Download Radial Net Force Wkst 3

Name
Date
Pd
Central Net Force Model Worksheet 3:
Circular Motion Examples
1. A woman flying aerobatics executes a maneuver as illustrated below. Construct a quantitative
force diagram of all relevant forces acting on the woman flying the airplane when upside-down
at the top of the loop.
N
Fg  mg  55kg(10 kg
)  550N
FN =
120N
Fg =
550N
mv 2
Fnet  mac 
r
2
m
1hr
55kg(235 km
hr (1000 km ) 3600 s )
 617N
380m
Fnet  Fg  FN  617N  550N  FN
r = 380
m
v = 235
km/hr
pilot
mass =
55 kg
FN  120N
2. Six children run on a track with equal speeds. Their masses are expressed in multiples of
mass “M” and their path radii are expressed in multiples of radius “R.”
a. Rank the centripetal acceleration of the lettered
from largest to smallest. (Ties are possible.)
largest -> _A=D_ ,_B=E_, _C=F_ <-smallest
Explain how you determined your ranking:
v2
Centripetal acceleration is
. Since the speeds are all
r
the same, the acceleration is inversely proportional to
the radius. A and D have the smallest radius and C and
F the largest.
D 2 F2
E
M M
4R 3
M
A4
M
B
M 5R
C6
M 6R
children
b. Rank the centripetal force needed for each child to remain in circular motion. (Ties
are possible.)
largest -> _A=C_, _E_, _D_, _F_, _B_ <-smallest
Explain how you determined your ranking:
m
mv 2
For centripetal force we use
. I looked at the ratio of
. For A and C it was 1,
r
r
for E it was 0.6, for D it was 0.5, for F it was 0.33 and for B it was 0.2.
©Modeling Instruction – AMTA 2013
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U7 Central Force Model - ws3 v3.1
3. Rollercoasters use a hill for riders to gain speed followed by an upside down loop. The loops
are designed with large radius bottoms and small radius tops, and such a shape is called a
clothoid. Answer the following questions in order to find out why the clothoid is used.
The speed of the rollercoaster is 22 m/s at the
bottom of the loop and 8 m/s at the top of the
loop.
r
1
a. Use quantitative force diagrams to
determine the size of the force on the rider
by the seat at the top and bottom of the
loop.
r
2
N
Fg  mg  1000kg(10 kg
)  10000N
FN =
58,000
N
Fg =
10,000
N
mv 2 1000kg(22 ms )2

 48400N
r
10m
 FN  Fg  48400N  FN  10,000N
Fnet 
Fnet
FN  58,000N
mv 2 1000kg (8 ms )2

 12,800N
r
5.0m
 FN  Fg  12,800N  FN  10,000
Fnet 
Fnet
FN =
2,800
N
Fg =
10,000N
FN  2,800N
b. Now suppose the rollercoaster had been designed with the small 5 meter radius at the
bottom and the large 10 meter radius at the top. Use quantitative force diagrams to
determine the size of the force on the rider by the seat at the top and bottom of the loop.
FN =
107,000
N
Fg =
10,000
N
mv 2 1000kg (22 ms )2

 96,800N
r
5m
 FN  Fg  96,800N  FN  10,000N
Fnet 
Fnet
FN  107,000N
mv 2 1000kg (8 ms )2

 6,400N
r
10m
 FN  Fg  6,400N  FN  10,000
Fnet 
Fnet
FN =
3,600N
Fg =
10,000N
FN  3,600N
c. Why is the clothoid shape used in rollercoasters?
As can be seen in part b. the radius of curvature at the top must be small enough that the
net force is larger than the force of gravity. If not, the rollercoaster won't make it over the
top. Also if the radius of curvature is too small at the bottom the upward force on the
person can be more than they can tolerate. In the example, the force is almost 11 times the
body weight.
©Modeling Instruction – AMTA 2013
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U7 Central Force Model - ws3 v3.1
4. 80 kg Tarzan grabs a vine to swing to another tree.
a. As Tarzan swings from point A to point B, describe qualitatively how the tension in the
vine changes and why.
8
m
A
2
BCm
Since the net force is inversely proportional to the radius (if the
speed does not change), one fourth the radius will give a net force 4
times larger.
b. At point B he is swinging at 7 m/s and the vine is 8 meters long. How hard does he have
to hang on to the vine to keep from slipping off?
F
T
F
g
mv 2
mv 2
 FT 
 Fg
r
r
80kg(7.0 ms )2
N
FT 
 80kg(10 kg
)  490N  800N  1300N
8.0m
Fnet  FT  Fg 
c. A moment later, at point C, the vine catches on a branch, reducing the radius of the swing to
2 m. If Tarzan is still traveling at 7 m/s, how hard does he now have to hold on to the vine?
80kg(7.0 ms )2
N
FT 
 80kg(10 kg
)  1,960N  800N  2800N
2.0m
He must hold on with a force 3.5 times his weight.
d. If Tarzan slips off at point C what will be the path he takes? (Sketch the path on the
diagram.) How does this path differ from circular motion?
Tarzan becomes a projectile. There is a constant downward force accelerating him
vertically and he moves to the right at a constant velocity.
©Modeling Instruction – AMTA 2013
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U7 Central Force Model - ws3 v3.1
5. Six children of equal masses run on a track. Their speeds are expressed in multiples of
velocity “V” and their path radii are expressed in multiples of radius “R.”
D 2 F2
from
E
V3 V
4R
V
A4
V
B
largest -> _C_, _A_, _E_,_D_,_F_, _B_<-smallest
V 5R
Explain how you determined your ranking:
C6
V 6R
v2
I looked at the ratio of
. For C it was 6, for A it was 4, for E it was 1.8,
for D it was 1,
r
for F it was 0.67 and for B it was 0.2.
a. Rank the centripetal acceleration of the lettered children
largest to smallest. (Ties are possible.)
b. Rank the centripetal force needed for each child to remain in circular motion. (Ties are
possible.)
largest -> _C_, _A_, _E_,_D_,_F_, _B_<-smallest
Explain how you determined your ranking:
Since they all have the same mass, the order will be the same as in part a.
6. An amusement park ride operates as follows: riders enter the cylindrical structure when it is
stationary with the floor at the point marked "a". They then stand against the wall as the
cylinder then begins to rotate. When it is up to speed, the floor is lowered to the position
marked "b", leaving the riders "suspended" against the wall high above the floor.
F
fs
F
FN
Side
g
view of
rider
Determine the period of rotation necessary to keep the riders from sliding down the wall when the
floor is lowered from point "a" to point "b". (Show all of your work and explain your reasoning.)
Vert : Ffs  Fg  0   FN  mg  FN 
mg

N
(10 kg
)1.5m
mv 2
mg mv 2
gr
2


 v2 
 v2 
 30 ms 2
r

r

0.50
2 r
2 (1.5m)


 1.7s

5.5 ms
Hor : Fnet  FN 
v  5.5 ms
©Modeling Instruction – AMTA 2013
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U7 Central Force Model - ws3 v3.1