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Physics 140 – Fall 2014
September 16
Projectile & Circular Motion
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Reminder:
• Mastering Physics assignment 1 due tonight at midnight
1
Questions on
today’s youtube
video?
3
Shooting a Monkey (or Bear)
• There was a clever monkey who, seeing that a hunter
was about to shoot, decided to drop from the tree at
the instant that he saw the flash from the gun as the
bullet left the barrel, and thus escape the bullet. The
hunter aimed the gun directly at the monkey at the
monkey and fired. What happened?
Let’s find out!
A. The bullet hit the monkey
B. The bullet missed the monkey and passed above it
C. The bullet missed the monkey and passed below it
Shooting a monkey
1 2
gt
2
Tilted Rocket problem – part 1 (based on P3.47)
A test rocket is launched by accelerating it along a 200.0-m
incline at 1.25 m/s2 starting from rest at point A on the diagram.
The incline rises at 35.0° above the horizontal, and at the instant
the rocket leaves the incline, its engines turn off and it is subject
only to gravity (ignore air resistance).
What is the speed of the rocket as it leaves the incline?
A)
B)
C)
D)
E)
11.2 m/s
22.4 m/s
33.6 m/s
44.8 m/s
56.0 m/s
v02  2as
 v0  2(1.25 m/s2 )(200. m)  22.4 m/s
v0
6
Tilted Rocket problem – part 2 (based on P3.47)
A test rocket is launched by accelerating it along a 200.0-m
incline at 1.25 m/s2 starting from rest at point A on the diagram.
The incline rises at 35.0° above the horizontal, and at the instant
the rocket leaves the incline, its engines turn off and it is subject
only to gravity (ignore air resistance). v0  22.4 m/s
What is the maximum height above the ground that the rocket
Reaches?
x0 = (200.0 m) cos(35.0°) = 163.8 m
A) 67 m
y0 = (200.0 m) sin(35.0°) = 114.7 m
B) 94 m
v0
C) 123 m
D) 156 m
E) 199 m
(x0, y0)
7
At the top:
 H  y0 
v  0  v  2ay (y  y0 )  v  2g(H  y0 )
2
y
v02y
2g
2
0y
 H  y0 
2
0y
v02y
2g
[(22.4 m/s)(sin(35o)]2
 H  114.7 m +
 123. m
2
(2)(9.8 m/s )
v0
(x0, y0)
H
R
x0 = 163.8 m
y0 = 114.7 m
8
Tilted Rocket problem – part 3 (try on your own)
A test rocket is launched by accelerating it along a 200.0-m
incline at 1.25 m/s2 starting from rest at point A on the diagram.
The incline rises at 35.0° above the horizontal, and at the instant
the rocket leaves the incline, its engines turn off and it is subject
only to gravity (ignore air resistance).?
What is the horizontal range of the rocket beyond point A?
A)
B)
C)
D)
E)
280 m
339 m
357 m
402 m
621 m
x0 = 163.8 m
y0 = 114.7 m
v0
v0  22.4 m/s
(x0, y0)
9
To find total range R, determine time to fall to ground:
1 2
y(t)  0  y0  v0 y t  gt
2
2
 v0 y 
2y0
t
   
g
g
 g 
v0y
2v0y
2y0
t 
t
0
g
g
2
t  1.31 s 
(1.31 s)2  23.4 s2
 6.32 s
v0
(x0, y0)
H
R
R  x0  v0x t  163.8 m + [(22.4 m/s)(cos(35o)(6.32 s)] = 280. 10m
Grinding wheel sparks problem
What do you think the sparks flying off of a grinding wheel
demonstrate most directly about the uniform circular motion?
A. Velocity is tangential to the path
B. Velocity is normal to the path
C. Acceleration is tangential to the path
D. Acceleration is normal to the path and directed towards the
center of the circle
E. Acceleration is normal to the path and directed away from the
center of the circle
Sparks fly off along the velocity vector of the particles and hence the
velocity of the outer rim of the wheel
Merry-Go-Round problem
A boy and a girl ride on a merry-go-round. The boy rides on a
“donkey” and the girl rides on a “horse”. The “donkey” is two
times as far from the platform axis as the “horse”. What is the ratio
of the centripetal acceleration experienced by the boy to the
centripetal acceleration experienced by the girl?
A.
B.
C.
D.
E.
1/4
1/2
1
2
4
2
v
a
R
2R
v
T
 2 R 
2


T
 2 
a
=  R
 T 
R
2
R doubles → v doubles → a doubles
Shortcut: (see youtube video):
a R
Where ω is angular frequency:
v 2
 
R T
R doubles → a doubles
2
Hovering in mid-air problem
How short would the day have to be in order for a person
at the equator for gravity to no longer hold a person to the
ground? The Earth’s radius is 6370 m.
A)
B)
C)
D)
E)
1.4 minutes
14 minutes
1.4 hours
14 hours
140 hours
a g R
2
 2 
g  R
 T 
2
R
 T  2
g
 T  2 (6.37  10 6 m)/(9.8 m/s2 )  5100 s = 1.4 hr
Neutron star problem
A neutron star has a mass 1.4 times that of the Sun, but has a radius
of only 10 km. As we will see later, that means its gravitational field
at the surface is about 2 × 1011 Earth “g’s”. The fastest known
rotation rate for a neutron star is 716 Hz. About how many of those
200 billion g’s are needed to keep an object on the equator from
flying off?
A)
B)
C)
D)
E)
2 thousand
2 million
2 billion
20 billion
200 billion
Necessary centripetal acceleration:
 R  2 f  R
2
2
 [(2 )(716 Hz)]2 (104 m) = 2.0 1011 m/s2
≈ 20 billion g’s

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