free-body diagram
TRANSCRIPT
Physics 211: Lecture 6, Pg 1
Physics 211: Lecture 6Physics 211: Lecture 6
Today’s AgendaToday’s Agenda
RecapRecap Problems...problems...problems!!Problems...problems...problems!!
Accelerometer Inclined plane Motion in a circle
Physics 211: Lecture 6, Pg 2
Something from Opportunity Set 2Something from Opportunity Set 2
Physics 211: Lecture 6, Pg 3
Total acceleration = tangential plus radialTotal acceleration = tangential plus radial
Tangentialdeceleration
blue arrowis velocity Look at this in polar
coordinates
Tangential motionTangential accel
Radial motion?Radial accel
2 2TANG RADIALa a a
Physics 211: Lecture 6, Pg 4
Total acceleration = tangential plus radialTotal acceleration = tangential plus radial
Tangentialdeceleration
blue arrowis velocity Tang accel is given
Radial accel depends on v and R
aR=v2/R
Figure out v at bottom, to figure out a there.
2 2TANG RADIALa a a
Physics 211: Lecture 6, Pg 5
Total acceleration = tangential plus radialTotal acceleration = tangential plus radial
2 20 2 TANGv v a path length
Tangentialdeceleration
blue arrowis velocity Tang accel is given
Radial accel depends on v and R
aR=v2/R
Figure out v at bottom, to figure out a there.
2 2TANG RADIALa a a
given
Centripetal =v2
R
path length = R
2
Physics 211: Lecture 6, Pg 6
Physics 211: Lecture 6, Pg 7
Physics 211: Lecture 6, Pg 8
Physics 211: Lecture 6, Pg 9
Physics 211: Lecture 6, Pg 10
ReviewReview
Discussion of dynamics.
Review Newton’s 3 Laws
The Free Body DiagramFree Body Diagram
The tools we have for making & solving problems:» Ropes & Pulleys (tension)» Hooke’s Law (springs)
Physics 211: Lecture 6, Pg 11
Review: Pegs & PulleysReview: Pegs & Pulleys
Used to change the direction of forces An ideal massless pulley or ideal smooth peg will
change the direction of an applied force without altering the magnitude: The tension is the same on both sides The tension is the same on both sides for a massless rope!for a massless rope!
FF1 = -T ii ideal peg or pulley
FF2 = T jj
| FF1 | = | FF2 |
massless rope
Physics 211: Lecture 6, Pg 12
Review: SpringsReview: Springs
Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
FX = -kx Where x is the displacement from the equilibrium and k is the constant of proportionality.
relaxed position
FX = 0
x
Spring demo
Physics 211: Lecture 6, Pg 13
Lecture 6, Lecture 6, Act 1Act 1Springs Springs
A spring with spring constant 40 N/m has a relaxed length of 1 m. When the spring is stretched so that it is 1.5 m long, what force is exerted on a block attached to the end of the spring?
x = 0
M
k kM
x = 0 x = 1.5
(a) -20 N (b) 60 N (c) -60 N
x = 1
Physics 211: Lecture 6, Pg 14
Lecture 6, Lecture 6, Act 1Act 1Solution Solution
Recall Hooke’s law:
FX = -kx Where x is the displacement from equilibrium.
FX = - (40) ( .5)
FX = - 20 N
(a) -20 N (b) 60 N (c) -60 N
Physics 211: Lecture 6, Pg 15
Problem: Problem: Pendulum AccelerometerPendulum Accelerometer
A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road (duh, it’s Illinois), and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g.
tells you what a is, then. The pendulum is an accelerometer
a
i i
Physics 211: Lecture 6, Pg 16
Accelerometer...Accelerometer...
Draw a free body diagram for the mass: What are all of the forces acting?
m
TT (string tension)
mgg (gravitational force)
i i
RememberFnet=ma
Physics 211: Lecture 6, Pg 17
Accelerometer...Accelerometer...
