Preparation problems

4-45. Estimate the acceleration of the moon, which completes a nearly circular orbit of 385,000 km radius in 27 days.

Solution:

To make an estimate of the moon's acceleration, we can start by taking the orbit to be circular, rather than nearly so. This being the case, we can use the equation for the radial acceleration of an object in uniform circular motion from our constants and equation sheet:

ar = v2
r
.
Since we do not know the velocity, but do know the period, we need to recall that the velocity of an object in circular orbit is the circumference of the circle divided by the period, or
v = 2 pr
T
.
To be successful, we need to be sure that all values are in MKS units, so we first convert:
r =
385,000  km =
3.85 ×108  m
T =
27  days =
2.33 ×106  s
Now we plug these values into our equation to find the velocity of the moon
v
=
2 p(3.85 ×108  m)
2.33 ×106  s
v
=
1037  m/s
Finally, we put this into our equation for the acceleration:
ar
=
v2
r
ar
=
(1037  m/s)2
3.85 ×108  m
ar
=
2.79 ×10-3  m/s2.

4-48. A 10-in-diameter circular saw blade rotates at 3500 revolutions per minute. What is the acceleration of one of the saw blades? Compare with the acceleration of gravity.

Solution:

Since the saw blade teeth are in uniform circular motion, we will use our equation for the radial acceleration from our constants and equation sheet:

ar = v2
r
.

We do not know the velocity, but do know how many times around the tooth goes per minute. Since each time around, the tooth has traveled a circumference, we can use this information to determine the velocity of the tooth. We start by recalling that

v = 2 pr  m/rev
60  s/min
 · rev
min
We know the saw blade has a 10 inch diameter. This means it has a 5 inch radius, or r = 0.127 m. Now we can find the velocity of a tooth:
v
=
2 p(0.127  m/rev)
60  s/min
 ·3500  rev/min
v
=
46.55  m/s
Finally, we put this into our equation for the acceleration:
ar
=
v2
r
ar
=
(46.55  m/s)2
0.127  m
ar
=
17,061  m/s2.
Remembering that the acceleration due to gravity is g = 9.8 m/s2, we can write this acceleration as
ar = 1741  g



Follow-up problems

4-29. A submarine-launched missile has a range of 4500 km. (a) What launch speed is needed for this range when the launch angle is 45°? (Neglect the distance over which the missile accelerates) (b) What is the total flight time? (c) What would be the minimum launch speed at a 20° launch angle, used to ``depress'' the trajectory so as to foil a space-based antimissile defense?

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 = 45°
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 = 4.5 ×106 m y = 0 m

(a) We have two equations possible (one in the x-direction, and one in the y-direction), and two unknowns. 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 y-direction, we get

y = y0 + vy0 t + 1
2
 ay t2 .
Recalling that y = y0 = 0 m, we get:
(0  m)
=
(0  m) + v0 sinqt - 1
2
 g t2
1
2
 g t2
=
v0 sinqt
g t2
=
2 v0 sinqt
t
=
2 v0 sinq
g
(1)
We can use this result in our equation in the x-direction, which looks like
x = x0 + vx0 t + 1
2
 ax t2 .
Recalling that ax = 0 m/s2, and substututing in for t, we get:
x
=
(0  m) + v0 cosqt
x
=
v0 cosq é
ê
ë 		2 v0 sinq
g
 ù
ú
û
x
=
2 v02 sinqcosq
g
v02
=
gx
2  sinqcosq
v0
=
  æ
Ö
gx
2  sinqcosq
 
(2)
v0
=
  æ
Ö
(9.8  m/s2)(4.5 ×106  m)
2 sin(45°) cos(45°)
 
v0
=
6641  m/s

(b) To find the time of flight, simply use Equation (1) from above, and plug in the known values.

t
=
2 v0 sinq
g
t
=
2 (6641  m/s) sin(45°)
9.8  m/s2
t
=
958  s

(c) To find the necessary velocity for a 20° launch angle, use Equation (2) above with the new angle q = 20°:

v0
=
  æ
Ö
gx
2  sinqcosq
 
v0
=
  æ
Ö
(9.8  m/s2)(4.5 ×106  m)
2  sin(20°) cos(20°)
 
v0
=
8283  m/s

4-36. You're 5.0 m from the left-hand wall of the house shown in Fig. 4-27 (on page 86), and you want to throw a ball to a friend 5.0 m from the right-hand wall. (a) What is the minimum speed that will allow the ball to clear the roof? (b) At what angle should it be thrown? Assume the throw and the catch both occur 1.0 m above the ground.

