classical mechanics lecture 9 - sfu.ca · mechanics((lecture(9,(slide(14 h....

Classical Mechanics Lecture 9

Today's Concepts:a) Energy and Fric6on

b) Poten6al energy & force

Mechanics Lecture 9, Slide 1

Joke of the dayWhen a third grader was asked to cite Newton's first law, she said, "Bodies in mo9on remain in mo9on, and bodies at rest stay in bed unless their

mothers call them to get up."


Some comments about the courseI understand the concepts and can do the math by thinking logically about the

problem, ge@ng the right answer. However, when I watch FlipItPhysics prelectures, I tend to get lost with all the nota9on and integrals and end up confused even though I knew what I was doing before. A more logical explana9ons of the math involved would benefit me way more than the complex mathema9cal explana9ons that I've learned to just tune out.


HW is harder than lecture

Need more prac4ce materials

Don’t know how to prepare for exam

Material is ge=ng harder

Common Issues:

Things we can do:

Do some HW problems in class

Provide prac4ce materials to prepare for exam

Post formula sheets

Make sure office hours are u4lized



Macroscopic Work done by Friction

Macroscopic Work:

This is not a new idea – it’s the same “work” you are used to.

Applied to big (i.e. macroscopic) objectsrather than point par4cles (picky detail)

We call it “macroscopic” to dis4nguish it from “microscopic”.


W =Z b

a~F · d~l

f f


Do spinning Heat Demo

“Heat” is just the kine9c energy of the atoms!


H

N1

mg

N2

mg

µ mg

must be nega6ve

m


A block of mass m, ini4ally held at rest on a fric4onless ramp a ver4cal distance H above the floor, slides down the ramp and onto a floor where fric4on causes it to stop a distance r from the boMom of the ramp. The coefficient of kine4c fric4on between the box and the floor is µk. What is the macroscopic work done on the block by fric4on during this process?

A) mgH B) –mgH C) µk mgD D) 0

D

m

CheckPoint


H

What is the macroscopic work done on the block by fric4on during this process?


B) All of the poten6al energy goes to kine6c as it slides down the ramp, then the fric6on does nega6ve work to slow the box to stop

C) Since the floor has fric6on, the work done by the block by fric6on is the normal force 6mes the coefficient of kine6c fric6on 6mes the distance.


CheckPoint

D

m

H

A block of mass m, ini4ally held at rest on a fric4onless ramp a ver4cal distance H above the floor, slides down the ramp and onto a floor where fric4on causes it to stop a distance D from the boMom of the ramp. The coefficient of kine4c fric4on between the box and the floor is µk. What is the total macroscopic work done on the block by all forces during this process?


CheckPoint


D

m

H

What is the total macroscopic work done on the block by all forces during this process?


C) The only work being done on the object is by the fric6on force. D) total work is equal to the change in kine6c energy. since the box starts and ends at rest, the change in kine6c energy is zero.

B) work= change in poten6al energy


CheckPoint

D

m

H

Potential Energy vs. Force


Demo

Potential Energy vs. Force


Suppose the poten4al energy of some object U as a func4on of x looks like the plot shown below.

Where is the force on the object zero?A) (a) B) (b) C) (c) D) (d)

U(x)

x

(a) (b) (c) (d)

Clicker Question



Where is the force on the object in the +x direc4on?A) To the leS of (b) B) To the right of (b) C) Nowhere

U(x)

x

(a) (b) (c) (d)

Clicker Question



Where is the force on the object biggest in the –x direc4on?A) (a) B) (b) C) (c) D) (d)

U(x)

x

(a) (b) (c) (d)

CheckPoint


L

v

Conserve Energy from ini6al to final posi6on


Homework

L

mg

T


Homework

mg

T


Homework

Conserve Energy from ini6al to final posi6on.

hh


Homework

h

mgT

r


Challenge Problem

Problem Solving: Problems with Two or More Objects S E C T I O N 4 - 8 | 113

FA1

m2m1

F I G U R E 4 - 3 1

5. Because the rope is of negligible mass and slides over the icewith negligible friction, the forces and are simply related.Express this relation:

TS

2TS

1

T2 ! T1 ! T

6. Substitute the steps-4 and -5 results into the step-2 and step-3equations: "T # mPg ! mPat

T # mSg sinu ! mSat

CHECK If is very much greater than we expect the acceleration to be approximately gand the tension to be approximately zero. Taking the limit as approaches 0 does indeed give

and for this case. If is much less than we expect the acceleration to beapproximately g (see step 3 of Example 4-7) and the tension to be zero. Taking the limit as

approaches 0 in steps 7 and 8, we indeed obtain at and At an extreme valueof the inclination we again check our answers. Substituting in steps 7 and 8, weobtain and This seems right since Steve and Paul would be in free-fall for

TAKING IT FURTHER In Step 1 we chose down the incline and straight down to be positiveto keep the solution as simple as possible. With this choice, when Steve moves in the di-rection (down the surface of the glacier), Paul moves in the direction (straight downward).

PRACTICE PROBLEM 4-10 (a) Find the acceleration if and if the masses areand (b) Find the acceleration if these two masses are interchanged.mP ! 92kg.mS ! 78kg

u ! 15°

#x$#x

u! 90°.T! 0.at! gu! 90°(u! 90°)T! 0.at! g sinumP

sinumS ,mPT ! 0at ! gmS

mS ,mP

7. Solve the step-6 equations for the acceleration by eliminating Tand solving for at :

mS sinu # mP

mS # mPgat !

8. Substitute the step-7 result into either step-6 equation and solvefor T:

mSmP

mS # mP(1 " sinu)gT !

FA1

y

F21

m1

F12

m2

x

F I G U R E 4 - 3 2

2. Apply to box 1.©FS

! maS FA1 " F21 ! m1a1x


! maS F12 ! m2a2x

4. Express both the relation between the two accelerations andthe relation between the magnitudes of the forces the blocksexert on each other. The accelerations are equal because thespeeds are equal at all times, so the rate of change of thespeeds are equal. The forces are equal in magnitude becausethe forces constitute a N3L force pair:

F21 ! F12 ! Fa2x ! a1x ! ax

CHECK Note that the result in step 5 is the same as if the force had acted on a singlemass equal to the sum of the masses of the two boxes. In fact, because the two boxes havethe same acceleration, we can consider them to be a single particle with mass m1 # m2 .

FS

A1

5. Substitute these back into the step-2 and step-3 results andsolve for ax .

FA1

m1 # m2ax !

(b) Substitute your expression for into either the step-2 or thestep-3 result and solve for F.

axm2

m1 # m2FA1F !

Example 4-13 Building a Space Station

You are an astronaut constructing a space station, and you push on a box of mass with forceThe box is in direct contact with a second box of mass (Figure 4-31). (a) What is the

acceleration of the boxes? (b) What is the magnitude of the force each box exerts on the other?

PICTURE Force is a contact force and only acts on box 1. Let be the force exerted by box2 on box 1, and be the force exerted by box 1 on box 2. In accord with Newton’s third law,these forces are equal and opposite , so Apply Newton’s second law to eachbox separately. The motions of the two boxes are identical, so the accelerations and are equal.

SOLVE

(a) 1. Draw free-body diagrams for the two boxes (Figure 4-32).

aS2aS1

F21 !F12 .(FS

21 ! "FS

12)FS

12

FS

21FS

A1

m2FS

A1.

m1

Problem Solving: Problems with Two or More Objects S E C T I O N 4 - 8 | 113

FA1

m2m1

F I G U R E 4 - 3 1

5. Because the rope is of negligible mass and slides over the icewith negligible friction, the forces and are simply related.Express this relation:

TS

2TS

1

T2 ! T1 ! T

6. Substitute the steps-4 and -5 results into the step-2 and step-3equations: "T # mPg ! mPat

T # mSg sinu ! mSat

CHECK If is very much greater than we expect the acceleration to be approximately gand the tension to be approximately zero. Taking the limit as approaches 0 does indeed give

and for this case. If is much less than we expect the acceleration to beapproximately g (see step 3 of Example 4-7) and the tension to be zero. Taking the limit as

approaches 0 in steps 7 and 8, we indeed obtain at and At an extreme valueof the inclination we again check our answers. Substituting in steps 7 and 8, weobtain and This seems right since Steve and Paul would be in free-fall for

TAKING IT FURTHER In Step 1 we chose down the incline and straight down to be positiveto keep the solution as simple as possible. With this choice, when Steve moves in the di-rection (down the surface of the glacier), Paul moves in the direction (straight downward).

