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Lecture

chapter 5 this is kind of tutorial that help you with chap 5


Department
Physics and Astronomy
Course Code
PHYS 1010
Professor
Marko Horbatsch

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Phys 205 Chapter 5 Solutions
5.3 The Earth has a radius of 6.4×106m ad completes one revolution about its axis in 24 h. (a) Find the speed
of a point on the equator. (b) Find the speed of New York City.
This is a kinematics of uniform circular motion problem.
Given: Radius of the Earth: rE= 6.4×106m
Period of rotation: T= 24h= 86,400 s
(a) We want to find the tangential speed of a point at the equator. Use Equation 5.1.
vEq =2πrE
TvEq =2π6.4×106m
86,400 s = 465 m/s
(b) New York City has a latitude of 40.7so the radius of its orbit will not be the full radius of the Earth (see
the figure for Problem 5.9 below). The radius of its orbit will be: r=rEcos 40.7= 4.85 ×106m.
vNY =2πr
TvNY =2π4.85 ×106m
86,400 s = 353 m/s
5.6 Consider the motion of the hand of a mechanical clock. If the minute hand of the clock has a length of 6.0 cm,
what is the centripetal acceleration of a point at the end of the hand?
This is a kinematics of uniform circular motion problem.
Given: Period T= 60 min = 3600 s
Radius: r= 0.060 m
We want to find the centripetal acceleration, ac. Use Equations 5.5 and 5.1.
ac=v2
r=1
r2πr
T2
ac=4π2r
T2
This last equation is useful to use for problems like this one when you are given the period & radius and want
the centripetal acceleration.
ac=4π2(0.060 m)
(3600 s)2= 1.8×107m/s2
5.9 Consider points on the Earth’s surface as sketched in the figure. Because of the Earth’s
rotation, these points undergo uniform circular motion. Compute the centripetal acceler-
ation of (a) a point at the equator, and (b) at latitude of 30.
This is a kinematics of uniform circular motion problem.
Given: Radius of the Earth: rE= 6.4×106m
Period of rotation: T= 24h= 86,400 s
(a) We want to find the centripetal acceleration, ac, of a point at the equator. Use the
equation from the previous problem.
!"#$
ac=4π2r
T2ac=4π26.4×106m
(86,400 s)2= 0.034 m/s2
(b) At a latitude of 30the radius of orbit will be: r=rEcos 30= 5.54 ×106m.
ac=4π2r
T2ac=4π25.54 ×106m
(86,400 s)2= 0.029 m/s2
1

