Circular Motion s1

Lyzinski Physics Circular Motion

In order to understand “how” and “why” object’s travel in circles, we need to first go back to Newton’s 1 st Law. Please state this law fully on the lines below. An object in motion (at a constant velocity) or at rest, will remain in motion (at a ______constant velocity) or at rest unless acted upon by an external, unbalanced force. ______

Remember, this is often called the “Law of Inertia”. Objects naturally want to do one of two things. They either want to stay are rest (the lazy bums) or they want to keep moving at a constant velocity IN A STRAIGHT LINE. If they don’t move in a straight line, it’s because “something” is forcing them to change direction.

The most basic problem that we can look at when studying circular motion is the motion of a ball on a string being swung in a horizontal circle above a person’s head. This can be seen in the picture at the right. If the ball is  being swung at a constant speed, does the ball accelerate? Some students will say “No, because constant speed means a = 0”. However, the deep-thinking student will say, “Yes, it does accelerate, because its velocity changes, not in magnitude, but in DIRECTION”.

Look at the pictures below, which show snapshots of the swinging ball, taken from a view above the man’s head. You can see that the magnitude of the velocity vector v doesn’t change, but that the direction does. Since a  , and since the velocity changes t (hence v = something), then the ball does in fact accelerate.

The next question is, “in which direction does the ball accelerate?” This is easy to see if you imagine a little boy’s motorized toy car. Imagine if the car is of the simplest variety: it can only move forward in a straight line. What would you have to do to make the car travel in a circle? The picture at the right (which again, shows a view from above the car) shows that you would have to push the car INTO a circular path. Imagine doing so with little “nudges”. Continuously “tapping” on the side of the car, nudging it, you will get it to move in a circular path. Since the tapping force is always directed towards the center of the circle, this is the direction of the acceleration.

direction of the “nudge’ Finally, our last question is the most difficult. If something moving in a circle DOES, in fact, accelerate, and if the acceleration is ALWAYS towards the center of the circle, then what is the value of this acceleration? The answer is given by the equation

v2 a  c R

The variable “ac“ stands for “centripetal” acceleration. Centripetal means “center-seeking”. The variable “v” is the velocity of the object that is moving in the circular path. And finally, the variable “r” is the radius of the circular path.

Example problems:

1) Find the centripetal acceleration of a car that travels around a circular turn of radius 50 m if it travels at a speed of 25 m/s. [12.5 m/s2 inward towards circle’s center] v 2 252 a    12.5 m c r 50 s2

2) Find the velocity of a ball on a string that is being swung in a horizontal circle above a man’s head if the circular path has a radius of 1m and if the centripetal acceleration is 100 m/s2. If the string were to suddenly snap, which direction would the ball travel? Explain. [10 m/s ; it will fly off in a straight line, tangent to the circle, because of its inertia] v 2 v 2 a   100 m   v  10 m c r s2 1 s

There are two more concepts to understand in our basic study of circular motion. Both will apply to objects that are moving at a constant speed in this circular path. Again, be careful to remember that the object will accelerate (inward) even though it is moving at a constant speed.

The first concept is the period of the motion. We will use the variable “T” to denote the period. The period of the object is the time that it takes to complete one full circular revolution. A large period corresponds to a slow moving object. A small period corresponds to a fast moving object. We will use the simple equation “distance equals rate times time” to derive an equation for the period. Why can we use “d = r t”? Because the speed is constant.

What distance does an object have to move to complete one circular revolution? Obviously, it needs to travel one “circumference”. Since Circumference = C = 2R , the distance that the object needs to travel to complete one full revolution is 2R . How fast will it be able to do this? Well, since the speed is a constant, we will use the variable “v”. Therefore, d = r t or 2R  vT . Rearranging, we get the equation 2R T  (measured in seconds) v Since the period is simply the time needed to complete one revolution, the period will be measured in “time” units, or seconds.

Example problems:

3) A man swings a ball in on a string in a horizontal circle above his head. If the ball is traveling at a speed of 5 m/s and the circle has a radius of 50 cm, find the time needed for the ball to complete one full revolution. [.628 sec]

2r 2 (.50)     .628 m v 5 s

4) A man holds his son by the arms and starts to spin him around in a circle of radius 1.5m. If the man spins at a constant velocity, and if he completes 5 full revolutions in 10 seconds, find

a) the period of the boys motion. [2 sec] b) the speed of the little boy’s feet. [4.71 m/s] c) the speed of the little boy’s belt, which is only .75 m from the center of the circle. [2.36 m/s]

10sec 2r 2 (1.5) 2r 2 (.75)    2 sec  2sec v    4.71 m v    2.36 m 5rev rev  2 s  2 s

The second concept that we need to understand is the frequency of motion. We could say that a fast moving ball completes a full circle more frequently than a ball that is moving very slow. The frequency depends on how fast the object moves around the circle. We will measure frequency in “revolutions per second” using the equation 1 v f  or f  (measured in revolutions per second) T 2R

Example problems:

5) On a dare, a softball player pus her forehead on the end of a bat and begins to spin around in a circle. Amazingly, she maintains a constant velocity and competes 10 full turns in 12 seconds. Find: a) The period of the motion. [1.2 sec (per full turn)] b) The frequency of the motion. [0.833 turns per second]

12sec 1 1   1.2 sec 1.2sec f    .83Hz 10rev rev  1.2sec Finally, let’s recap

 If an object is moving in a circular path (even it is moving at a constant speed), it is definitely accelerating.  The direction of the acceleration is always INWARD, toward the center of the circle.  This inward, center-seeking acceleration is called “centripetal” acceleration and we v2 can find it using a  c R  The time necessary to complete one revolution is known as the Period (T), which 2R 1 can be found using T  or T  v f  The # of times that an object competes a full revolution per second is known as the 1 v frequency, which can be found using f  or f  T 2R Accelerated Physics Circular Motion Worksheet (Basic)

1. A washing machine drum rotates at 1200 revolutions per minute. What is its period (in seconds)?

1200rev 1200rev rev 1 1 f    20     .05sec/ rev  .05ec 1min 60sec sec f 20

2. An atom of hydrogen consists of an electron of mass 9.1 x 10 31 kg in circular orbit at uniform speed v. The orbital radius is 5.3 x 10  11 m and the period is 1.4 x 10  16 sec. Calculate the speed of the electron.