PRACTICE PROBLEM 4-10 (a) Find the acceleration if and if the masses areand (b) Find the acceleration if these two masses are interchanged.mP ! 92kg.mS ! 78kg

u ! 15°

#x$#x

u! 90°.T! 0.at! gu! 90°(u! 90°)T! 0.at! g sinumP

sinumS ,mPT ! 0at ! gmS

mS ,mP

7. Solve the step-6 equations for the acceleration by eliminating Tand solving for at :

mS sinu # mP

mS # mPgat !

8. Substitute the step-7 result into either step-6 equation and solvefor T:

mSmP

mS # mP(1 " sinu)gT !

FA1

y

F21

m1

F12

m2

x

F I G U R E 4 - 3 2


! maS FA1 " F21 ! m1a1x


! maS F12 ! m2a2x

4. Express both the relation between the two accelerations andthe relation between the magnitudes of the forces the blocksexert on each other. The accelerations are equal because thespeeds are equal at all times, so the rate of change of thespeeds are equal. The forces are equal in magnitude becausethe forces constitute a N3L force pair:

F21 ! F12 ! Fa2x ! a1x ! ax

CHECK Note that the result in step 5 is the same as if the force had acted on a singlemass equal to the sum of the masses of the two boxes. In fact, because the two boxes havethe same acceleration, we can consider them to be a single particle with mass m1 # m2 .

FS

A1

5. Substitute these back into the step-2 and step-3 results andsolve for ax .

FA1

m1 # m2ax !

(b) Substitute your expression for into either the step-2 or thestep-3 result and solve for F.

axm2

m1 # m2FA1F !

Example 4-13 Building a Space Station

You are an astronaut constructing a space station, and you push on a box of mass with forceThe box is in direct contact with a second box of mass (Figure 4-31). (a) What is the

acceleration of the boxes? (b) What is the magnitude of the force each box exerts on the other?

PICTURE Force is a contact force and only acts on box 1. Let be the force exerted by box2 on box 1, and be the force exerted by box 1 on box 2. In accord with Newton’s third law,these forces are equal and opposite , so Apply Newton’s second law to eachbox separately. The motions of the two boxes are identical, so the accelerations and are equal.

SOLVE

(a) 1. Draw free-body diagrams for the two boxes (Figure 4-32).

aS2aS1

F21 !F12 .(FS

21 ! "FS

12)FS

12

FS

21FS

A1

m2FS

A1.

m1

Practise Problem

Blocks on InclineA frictionless surface is inclined at an angle of 30.0° to the horizontal. A 270-g block on the ramp is attached to a 75.0-g block using a pulley, as shown in the figure. The pulley is massless and frictionless.

(a) Draw two free-body diagrams, one for the 270-g block and the other for the 75.0-g block.

(b) Find the tension in the string and the acceleration of the 270-g block.

(c) The 270-g block is released from rest. How long does it take for it to slide a distance of 1.00 m along the surface? Will it slide up the incline, or down the incline?

Block on Incline Solution

Chapter 4

364

Substitute numerical values and

evaluate Δx:

( )( ) cm 1.6

km/s 10.02

m/s 5.3Δ2

2

=−−

=x

(c) Apply a constant-acceleration equation to relate the stopping time of the head to its initial and final speeds and to its acceleration:

tavv xxx Δ0 += ⇒x

xx

avv

t 0Δ −=


evaluate Δt: ms 35km/s10.0

m/s 5.30Δ2=

−−

=t

90 •• A frictionless surface is inclined at an angle of 30.0º to the horizontal. A 270-g block on the ramp is attached to a 75.0-g block using a pulley, as shown in Figure 4-62. (a) Draw two free-body diagrams, one for the 270-g block and the other for the 75.0-g block. (b) Find the tension in the string and the acceleration of the 270-g block. (c) The 270-g block is released from rest. How long does it take for it to slide a distance of 1.00 m along the surface? Will it slide up the incline, or down the incline?

Picture the Problem The application of Newton’s 2nd law to the block and the hanging weight will lead to simultaneous equations in their common acceleration a and the tension T in the cord that connects them. Once we know the acceleration of this system, we can use a constant-acceleration equation to predict how long it

takes the block to travel 1.00 m from rest. Note that the magnitudes of T and

'T are equal. (a) The free-body diagrams are shown to the right. m270 represents the mass of the 270-g block and m75 the mass of the 75.0-g block.