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5.12 When a fighter pilot makes a quick turn, he experiences a centripetal acceleration. When this acceleration
is greater than about 8 ×g, the pilot will usually lose consciousness (“black out”). consider a pilot flying at
a speed of 900 m/s who wants to make a sharp turn. What is the minimum radius of curvature she can take
without blacking out?
This is a kinematics of uniform circular motion problem.
(a) Given: Maximum centripetal acceleration: ac= 8 ×g= 78.4 m/s2
Tangential speed: v= 900 m/s
We want to find the minimum radius of curvature which corresponds to the maximum centripetal acceleration.
Use Equation 5.5.
ac=v2
rr=v2
ac
r=(900 m/s)2
78.4 m/s2= 1.0×104m
The minimum radius is roughly 10 km.
5.14 The Daytona 500 stock car race is held on a track that is approximately 2.5 mi long, and the turns are banked
at an angle of 31. It is currently possible for cars to travel through the turns at a speed of 180 mi/h. Assuming
these cars are on the verge of slipping into the outer wall of the racetrack (because they are racing!), find the
coefficient of static friction between the tires and the track.
This is a dynamics of uniform circular motion problem.
Given: Track length: L= 2.5 mi = 4025 m
Tangential velocity: v= 180 mi/h = 80.5 m/s
Bank angle: θ= 31
We want to find the coefficient of static friction, µs, assuming the friction force is
at a maximum at this speed. A diagram will help (see right). In order to find the
required centripetal acceleration, we need to know the radius of curvature of the
banked turns. If we assume a circular track, the radius would be L/2π= 0.398 mi.
The actual track shape is much more oval so I’ll estimate r= 0.20 mi = 322 m.
The centripetal acceleration is:
ac=v2
r=(80.5 m/s)2
322 m = 20.1 m/s2= 2.05g
!"#$
!
!
!
!
F
g
!
!
F
fr
!
!
F
N
!
y
!
x
Applying Newton’s 2nd Law:
X~
F=m~ac
xcomponent: FNsin θFfr cos θ=mac
ycomponent: FNcos θFfr sin θmg = 0
If the static friction force is at a maximum, we have Ffr =µsFN, and this gives us:
FNsin θ+µsFNcos θ=macFNcos θµsFNsin θ=mg
FN(sin θ+µscos θ) = macFN(cos θµssin θ) = mg
This gives us two equations with three unknowns (FN,m, and µs). We want to solve this for µs. In order to
isolate µs, I’m going to divide the equation on the left by the equation on the right:
FN(sin θ+µscos θ)
FN(cos θµssin θ)=mac
mg (sin θ+µscos θ)
(cos θµssin θ)=ac
gsin θ+µscos θ= (cos θµssin θ)ac
g
Both the mass and the normal force cancel out and I’m left with one equation for µs. Isolate µson one side:
µs(cos θ+ 2.05 sin θ) = 2.05 cos θsin θµs=2.05 cos 31sin 31
cos 31+ 2.05 sin 31= 0.65
Your answer will vary depending on your estimate of the radius of curvature.
2
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5.16 Consider the motion of a rock tied to a string of length 0.50 m. The string is spun
so that te rock travels in a vertical circle as shown in figure. The mass of the rock
is 1.5 kg, and it is twirling at constant speed with a period of 0.33 s.
(a) Draw free-body diagrams for the rock when it is at the top and when it is at
the bottom of the circle. Your value should include the tension in the string
but the value of FTis not yet known.
(b) What is the total force on the rock directed towards the center of the circle?
(c) Find the tension in the string when the rock is at the top and when it is at
the bottom of the circle.
!"#$
!
!
!
!
F
g
!
!
F
N
!
y
!
x
!
r=0.50 m
!
m=1.5 kg
This is a dynamics of uniform circular motion problem.
Given: Radius: r= 0.50 m
Period: T= 0.33 s
Mass: m= 1.5 kg
Calculate:
Centripetal acceleration: ac=4π2r
T2= 181 m/s2
Centripetal force: mac= 272 N
(a) Free body diagrams for the rock at the top and at the bottom of the circle.
Free body diagram at the top:
h
v0
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FN
FT
mg
m2
m3
m1
F
N
Ff,max
mg
Fyou
Fleg
T
T
m
FT
mg
Free-body diagram at the bottom:
h
v0
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FS
FN
FT
mg
m2
m3
m1
F
N
Ff,max
mg
Fyou
Fleg
T
T
m
FT
mg
(b) If the rock is in uniform circular motion, then the force directed towards the center of the circle is always
mac= 272 N.
(c) At the top both FTand Fgcontribute to the centripetal force. At the bottom, FThas to overcome Fgto
provide the centripetal force.
Top: FT+mg =macFT=macmg FT= (1.5 kg) 181 m/s29.80 m/s2= 257 N
Bottom: FTmg =macFT=mac+mg FT= (1.5 kg) 181 m/s2+ 9.80 m/s2= 286 N
5.20 A roller coaster track is designed so that the car travels upside down on a certain
portion of the track as shown in the figure. What is the minimum speed the roller
coaster can have without falling from the track? Assume the track has a radius of
curvature of 30 m.
This is a dynamics of uniform circular motion problem.
!
r
!
!
v
Given: Radius of curvature: r= 30 m
We want to find the minimum speed that keeps the roller coaster car on the track. At high speeds, the normal
force from the track on the car keeps the car in circular motion. At slower speeds, the normal force becomes
smaller. At the minimum speed, gravity provides all of the centripetal acceleration at the top. At slower
speeds, the car would fall.
X~
F=m~a mg =macg=v2
r
v=gr v =q(9.80 m/s2)(30 m) = 17 m/s
3
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