Using components Using components (recommended) you decompose or resolve (same thing) F=ma into components:
ii: FX = TX = T sin = ma
jj: FY = TY mg = T cos mg = 0
TT
mgg
mmaa
jj
ii
TX
TY
Physics 211: Lecture 6, Pg 18
Accelerometer...Accelerometer...
Using components Using components :
ii: T sin = ma
jj: T cos - mg = 0
Eliminate T :
mgg
mmaaT sin = ma
T cos = mgtan
ag
TX
TY jj
ii
TT
It’s an accelerometer!For every there is a particular a
Physics 211: Lecture 6, Pg 19
Accelerometer...Accelerometer...
Let’s put in some numbers:
Say the car goes from 0 to 60 mph in 10 seconds: 60 mph = 60 x 0.45 m/s = 27 m/s. Acceleration a = Δv/Δt = 2.7 m/s2. So a/g = 2.7 / 9.8 = 0.28 (no units).
= arctan (a/g) = 15.6 deg
tan ag
a
Cart w/ accelerometer
helium accelerometer
Physics 211: Lecture 6, Pg 20
Another Problem: Inclined planeAnother Problem: Inclined plane
A block of mass m slides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration a ?
ma
Physics 211: Lecture 6, Pg 21
Inclined plane...Inclined plane...
Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only.
ma
ii
jj
What’s the “physics” here? Gravity provides downward force.Ramp constrains the motionWhere’s block going to go?
g
Physics 211: Lecture 6, Pg 22
Inclined plane...Inclined plane...Identify the forces in the problemIdentify the forces in the problem
free body diagramfree body diagram
Force components: consider x and y components separately: just draw forces acting on box
ii: mg sin =ma. a = g sin
jj: N - mg cos = 0. N = mg cos
mgg
NN
maa
ii
jj
mg cos
mg sin
Incline
Physics 211: Lecture 6, Pg 23
Angles of an Inclined planeAngles of an Inclined plane
ma = mg sin
mgN
The triangles are similar, so the angles are the same!
Prove the triangles are similar
Physics 211: Lecture 6, Pg 24
Lecture 6, Lecture 6, Act 2Act 2Forces and MotionForces and Motion
A block of mass M = 5.1 kg is supported on a frictionless ramp by a spring having constant k = 125 N/m. When the ramp is horizontal the equilibrium position of the mass is at x = 0. When the angle of the ramp is changed to 30o what is the new equilibrium position of the block x1?
(a) x1 = 20cm (b) x1 = 25cm (c) x1 = 30cm
x = 0
M
k
x 1 = ?
Mk
= 30o
Physics 211: Lecture 6, Pg 25
Lecture 6, Lecture 6, Act 2Act 2SolutionSolution
x 1
Mk
x
y
Choose the x-axis to be along downward direction of ramp. (smart)
Mg
FBD: The total force on the block is zero since it’s at rest. (aha!)
N
Fx,g = Mg sin
Force of gravity on block is Fx,g = Mg sin Consider x-direction:
Force of spring on block is Fx,s = -kx1
F x,s = -kx 1convenient coord system
0
Physics 211: Lecture 6, Pg 26
Lecture 6, Lecture 6, Act 2Act 2SolutionSolution
x 1
Mk
x
y
Since the total force in the x-direction must be 0:
F x,g = Mg sin
F x,s = -kx 1
Mg sinkx1 κsin Μgx1
θ
m20mN125
50sm189kg15x2
1 ....
Physics 211: Lecture 6, Pg 27
Problem: Two Blocks Problem: Two Blocks
Two blocks of masses m1 and m2 are placed in contact on a horizontal frictionless surface. If a force of magnitude F is applied to the box of mass m1, what is the force on the block of mass m2?
