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Circular Motion, Quantitative Description

An object moving in a circle has an acceleration which has, in general, two components: the centripetal acceleration and the tangential acceleration. The former is directed toward the center and is responsible for changing the direction of the object. The latter is responsible for changing the speed. The acceleration is given by



A bombardier beetle sits on a blade of a windmill which is going counterclockwise and slowing down. The blade is at the top of its cycle. Sketch the acceleration vector.


In Figure 5-10, if the beetle is going left and slowing down, then the tangential acceleration must be to the right. The centripetal acceleration vector is down, and the total acceleration a is shown.

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Figure 5-10



Use the following information to find the mass of the sun:

G = 6.67 x 10–11 m3/kg s2
R = 1.5 x 1011 m (distance from the Sun to the Earth)

First, we DRAW A DIAGRAM showing all the forces (Figure 5-11). Some students are tempted to draw two forces on the Earth: a gravitational and a centripetal force. But the only force is gravity, and this provides the centripetal force in this problem. This last sentence provides the clue for solving the problem. We know expressions for gravitation and for centripetal acceleration, so that we have


We use MEarth on the right-hand side of the equation, because it is the Earth's acceleration we are concerned about. The Sun's acceleration is much smaller because its mass is larger. (Recall that the force the Earth exerts on the Sun is the same as that which the Sun exerts on the Earth.) Note that MEarth cancels.

Figure 5-11

What expression shall we use for v? What is the velocity of the Earth? If the Earth travels a full circuit in a year, then velocity is simply distance per time, where the distance is the circumference of the circle. Thus we write


where T is 1 year. Do not simply memorize this, but take a minute to think about why this equation is true, so it will come immediately to mind in any similar situations. Substituting this into equation (7) and doing some algebra gives


into which we can substitute the values given in the problem, along with


to yield
Msun = 2 x 1030 kg
The importance of this example does not lie in the arithmetic. The important parts are the method of setting two expressions for the same force equal to each other and the use of equation (8).



A fan spins at a frequency of 50 cycles per second, and its plastic blades are 0.4 meters long. What is the centripetal acceleration of a piece of plastic at the tip of one of the blades?



You should try to work this out before you read the solution.

If the fan spins at 50 cycles per second, it must undergo one cycle in one fiftieth of a second, so T = 1/50 s = 0.02 s. The velocity is given by


so the acceleration is given by





A space warrior must fly his spacecraft at constant speed around a spherical space station "Bad Star". Bad Star is large but not large enough to have an atmosphere or gravity worth considering. It is important for the warrior to stay close to the surface and maintain constant speed. A conventional rocket provides the thrust to maintain course, so the plume appears in the opposite direction of the desired thrust. In which direction (Figure 5-12) would we see the plume?


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Figure 5-12

No one chooses B. Few people choose A. In fact, almost everyone chooses E, because they are thinking about a car driving on the surface of the Earth. But a car on the Earth's surface encounters the force of drag, both by friction and by air resistance, whereas the space warrior encounters neither. No force is required to maintain constant speed. If you chose E, it means you need to study the first law of motion again.

Okay, what is the correct way to think of this problem? The net force of the spacecraft is down, toward the center of Bad Star, because the craft is moving at constant speed in a circle. there is no gravity, no friction, no air drag, so the only force on the craft is due to the rockets. Therefore the rocket plume points in the direction of A.

Some students object that a rocket firing in the direction of A would push the craft into the Bad Star. This is the same as objecting that if the Earth pulls on the Moon, it ought to fall down. What holds up the Moon? The answer in both cases is that the centripetal force is large enough to keep the object (spacecraft or Moon) from moving away but not so large as to pull them into the ground. That is, in Figure 5-13, path 1 is the path the spacecraft takes if the warrior does not fire his rockets at all (no force). Path 2 is the path if he does not fire the rockets enough; and path 4, if he fires them too much. Path 3 is just right.

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Figure 5-13


The most important concept in this chapter is the independence of horizontal and vertical motion. We saw this concept for the first time in Chapter 4. In this chapter we have used the concept to solve problems involving inclined planes and oblique forces, that is, forces which are neither horizontal nor vertical.

We also looked at circular motion. If an object is moving in a circle, its velocity vector is not constant, and the object must be accelerating. If the object is moving at constant speed, then the direction of the acceleration vector is toward the center, called centripetal acceleration, and its magnitude is acent = v2/r.


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