Preparation problems

4-59. An alpine rescue team is using a slingshot to send an emergency medical packet to climbers stranded on a ledge, as shown in Fig. 4-31 (on page 88). What should be the launch speed from the slingshot.

Solution:

As is always the case with 2-D kinematics problems, we start by making a table of our knowns, which looks something like this:

Final Time t =  ?
Initial Speed v0 =  ?
Angle q = 70°
x-direction y-direction
Initial Positions x0 = 0 m y0 = 0 m
Init. Velocities vx0 = v0 cosq vy0 = v0 sinq
Accelerations ax = 0 m/s2 ay = -g = -9.8 m/s2
Final Positions x = 390 m y = 270 m

Since we are looking for the initial speed v0, we will solve first for the time, and then substitute that into our remaining equation. Starting with the x-direction, we get

x = x0 + vx0 t + 1
2
 ax t2 .
Recalling that ax = 0 m/s2, we can solve for t, getting:
x
=
(0  m) + v0 cosqt
t
=
x
v0 cosq
(1)

This equation has two unknowns, so we need to look for another equation to solve. The best candidate is the position equation in the y-direction:

y = y0 + vy0 t + 1
2
 ay t2 .
Recalling that y = 0 m, and substituting in for t from Equation (1), we get:
y
=
(0  m) + v0 sinqt - 1
2
 g t2
y
=
v0 sinq é
ê
ë 		x
v0 cosq
 ù
ú
û 		- 1
2
 g é
ê
ë 		x
v0 cosq
 ù
ú
û 		2

 
y
=
x sinq
cosq
 - gx2
2 v02 cos2 q
We need to solve this equation for v0, which gets us:
y
=
x tanq- gx2
2 v02 cos2 q
(y - x tanq)
=
gx2
2 v02 cos2 q
v02(y - x tanq)
=
gx2
2 cos2 q
v02
=
gx2
2 cos2 q (y - x tanq)
v0
=
  æ
Ö
gx2
2 cos2 q (y - x tanq)
 
(2)
Plugging our known values into Equation (2), we get
v0
=
  æ
Ö
(9.8  m/s2)(390 m)2
2 cos2 (70°)  [(270  m) - (390 m) tan(70°) ]
 
v0
=
89.2  m/s



Follow-up problems

4-51. Electrons in a TV tube are deflected through a 55° angle, as shown in Fig. 4-30 (on page 87). During the deflection they move at constant speed in a circular path of radius 4.3 cm. If they experience an acceleration of 3.35 ×1017 m/s2, how long does the deflection take?

Solution:

We are given the radial acceleration of the electrons, and told that they move along part of a circular path. We can use the equation for radial acceleration in circular motion from our constants and equation sheet

ar = v2
r
.
We do not know the velocity here, but we are asked for a time. This probably means that we need to recall how we can find the velocity from the period of circular motion.
v = 2 pr
T
.
We can substitute this into our acceleration equation and solve for T to get
ar
=
(2 pr)2
(T)2 r
ar
=
4 p2 r
T2
T2
=
4 p2 r
ar
T
=
  æ
Ö
4 p2 r
ar
 
Plugging in our numbers, we get
T
=
  æ
Ö
4 p2  (0.043  m)
3.35 ×1017  m/s2
 
T
=
2.25 ×10-9  s
We have solved for the time it takes for the electron to go all the way around a circle of radius 4.3 cm, but we know that the electron only turns through 55°. So it must only take 55/360 of the time it takes to go all the way around the circle to accelerate through the 55° deflection. This gives us a final time of
t
=
T · 55°
360°
t
=
2.25 ×10-9  s · 55°
360°
t
=
3.44 ×10-10  s .

4-54. A space station 120 m in diameter is set rotating in order to give its occupants ``artificial gravity.'' Over a period of 5.0 min, small rockets bring the station steadily to its final rotation rate of 1 revolution every 20 s. What are the radial and tangential accelerations of a point on the rim of the station 2.0 min after the rockets start firing?

Solution:

There are several ways to approach this problem. One method is to simply look at the point on the rim and ask what were the tangential velocities of the point at 0 min, 2 min, and 5 min? If we know these velocities, we can determine the radial acceleration at these times. We can also use the change in the tangential velocities to determine the tangential acceleration.

We will start by finding the final velocity (at 5 min). Since the radius of the station is 60 m, and its period of rotation is 20 s, we see that

vf
=
2 pr
T
=
2 p(60  m
20  s
vf
=
18.85  m/s
Since the initial velocity was 0 m/s, we can find the average tangential acceleration by recalling
at
=
Dvt
Dt
=
vf - v0
(5  min)(60  s/min)
at
=
0.063  m/s2
Now we can find the velocity at t = 2 min, or t = 120 s, using the kinematics equation from our constants and equation sheet
v
=
v0 + at t
=
0  m/s + (0.063  m/s2)(120  s)
v
=
7.54  m/s
Finally, we can find the radial acceleration for this velocity using
ar
=
v2
r
=
(7.54  m/s)2
60  m
ar
=
0.960  m/s2

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Solutions translated from TEX by TTH, version 1.57.