Transcript Dynamics II Motion in a Plane

Dynamics II
Motion in a Plane
Review Problems
Problem 1
A 500 g model rocket is on a cart that is rolling
to the right at a speed of 3.0 m/s. The rocket
engine, when it is fired , exerts an 8.0 N thrust
on the rocket. Your goal is to have the rocket
pass through a small horizontal loop that is 20
m above the launch point. At what horizontal
point to the left of the hoop should you
launch?
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Problem 1 con’t
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Problem 1 con’t
Analysis: Essentially a projectile motion problem with
positive vertical acceleration and constant horizontal
velocity. Assume a particle model. Compute the time
required to rise 20 m and use that to determine the
horizontal distance.
8N- .5kg(9.8m / s2 )
= 6.2m / ss
FT - mg = ma a =
.5kg
 1
20 =    6.2  t 2
2
t = 2.54 sec
d = 2.54s(3m / s) = 7.62m
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Problem 2
A highway curve of radius 500 m is designed for traffic
moving at speeds of 90 kph. What is the correct
banking angle of the road?
Analysis: Nr provides the centripetal acceleration and Nz counters the
gravitational force.
mv2
Nsinθ =
Ncosθ = mg
r
θ
θ
v2
tanθ =
rg
252
=
 0.128    7.3
500(9.8)
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Problem 3
A 30 g ball rolls around a 40 cm diameter horizontal
track at 60 rpm. What is the magnitude of the net force
that the track exerts on the ball? Neglect rolling
friction.
F
z
Analysis: Force against the vertical
element of the track provides centripetal
acceleration and the horizontal supports
the weight.
Fr
2
 60 rev 2π rad 1min 
Fr = mr 2 =  0.030 kg  0.20 m  
×
×
= 0.24 N