2r 2r 2 (5.3E 11) T   v    2.38E6 m v  (1.4E 16) s

3. A mass is moving at constant speed in a circle. How many revolutions per second will it perform if the period is:

a. 0.50 sec b. 0.01 sec c. 5.0 sec d. If it performs 4.0 rev/sec, what is the period? e. What is the relation between the period and the number of revolutions per second? f. The number of revolutions per second is called frequency f. Express algebraically f in terms of T.

1 1 1 1 f    20rev / sec  20Hz f  100Hz f   .20Hz T .05sec .01sec 5sec 1 T   0.25sec rev 4 sec f and T are reciprocals 1 1 f  and T  T f

4. Old fashion record turntables rotate at frequencies of 16.6 turns per min., 33.34 turns per min., 45.00 turns per min., and 78.00 turns per min. What is the period corresponding to each of these frequencies?

1 1 1 1 T    3.61sec T   1.8sec rev 1min rev 1min f 16.6 min 60sec  f 33.64 min 60sec  1 1 1 1 T   1.33sec T    0.77sec rev 1min rev 1min f 45 min 60sec  f 78 min 60 sec  5. A merry-go-round revolves at a frequency of 0.1 / sec. What is the period of this motion? The horses on this merry-go-round are 3 abreast with 1.5 m between each. The horse on the inside is 3.5 m from the center of rotation. What is the velocity of each horse?

1 1 T   10sec rev f 0.1 sec 2r 2 (3.5m) 2r 2 (5m) 2r 2 (6.5m) v    2.2 m v    3.14 m v    4.08 m  10sec s  10sec s  10sec s

6. At constant speed, a Porsche can make one revolution of a test track in 160 sec. The circumference of the track is 10.0 km. What is the car’s period? What is the car’s centripetal acceleration?

C  2r  10,000m  2r  r 1,591.55m m d  rt  10,000m  v(160sec)  v  62.5 s 2r 2 (1,591.55m) T   160sec m v 62.5 s v 2 62.52 a    2.45 m c r 1,591.55 s 2

7. What is the centripetal acceleration of an object moving along a horizontal circular path of 16.0 m radius with a speed of 40.0 m/s?

v 2 402 a   100 m c r 16 s2

8. In aviation, a “standard turn” for level flight of a propeller-type plane is one in which the plane makes a complete circular turn in 2.0 minutes. If the speed of the plane is 170 m/sec.

a. What is the radius of the circle? b. What is the centripetal acceleration of the plane?

120sec   120sec 1rev 2r Tv (120)(170) T   r    3,246.76m v 2 2 v 2 170 2 a    8.90 m c r 3,246.76 s2 9. HONORS ONLY: The earth is spinning…..really fast!!!! If someone stands on the equator, find the centripetal acceleration of this person. You should use the information from the back cover of your Wilson, Buffa, Lou Physics book to assist you. Then, decide whether this acceleration would make the person feel heavier or lighter. Then, decide what percentage of their regular weight a person at the equator would feel due to this spinning motion.

3 6 r = 6.378 ҙ 10 km = 6.378 ҙ 10 m T = 24hrs = 86,400sec 2pr v = = 463.8 m T s 2 2 v 463.8 m ac = = 6 = 0.034 2 r 6.378 ҙ 10 s

Since the earth is spinning in a circular path, people on this path always want to move in a straight line…OUT of the circle, thus lifting them off the earth. Thus, the person will “feel” a gravitational acceleration of (9.8 – 0.034) = 9.766 m/s2, and thus they will be (9.766/9.8)*100% = 99.7% of their normal weight.

10. HONORS ONLY: As an extension of problem #9 above, how fast would the earth need to rotate (and it currently rotates at a rate of 1 rev every 24 hours) in order for someone at the equator to feel “weightless”? 3 6 r = 6.378 ҙ 10 km = 6.378 ҙ 10 m T = ??? 2 2 v v m ac = ® 9.8 = 6 ® v = 7,905.97 2 r 6.378 ҙ 10 s 2pr 2p(6.378 106 ) T = = ҙ = 5068.85sec =1.41hrs v 7,905.97

Answers

1) 0.05 sec 2) 2.38 x 106 m/sec 3) 2.0; 100; 0.2; 0.25; T = 1/N; f = 1/T 4) 0.0602 min (3.6 sec); 0.030 min (1.8 sec); 0.02 min (1.33 sec); 0.013 min (0.77 sec) 5) T = 10 sec/rev; VINSIDE = 2.2 m/sec; VMIDDLE = 3.1 m/sec; VOUTSIDE = 4.1 m/s 6) T = 160 sec; a = 2.45 m/s2 7) 100 m/s2 8) 3..246 km; 8.9 m/s2 9) 463.8 m/s, 0.034 m/s2, 99.7% of regular weight 10) v = 7905.97 m/s, T = 1.41 hrs = 5068.85 sec