2

vf = 4gho

3 = 1.15 gho

Angular Momentum

Conservation of Angular Momentum

¨  Conservation of Angular Momentum – the total angular momentum of a system remains constant (is conserved) if the net external torque acting on the system is zero.


¨  An artificial satellite is placed into an elliptical orbit about the earth, as shown in the figure. As the satellite travels from perigee to apogee determine how the following quantities will change: ¤ gravitational force: ¤  tangential speed: ¤ kinetic energy: ¤  linear momentum: ¤ angular momentum:


¨  An artificial satellite is placed into an elliptical orbit about the earth, as shown in the figure. As the satellite travels from perigee to apogee determine how the following quantities will change:

¤ gravitational force:

¤  tangential speed:

¤ kinetic energy:

¤  linear momentum:

¤ angular momentum:

Fg =

GmEms

r2 = (−)(−)(−)

(↓)2 = ↓Fg

⎛⎝⎜

⎞⎠⎟

K = 1/2 mv2 = (-)(-)( ↓ )2 = ↓K( )

v =

F mE

r =

( ↓ )(-)( ↑ )

= ↓ v⎛

⎝⎜

⎞

⎠⎟

p = mv = (-)( ↓ ) = ↓ p( )

L = mvr = (-)( ↓ )( ↑ ) = same L( )

Fc = Fg

msv2

r =

GmEms

r2

Can also use Lo = Lf.


¨  A 34.0 kg child runs with a speed of 2.80 m/s tangential to the rim of a stationary merry-go-round. The merry-go-round has a moment of inertia of 510 kg m2 and a radius of 2.31 m. When the child jumps onto the merry-go-round, the entire system begins to rotate. What is the angular speed of the system?


¨  A 34.0 kg child runs with a speed of 2.80 m/s tangential to the rim of a stationary merry-go-round. The merry-go-round has a moment of inertia of 510 kg m2 and a radius of 2.31 m. When the child jumps onto the merry-go-round, the entire system begins to rotate. What is the angular speed of the system?

Lo = Lf

mcvr = (I + mcr2)ω

ω = mcvr

I + mcr2

ω = (34.0 kg)(2.80 m/s)(2.31 m)

(510 kg m2 + (34.0 kg)(2.31 m)2

ω = 3.18 rad/s

Angular Impulse

¨  Angular Impulse – Angular Momentum theorem – The angular impulse on an object is equal to the object’s final angular momentum minus the object’s initial angular momentum.

τnet = I α

τnet = I ω f - ω i

tτnet t = I (ω f - ω i )

τnet t = Lf - Li

ΔL = τ Δt

Documents

Topic 1: Newtonian Mechanics Energy & Momentummrslibretto.weebly.com/uploads/1/3/3/1/13312103/unit_15... · 2019-08-24 · Topic 1: Newtonian Mechanics ! In the previous example,