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 = ?
x-direction y-direction
Initial Positions x0 = 0 m y0 = 1 m
Init. Velocities vx0 = v0 cosq vy0 = v0 sinq
Accelerations ax = 0 m/s2 ay = -g = -9.8 m/s2
Positions at top of roofx1 = 8 m y1 = 6 m
Velocities at top of roofvx1 = v0 cosq vy1 = 0 m/s
Final Positions x = 16 m y = 1 m

(a) This problem looks a little different from the other 2-D kinematics problems we have been working, in that we have 3 unknowns. In order to solve this problem, we will need the information about the trajectory at the top of the roof. We can find the initial velocity in the y-direction that will let us just clear the roof of the house using the third kinematics equation from our constants and equations sheet for the first half of the ball's flight:

vy12 = vy02 + 2 ay (y1 - y0) .
We see that our only unknown is vy0. Solving this gives us
(0  m/s)2
=
vy02 - 2 g (y1 - y0)
(1)
vy02
=
2 g (y1 - y0)
(2)
vy0
=
  ___________
Ö 2 g (y1 - y0)
 
(3)
vy0
=
  æ
Ö
2 (9.8  m/s2)(6  m - 1 m)
 
(4)
vy0
=
9.9  m/s
(5)
Now we need to find the component of velocity in the x-direction to finish solving our problem. The easiest way to do this is to find the time of flight, and use that to determine the velocity in the x-direction. To find the time of flight, we can use the equation:
y = y0 + vy0 t + 1
2
 ay t2 .
Recalling that y = y0 = 1 m, we get:
(1  m)
=
(1  m) + vy0 t - 1
2
 g t2
1
2
 g t2
=
vy0 t
g t2
=
2 vy0 t
t
=
2 vy0
g
t
=
2.02  s
Next, we use the first kinematics equation in the x-direction to solve for the component of velocity in that direction:
x = x0 + vx0 t + 1
2
 ax t2 .
Recalling that ax = 0 m/s2, and substututing in for our known values, we get:
16  m
=
(0  m) + vx0 (2.02  s)
vx0
=
16  m
2.02  s
vx0
=
7.91  m/s
Now that we know each component of v0, we can calculate its magnitude easily using the Pythagorean Theorem:
v0
=
  æ
Ö
vx02 + vy02
 
v0
=
  æ
Ö
(7.91  m/s)2 + (9.9  m/s)2
 
v0
=
12.7  m/s

(b) To find the angle at which the ball must be thrown, we recall

q
=
tan-1 é
ê
ë 		vy0
vx0
 ù
ú
û
q
=
tan-1 é
ê
ë 		9.9  m/s
7.91  m/s
 ù
ú
û
q
=
51.6°

4-39. In 1991, Mike Powell shattered Bob Beamon's 1968 world long jump record with a leap of 8.95 m (see Fig. 4-28 on page 87). Studies show that Powell jumps at 22° to the vertical. Treating him as a projectile, at what speed did Powell begin his jump?

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 = (90°-22°) = 68°
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 = 8.95 m y = 0 m

We see that we have the same set of unknowns here as we do in Problem 4-29 above. We can therefore proceed to solve the problem in a similar manner. 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 y-direction, we get

y = y0 + vy0 t + 1
2
 ay t2 .
Recalling that y = y0 = 0 m, we get:
(0  m)
=
(0  m) + v0 sinqt - 1
2
 g t2
1
2
 g t2
=
v0 sinqt
g t2
=
2 v0 sinqt
t
=
2 v0 sinq
g
(1)
We can use this result in our equation in the x-direction, which looks like
x = x0 + vx0 t + 1
2
 ax t2 .
Recalling that ax = 0 m/s2, and substututing in for t, we get:
x
=
(0  m) + v0 cosqt
x
=
v0 cosq é
ê
ë 		2 v0 sinq
g
 ù
ú
û
x
=
2 v02 sinqcosq
g
v02
=
gx
2  sinqcosq
v0
=
  æ
Ö
gx
2  sinqcosq
 
(2)
v0
=
  æ
Ö
(9.8  m/s2)(8.95  m)
2 sin(68°) cos(68°)
 
v0
=
11.24  m/s

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