Intuition Technique (free body diagram or FBD)
mm11 mm22
F F2,1F1,2
Only force acting on 2Is F2,1
Physics 211: Lecture 6, Pg 28
Problem: Two BlocksProblem: Two Blocks
Realize that F = (mm11+ mm22) a :
Draw FBD of block mm22 and apply FNET = ma:
F2,1F2,1 = mm22 a
F / (mm11+ mm22) = a
m22,1
m2m1
FF
Substitute for a :
mm22
(m1 + m2)m2F2,1 F
Physics 211: Lecture 6, Pg 29
Problem: Two BlocksProblem: Two Blocks
What about the net force on mass 1?
F acts in +x-direction, but F1,2 acts in –x-direction
The net force on mass 1 matters
11
1 2NET onm
mF Fm m
Physics 211: Lecture 6, Pg 30
Problem: Tension and AnglesProblem: Tension and Angles
A box is suspended from the ceiling by two ropes making an angle with the horizontal. What is the tension in each rope?
mm
Physics 211: Lecture 6, Pg 31
Problem: Tension and AnglesProblem: Tension and Angles
Draw a FBD:
T1 T2
mg
Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0
T1sin T2sin
T2cos T1cos
jj
ii
Fx,NET = T1cos - T2cos = 0 T1 = T2
2 sin mg
T1 = T2 =Fy,NET = T1sin + T2sin - mg = 0
Physics 211: Lecture 6, Pg 32
Problem: Motion in a CircleProblem: Motion in a Circle
A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v.
What is the tension T in the string at the top of the rock’s trajectory?
R
v
TT
Tetherball
Physics 211: Lecture 6, Pg 33
Motion in a Circle...Motion in a Circle...
Draw a Free Body Diagram (pick y-direction to be down): We will use FFNET = maa (surprise) First find FFNET in y direction:
FFNET = mg +T
TTmgg
y y
Physics 211: Lecture 6, Pg 34
Motion in a Circle...Motion in a Circle...
FFNET = mg +T Acceleration in y direction is known: It’s centripetal…
maa = mv2 / R
mg + T = mv2 / R
T = mv2 / R - mg
R
TT
v
mgg
y y
F = ma
Physics 211: Lecture 6, Pg 35
Motion in a Circle...Motion in a Circle... What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp?
i.e. find v such that T = 0.
mv2 / R = mg + T
v2 / R = g
Notice that this doesnot depend on m.
R
mgg
v
T= 0
v Rg
Bucket
Physics 211: Lecture 6, Pg 36
Lecture 6Lecture 6, , Act 3Act 3Motion in a CircleMotion in a Circle
A skier of mass m goes over a mogul having a radius of curvature R. How fast can she go without leaving the ground?
R
mgg NN
vv
(a) (b) (c) Rgv =mRgv =mRg
v =
Trackw/ bump
Physics 211: Lecture 6, Pg 37
Lecture 6Lecture 6, , Act 3Act 3SolutionSolution
mv2 / R = mg - N
For N = 0:
R
v
mgg NN
v Rg
Physics 211: Lecture 6, Pg 38
Recap of Today’s lecture:Recap of Today’s lecture:
Example ProblemsExample Problems
Accelerometer Inclined plane (Text: example 6-1) Motion in a circle (Text: 3-5, 5-2, 9-1)
Look at textbook problems Chapter 4: # 49 Chapter 5: # 65, 69
Next time: Friction…
Physics 211: Lecture 6, Pg 39
Accelerometer...Accelerometer...
Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as straightforward):straightforward):
Find the total vector force FFNET:
TTmgg
FFTOT
m
TT (string tension)
mgg (gravitational force)
Physics 211: Lecture 6, Pg 40
Accelerometer...Accelerometer...
Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as systematic):systematic):
Find the total vector force FFNET: Recall that FFNET = ma:
So
maa
tan mamg
ag
TTmgg
tan ag
m
TT (string tension)
mgg (gravitational force)
Physics 211: Lecture 6, Pg 41
Inclined plane...Inclined plane...
Alternative solution using vectors:
m
mgN
a = g sin iiN = mg cos jj
ii
jj
How do we know what is?