1 rev
60 s 
 min
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Problem 4
A student has 65 cm arm length. What is the minimum angular velocity,
in rpm, for swinging bucket of water in a vertical circle without spilling
any? The distance from the handle to the bottom of the bucket is 35 cm.
Analysis: The minimum angular velocity for swinging a
bucket of water in a vertical circle without spilling any
water corresponds to the case when the speed of the
bucket is critical. In this case, n = 0 N when the bucket is
in the top position of the circular motion.
mvc2
FrG = 0 N+mg = r = mrωc2
9.8 m / s2
ωc = g / r =
= 3.13 rad / s = 30 rpm
1.00 m
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Problem 5
A car is tested on a 200 m diameter track. If the car speeds up at a
steady 1.5 m/s2, how long after starting is the magnitude of its
centripetal accelerations equal to the tangential acceleration?
Analysis: NUCM; we know at = 1.5 m/s2 from that we should be able to
determine the angular velocity and acceleration, ω and α, and then the time.
ar = ω r = 1.5 m / s
2
2
1.5 m / s2
1.5 m / s2
ω=
=
= 0.122 rad / s
r
100 m
at 1.5 m / s2
α= =
= 1.5 ×10-2 s-2 .
r
100 m
ω = ωi + αΔt
ω - ωi 0.122 s-1 - 0 s-1
Δt =
=
= 8.2 s
α
0.015 s-2
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Problem 6
A popular pastime is to see who ca push an object closest to the
edge of a table without its going off. You push the 100 g object and
release it 2.0 m from the table edge. Unfortunately you push a little
too hard. The object slides across, sails off the edge, falls 1.0 m to
the floor, and lands 30 cm from the edge of the table. If the
coefficient of kinetic friction is 0.50, what was the object’s speed as
you released it?
Analysis: After release velocity is
affected by friction only. When it
reaches the edge of the table it
becomes a projectile motion
problem with horizontal velocity
constant. Best approaches appear
to be to work backward from the
projectile motion to the release
velocity.
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Problem 6 con’t
Δy = 1 m and v1y = 0
y2 = y1 + v1y  t2 - t1  + 21 -g t2 - t1 
2
0 m = 1.0 m + 0 m-
g
2
 t2 - t1    t2 - t1  = 0.4518 s
2
Similarly in the x direction
x2 = x1 + v1x  t 2 - t1 
2.30 m = 2.0 m + v1x  0.4518 s   v1x = 0.664 m/s
Given v1x we can find v0x given the acceleration.
F
x
= -fk = max = -μkmg
ax = -μkg = - 0.50 9.8 m/s2  = -4.9 m/s2
2
2
v1x
= v0x
+ 2ax  x1 - x0  :
0.664 m/s
2
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= v20x +2-4.9 m/s2  2.0 m  v0x = 4.5 m/ s
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Problem 7
A motorcycle daredevil plans to ride up a 2.0 m high, 20° ramp, sail
across a 10 m wide pool filled with crocodiles, and land at ground
level on the other side. Unfortunately, the motorcycle engine dies
just as he starts up the ramp. He is going 11 m/s at that instant, and
the coefficient of rolling friction is 0.02. Does he make it?
Analysis: Two part problem: First is a
ramp problem to determine velocity at
end of ramp. Second part is a
projectile motion problem determining
how far right he will travel while
dropping 2.0 m. Different sets of axes
will be required for each part.
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Problem 7 con’t
First find the acceleration parallel to the ramp:
(Fnet )|| = -fr - mgsin20°
= -μrn - mgsin20°
= -μrmgcos20° - mgsin20° = ma0
Dividing by M and solving for a0 gives:
a0 = -g(μr cos20° + sin20°)
= -(9.8 m / s2 )((0.02)cos20° + sin20°) = -3.536 m / s2
The length of the ramp is 2 m/ sin 20° = 5.85 m
Using v12 = v02 + 2a0 (s1 - s0 )
v1 = (11.0 m / s)2 + 2(-3.536 m / s2 )(5.85 m) = 8.92 m / s
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Problem 7 con’t
Now we have the velocity at the beginning
of projectile motion. Axes are now
horizontal and vertical. Velocity
components are:
v1x = v1cos20° = 8.38 m s
v1y = v1sin20° = 3.05 m S
Time in the air is given by:
y2 = 0 m = y1 + v1y t + 21 a1y t 2 For Δt = t 2 - t1
= 2.0 m + (3.05 m / s) t - (4.90 m / s2 ) t 2
t = 1.021s
X2 is given by: 8.38 m/s (Δt) = 8.56 m.
Looks like crocodile food!!
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Problem 8
A concrete highway curve of radius 70 m is banked at a 15° angle.
What is the maximum speed with which a 1500 kg rubber tired car can
take this curve without sliding?
The summation of r and z forces gives:
mv 2
 Fr = fkcosθ + nsinθ = r
Fz = ncosθ - fk sinθ - FG = 0
Maximum speed is when the static friction force
reaches its maximum value  fk max = μkn.
Substituting this into the above equations gives:
mv 2
n μscos15° + sin15°  =
r
n cos15° - μssin15° = mg
Dividing the bottom into the top gives:
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μs + tan15° v2
μ + tan15°
=
 v = gr s
= 34 m s
1- μs tan15° gr
1- μs tan15°
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Problem 9
A conical pendulum is formed by attaching a 500g ball
to a 1.0 m long string, then allowing the mass to move
in a horizontal circle of radius 20 cm. a. Find the
tension is the string and b. the angular speed of the
ball in rpm.
Analysis: The mass moves in a horizontal circle of radius
The acceleration and the net force vector point to the
center of the circle, not along the string. The only two
forces are the string tension, which does point along the
string, and the gravitational force. These are shown in
the free-body diagram. Newton’s second law for circular
motion is
F = Tcosθ - F = Tcosθ - mg = 0 N
mv
F
=
Tsinθ
=
ma
=

r
z
G
2
r
From FZ:
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r
 0.500 kg 9.8 m / s
mg
T=
=
cosθ
cos11.54°
2
 = 5.00 N
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Problem 10
A 500 g ball moves in a vertical circle on a 102
cm long string. If the speed at the top of the
4.0m/s, then the speed at the bottom will be 7.5
m/s. Find (a) the gravitational force acting on
the ball, (b) The tension in the string at the top
of the circle and (c) the tension in the string at
the bottom of the circle.
(a) The gravitational force is:
FG = mg =  0.500 kg   9.8 m / s2  = 4.9 N.
(b) The tension in the string at the top of the circle is:
  4.0 m / s 2

 v2

2
T1 = m  - g  =  0.500 kg 
- 9.8 m / s  = 2.9N
 1.02 m

 r

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Problem 10 con’t
(c) Similarily at the bottom of the circle the tension is:
mv 2
 Fr = T2 - FG = r
2


7.5
m
/
s



v2 
2
T2 = m  g +  =  0.500 kg 9.8 m / s +
 = 32 N
r 
1.02 m 


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