Physics Assignment Answers

Ch 9 - Circular Motion

Ch 9 Review Questions:

  1. The action/reaction forces are "Can pushes bug" and "Bug pushes can." 
  2. There is no object that supplies the "centrifugal force" that the ladybug feels. Nothing is pushing the ladybug outward. Therefore, this so-called "centrifugal force" is not part of an action/reaction force pair. The "centrifugal force" is not a real force at all - it is a fictitious force. 
  3.  The "centrifugal force" that the ladybug feels is called a fictitious force because a real force is an interaction between two objects, and there is no material object exerting an outward force on the ladybug. For a force to act, Newton's Third Law says that we must have both "Object X pulls ladybug" and "ladybug pulls Object X" - and there is no Object X! The pull that the ladybug feels is due to her inertia - she "wants" to move in a straight line, but is being pushed toward the center of the circle by the bottom of the can.
  4. In a rotating space habitat, the magnitude of its simulated gravity is directly proportional to the square of its rotational velocity (or proportional to the square of the tangential velocity of a point on its rim).  
  5. For space habitats rotating at the same rate, the magnitude of the simulated gravitational force is directly proportional to the habitat's radius.  
  6. Space habitats that simulate gravity would need to be large structures in order to have a low rotational velocity (see review questions 16 and 17). At high rotational velocities, the habitat would feel like an amusement park ride to its occupants! They would feel dizzy, nauseous, at best. At low rotational velocities, the occupants of the habitat could adjust to the rotation and eventually not even notice.

Ch 9 Think & Explain Answers:

  1. No, the centripetal force never does any work on a rotating object, since the centripetal force points toward the center of the circle, and the velocity of the object is tangent to the circle, the direction of the force is perpendicular to the direction of the object's motion - work done = 0.
  2. First, the Earth:
    1. Suppose you are standing on a scale on a non-rotating Earth. Two forces act on you- the Earth is pulling you down with a force we call weight (= mg), and the scale is pushing up. Since you are at rest, Newton's First Law tells us that the net force on you is zero, so the scale force must equal the weight force. The scale reads your weight, in other words.
    2. Now, let the Earth start rotating. The same two forces push/pull on you - the Earth still pulls down (the same as before, W = mg) and the scale still pushes up. Now though, you are not at rest, you are revolving around the center of the Earth, so there must be a net force (the centripetal force) on you that points toward the center of the Earth. Therefore the weight force and the scale force are no longer equal! The downward force the Earth exerts is still W = mg - it doesn't change. In order for there to be a net force toward the center of the Earth, the upward scale force must be less. In other words, you weigh less (according to a scale, at least) since the Earth is rotating.
    3. Now let the Earth start rotating faster. This means that your rotational and linear speeds increase since you are attached to the Earth. Since the centripetal force on you is proportional to the square of your linear velocity, as your velocity increases the centripetal force needed to keep you moving in a circle must also increase. How can this happen? The Earth's downward pull on you = mg, and this doesn't change. In order for the downward (toward the Earth's center) force to increase, the upward scale force must decrease! Therefore, if the Earth were to spin faster you would weigh less - at least according to a scale. It is important to notice that your mg weight would not change.

    Now, the space station:
    1. space station diagramIf you were standing on a scale on the rim of a non-rotating space station (far from any planet), no forces would act on you - you would be weightless and the scale would read zero.
    2. Now let the space station start rotating. If you are to revolve around the central axis of the space station, there must be a net force (the centripetal force) on you that points toward the center of the space station. The only object that can exert that force is the scale that you are standing on. Therefore, in a rotating space station the scale would read some force - apparently, you would have weight.
    3. Now let the space station start rotating faster. This means that your rotational and linear speeds increase since you are revolving with the space station. Since the centripetal force on you is proportional to the square of your linear velocity, as your velocity increases the centripetal force needed to keep you moving in a circle must also increase. How can this happen? There is only one force pushing on you - the scale's force. So as the space station rotates faster, the scale must exert more force on you. According to the scale, you would be heavier.

      So, if the Earth were to rotate faster, you would (seem to) weigh less, but if a space station were to rotate faster, you would (seem to) weigh more.

Ch 9 Think & Solve Answers:

  1. (a) If the turntable turns at 10 revolutions/second, its period (time to rotate once) is 0.1 second. During this time, the tip of the light beam travels in a circle of circumference C = 2 pi r, so its speed is: v = 630 km/s.
    (b) Doubling the distance would double the speed, or you can just recalculate: v = 1260 km/s
    (c) r = 4800 km
  2. There are two forces on the rock as it moves in a circular path: its weight (mg) is pulling it directly downward, and the tension in the string (T) is pulling it toward the center of the circle. When the rock is at the top of its path, both of these forces point directly downward - toward the center of the circle - so the centripetal force on the rock at the top of its path equals its weight plus the tension in the string (Fnet = mg + T).
    Now, as you whirl the rock more and more slowly, the tension in the string decreases as the rock passes through its highest point. The minimum speed at which the rock can "just get around" the top of the circle is the speed at which the tension in the string is zero. At this speed, the centripetal force needed to move the rock in a circle equals its weight, mg. So:
    v = 3.1 m/s
  3. The man's head would experience 0.5g. Since his head is located halfway to the axis of rotation of the space station, its linear velocity would be half the linear velocity of his feet and so his centripetal acceleration (= v2/r) would be half the centripetal acceleration of his feet, too.
    This would be an extremely uncomfortable situation! If you have ever felt sick when riding a rotating amusement-park ride, you know the feeling. Imagine feeling like that 24 hours a day! Large rotational velocities make people very ill!
    We would like people in the space station to feel "normal gravity" = 1g at its rim. In order to do that, the space station must rotate at a speed such that the rim exerts an inward centripetal force on a person equal to their weight (check the answer to Think & Explain #29 for more details).
    We know that Fcent = mv2/r, but it is rotational velocity - not linear velocity - that is the culprit here. Large linear velocities have no effect on us. After all, your linear velocity relative to the center of the Earth is over 1000 mi/hr right now! However, spinning around really fast (with a linear velocity of zero) will make you dizzy, and if you do it long enough, ill. To see what is going on, we have to convert the centripetal force equation to tell us about rotational velocities.
    The relationship between linear and rotational velocity is . Substituting this into the equation for centripetal force gives:

    Now we can get down to business. We want the centripetal force that the rim of the space station exerts on an object to equal the object's weight. In symbols:

    You can cancel an "m" on both sides. Physically, this means that the mass of the object won't matter. You, your physics book, and your pencil will all feel normal weight at the same speed!

    Solving this equation for rotational velocity gives:

    This last set of equations tell us that the rotational velocity of a "1 g" space station would be inversely proportional to the square root of the station's radius. A station with four times the radius would rotate with half the rotational velocity. A station with nine times the radius would rotate with a third of the rotational velocity.
    So to get down to a numerical example, for a space station with a radius of 4m,

    but for a space station with a radius of 1000 m,

    So, the 4-meter-radius space station would need to revolve at 15 rpm, but a 1000-meter-radius station would revolve at a more-leisurely 1 rpm.
  4. Not assigned
  5. The centripetal force exerted on you by the space station's rim, Fcent = mv2/r, is the "weight" that you feel in the space station, mgstation. (It takes the place of the force that the floor exerts on you here on Earth.) So, mv2/r = mgstation. Canceling the "m" on both sides of the equation gives gstation= v2/r = (300 m/s)2/(9184 m) = 9.8 m/s2. This equals g here on Earth! In the space station, you would feel like you were experiencing normal Earth gravity.