Tr xy

θ

nFr 'T

r

x

gmr

270gmr

75

(b) Apply ∑ = xx maF to the block

and the suspended mass:

xamgmT 1270270 sin =− θ

and

xamTgm 27575 =−

Block on Incline solutionNewton’s Laws

365

Letting a represent the common

acceleration of the two objects,

eliminate T between the two

equations and solve a:

gmm

mma

75270

27075 sin

+−

=θ

Substitute numerical values and evaluate a:

( ) ( ) 222m/s71.1m/s 706.1m/s9.81

kg0.270kg0.0750

sin30kg0.270kg0.0750−=−=

+°−

=a

where the minus sign indicates that the acceleration is down the incline. Substitute for a in either of the

force equations to obtain:

N864.0=T

(c) Using a constant-acceleration

equation, relate the displacement of

the block down the incline to its

initial speed and acceleration:

( )221

0 ΔΔΔ tatvx xx +=

or, because v0x = 0,

( )221 ΔΔ tax x= ⇒

xax

tΔ2Δ =


evaluate Δt: ( )

s08.1m/s1.706

m1.002Δ2=

−−

=t

Because the block is released from rest and its acceleration is negative, it will

slide down the incline. 91 •• A box of mass m1 is pulled along a frictionless horizontal surface by a

horizontal force F that is applied to the end of a rope (see Figure 4-63). Neglect

any sag of the rope. (a) Find the acceleration of the rope and block, assuming

them to be one object. (b) What is the net force acting on the rope? (c) Find the

tension in the rope at the point where it is attached to the block.

Picture the Problem Note that, while the mass of the rope is distributed over its

length, the rope and the block have a common acceleration. Because the surface

is horizontal and smooth, the only force that influences our solution isFG

. The

figure misrepresents the situation in that each segment of the rope experiences a

gravitational force; the combined effect of which is that the rope must sag.

(a) Apply totnet / mFa

GG= to the rope-

block system to obtain:

21 mm

Fa+

=

Practise Problem: LoopLoop

You are designing a new roller-coaster. The main feature of this particular design is to be a vertical circular loop-the-loop where riders will feel like they are being squished into their seats even when they are in fact upside-down (at the top of the loop).

The coaster start at rest a height of 80m above the ground, speeds up as it descends to ground level, and then enters the loop which has a radius of 20m. Suppose a rider is sitting on a bathroom scale that initially reads W (when the coaster is horizontal and at rest). What will the scale read when the coaster is moving past the top of the loop?

(You can assume that the coaster rolls on the track without friction).

Loop Solution

Physics 211 Week 4

Work and Kinetic Energy: Loop (Solutions)

You are designing a new roller-coaster. The main feature of this particular design is to be a vertical circular loop-the-loop where riders will feel like they are being squished into their seats even when they are in fact upside-down (at the top of the loop). The coaster start at rest a height of 80m above the ground, speeds up as it descends to ground level, and then enters the loop which has a radius of 20m. Suppose a rider is sitting on a bathroom scale that initially reads W (when the coaster is horizontal and at rest). What will the scale read when the coaster is moving past the top of the loop? (You can assume that the coaster rolls on the track without friction). ______________________________________________________________________

You can use conservation of energy to find the velocity of the coaster at the top of the loop. The coaster will begin with gravitational potential energy, which will change to kinetic energy as the coaster begins moving. You can use the value of the kinetic energy to determine an expression for the velocity in terms of gravity, the initial height, and the radius of the loop. With the values given in the problem, you know that the initial height is 4 times the radius of the loop. You can use this relationship to simplify the velocity expression to be only in terms of the acceleration due to gravity and the radius. Next, you can use Newton’s second law to determine the normal force on the coaster at the top of the loop. This can be done by combining the normal force and force of gravity to obtain an expression for the mass times the centripetal acceleration. You then use your velocity from the conservation of energy and insert it into the centripetal acceleration equation. If you solve the expression for the normal force, you should obtain the value N = 3mg. The normal force is the value that the scale will read. A scale would usually read mg, so the rider experiences what we would call 3g’s.

H

R

mg N

classical mechanics lecture 9 - sfu.ca · mechanics((lecture(9,(slide(14 h....

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