2.
Acceleration Characteristics for Circular Motion
An object moving in uniform circular motion is moving in a circle with a uniform or constant speed.
3.
Explain why there is acceleration in circular motion if the SPEED is Constant.
An accelerating object is an object which is changing its velocity.
Since velocity is a vector which has both magnitude and direction, a change in either the magnitude or the direction constitutes a change in the velocity.
For this reason, it can be boldly declared that an object moving in a circle at constant speed is indeed accelerating.
It is accelerating because its velocity is changing its directions.
4.
In which direction does the velocity change (Acceleration) vector point?
The velocity change vector is directed towards the center.
An object moving in a circle at a constant speed from A to B experiences a velocity change and therefore an acceleration.
This acceleration is directed towards the center of the circle .
In this time, the velocity has changed from v i to v f . The process of subtracting v i from v f is shown in the vector diagram; this process yields the change in velocity.
There is a velocity change for an object moving in a circle with a constant speed.
Note that this velocity change vector is directed towards the center.
In the case of an object moving in a circle about point C, the object has moved from point A to point B.
6.
Give two real world demonstrations of this inward acceleration.
If a glass with a lit candle is held at the end of an outstretched arm as you spin in a circle at a constant rate (such that the flame experiences an acceleration), then the candle flame will no longer extend vertically upwards.
Instead the flame deflects from its upright position, signifying that there is an acceleration when the flame moves in a circular path at constant speed.
The deflection of the flame will be in the direction of the acceleration.
When you do this experiment, you find that the flame deflects towards the center of the circle, thus indicating that not only is there an acceleration; but that there is an inward acceleration.
7.
Give two real world demonstrations of this inward acceleration.
a. Use a cork accelerometer (a cork submerged in a sealed flask of water) held in an outstretched arm and move in a circle at a constant rate of turning.
As the cork-water combination spun in the circle, the cork leaned towards the center of the circle.
Once more, there is proof that an object moving in circular motion at constant speed experiences an acceleration which directed towards the center of the circle.
8.
Do the “C heck Your Understanding ” #1-8 on the site.
Go to the website to use the interactive questions.
9.
An object moving in a circle is experiencing acceleration. Even if moving around the perimeter of the circle with a constant speed, there is still a change in velocity and subsequently an acceleration. This acceleration is directed TOWARDS THE CENTER of the circle.
an object which experiences an acceleration must also be experiencing a net force
the direction of the net force is in the same direction as the acceleration.
11.
So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration.
12.
This is sometimes referred to as the CENTRIPETAL FORCE REQUIREMENT.
13.
The word "centripetal" means CENTER-SEEKING. For objects moving in circular motion, there is a net force acting towards the center which causes the object to seek the center.
14.
Newton's first law of motion - the law of inertia states that….
"... objects in motion tend to stay in motion with the same speed and the same direction unless acted upon by an unbalanced force."
15.
Objects will tend to naturally travel in straight lines; an unbalanced force is required to cause it to turn. The presence of THE UNBALANCED FORCE is required for objects to move in circles.
16.
There is an outward force and an outward acceleration acting on your body when you sit in the passenger seat of a car making a left hand turn.
We are not introducing a new type of force but rather describing the direction of the net force acting upon the object which moves in the circle.
Whatever the object, if it moves in a circle, there is some force acting upon it to cause it to deviate from its straight-line path, accelerate inwards and move along a circular path.
As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion.
You are carrying a tennis ball upon a flat, level board.
Once the tennis ball and the board are in motion, they will continue in motion in the same direction at the same speed unless acted upon by an unbalanced force.
This is in accord with Newton's first law of motion.
Now a block is secured to the board in such a manner that the block applies an unbalanced force to the ball that is directed towards the center of the circle.
The block provides a normal force directed inward.
What will happen to the ball as the flat board is moved in a circle?
43.
The acceleration of an object moving in a circle can be determined by the equations
44.
The net force is related to the acceleration of the object (as is always the case) and is given by the equations
45.
This set of circular motion equations can be used in two ways:
a. as a recipe for algebraic problem solving in order to solve for an unknown quantity.
b. as a "guide to thinking" about how an alteration in one quantity would effect a second quantity
.
46.
How can the equations for circular motion be used as a guide to thinking?
the equation relating the net force ( F net ) to the speed ( v ) of an object moving in uniform circular motion is F net = mv 2 /R
This equation shows that the net force required for an object to move in a circle is directly proportional to the square of the speed of the object. For a constant mass and radius, the F net is proportional to the speed 2 .
The factor by which the net force is altered is the square of the factor by which the speed is altered. Subsequently, if the speed of the object is doubled, the net force required for that object's circular motion is quadrupled. And if the speed of the object is halved (decreased by a factor of 2), the net force required is decreased by a factor of 4.
The process of solving a circular motion problem is much like any other problem in physics class.
The process involves:
a careful reading of the problem
the identification of the known and required information in variable form
the selection of the relevant equation(s)
substitution of known values into the equation
finally algebraic manipulation of the equation to determine the answer
49.
A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.
To determine the acceleration of the car, use the equation a = (v 2 )/R. The solution is as follows:
a = (v 2 )/R a = ((10.0 m/s) 2 )/(25.0 m)
a = (100 m 2 /s 2 )/(25.0 m)
a = 4 m/s 2
To determine the net force acting upon the car, use the equation Fnet = m*a. The solution is as follows.
F net = m*a F net = (900 kg)*(4 m/s 2 )
F net = 3600 N
50.
A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path which is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed of the halfback.
To determine the speed of the halfback, use the equation v = d/t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:
v = d/t v = (0.25 * 2 * pi * R)/t
v = (0.25 * 2 * 3.14 * 12.0 m)/(2.1 s)
v = 8.97 m/s
51.
A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path which is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the acceleration of and net force acting upon the halfback.
To determine the acceleration of the halfback, use the equation a = (v 2 )/R. The solution is as follows:
a = (v 2 )/R a = ((8.97 m/s) 2 )/(12.0 m)
a = (80.5 m 2 /s 2 )/(12.0 m)
a = 6.71 m/s 2
To determine the net force acting upon the halfback, use the equation Fnet = m*a. The solution is as follows.
F net = m*a F net = (95.0 kg)*(6.71 m/s 2 )
F net = 637 N
52.
Anna fills a bucket with water, ties it to a strong rope, and spins it in a circle. Anna spins the bucket when it is half-full of water and when it is quarter-full of water. In which case is more force required to spin the bucket in a circle?
Answer the question and then check at The Physics Classroom!
53.
A Lincoln Continental and a Yugo are making a turn. The Lincoln is four times more massive than the Yugo. If they make the turn at the same speed, then how do the centripetal forces acting upon the two cars compare. Explain.
Answer the question and then check at The Physics Classroom!
54.
The Gravitron at Six Flags is a ride in which occupants line the perimeter of a cylinder and spin in a circle at a high rate of turning. When the cylinder begins spinning very rapidly, the floor is removed from under the riders' feet. What effect does a doubling in speed have upon the centripetal force? Explain.
Answer the question and then check at The Physics Classroom!
55.
Determine the centripetal force acting upon a 40-kg child who makes 10 revolutions around the Cliffhanger in 29.3 seconds. The radius of the barrel is 2.90 meters.
Answer the question and then check at The Physics Classroom!
abc
test
sdfgds