Repeat (b) for Step 2.
Calculate the distance that the stunt driver should land from the bottom of the cliff.
You are pulling open a stuck drawer, but since you’re a physics geek you’re pulling it open with an electronic device that measures force! You measure the following behavior. The drawer has a weight of . Draw a graph of friction force vs. time.
Draw arrows representing the forces acting on the cannonball as it flies through the air. Assume that air resistance is small compared to gravity, but not negligible.
A tug of war erupts between you and your sweetie. Assume your mass is 60 kg and the coefficient of friction between your feet and the ground is (good shoes). Your sweetie’s mass is and the coefficient of friction between his/her feet and the ground is (socks). Who is going to win? Explain, making use of a calculation.
A block has a little block hanging out to its side, as shown: As you know, if the situation is left like this, the little block will just fall. But if we accelerate the leftmost block to the right, this will create a normal force between the little block and the big block, and if there is a coefficient of friction between them, then the little block won’t slide down! Clever, eh?
The mass of the little block is . What frictional force is required to keep it from falling? (State a magnitude and direction.)
If both blocks are accelerating to the right with an acceleration , what is the normal force on the little block provided by the big block?
What is the minimum coefficient of static friction required?
Answers to Selected Problems
.
.
.
Zero; weight of the hammer minus the air resistance.
forces
force
No
The towel’s inertia resists the acceleration
a. Same distance b. You go farther
c. Same amount of force
.
a. b.
.
.
.
Left picture: , right picture:
The rope will not break because his weight of is distributed between the two ropes.
Yes, because his weight of is greater than what the rope can hold.
Mass is and weight is
a. While accelerating down b.
c.
a. b.
b. c. d.
e. Eraser would slip down the wall
a. b.
c.
d. Friction between the tires and the ground
e. Fuel, engine, or equal and opposite reaction
b. c. no, the box is flat so the normal force doesn’t change
d.
e.
f. no
g.
h.
i.
j.
k.
.
a. zero b.
b. c. Ma
d. and
e. Solve by using and substituting for
a. Yes, because it is static and you know the angle and b. Yes, and the angle gives you and the angle and gives you and
a. seconds d.
.
.
.
a.
Chapter 7: Centripetal Forces Version 2
Forces so Far
In the absence of a net applied force, moving objects travel in a straight lines; this is Newton's First Law. If their velocity changes, even only in direction, there must be an applied force. When something experiences a net applied force, there are several possibilities:
The force is constant in magnitude and direction and points along the line of motion, or the object is at rest. In this case, the object will accelerate or decelerate in the direction of the force. The object's position and velocity can be found using the so called 'big three' equations.
The force is constant in magnitude and direction and acting at some non-right angle to the direction of motion. In this case, the object's velocity vector can be broken down into perpendicular components in such a way that on is parallel to the force. Then the problem is reduced to two one dimensional problems --- along the force and perpendicular to it --- which can be solved according to case 1 above. An example of this is parabolic motion under the influence of gravity.
The force is constant in magnitude, the object is in motion, and the force is always perpendicular to the velocity vector. In this case, the object will move in a circle. This is a kind of 'opposite' of case 1: then, the object's speed changed --- whether increased or decreased --- by the largest possible amount for a given force, while now the object's speed does not change at all.
The force is not constant The force is due to a compressed or extended spring. We will cover this later.
All other cases: beyond the scope of this book.
Centripetal Forces
Summary
Forces that cause objects to follow circular paths --- case 3 above --- are known as centripetal, or 'center seeking', forces. Such forces must continuously change direction to stay perpendicular to the velocity vector. We saw in the chapters on vectors and kinematics that vectors cannot impact motion in directions perpendicular to them. This is why the horizontal velocity of projectiles on earth does not change (as in two dimensional motion).
Speed vs. Direction
Think of a ball rolling horizontally off a cliff. At first, its velocity is perpendicular to the force of gravity. As it falls, its velocity in the x direction stays constant, but it accelerates downward due to gravity. This ball will not travel in circle, though, because gravity is only perpendicular to its velocity at the instant it leaves the cliff. Eventually, the ball's velocity components will be equal. After some time, if the cliff is tall enough, the ball's vertical velocity will dwarf its horizontal velocity.
Let's now compare how gravity affects the ball's speed at different 1 second intervals during its flight.
Let's say this ball initially had a horizontal velocity --- and therefore also speed --- of 100 m/s, and a vertical speed of 0. After the first second, its vertical velocity will about 10 m/s (assume . Using the pythagorean theorem, we find that the speed is now , a change of less than .5 m/s.
Now consider the ball after 10 seconds, when its velocity components are equal. Between the 10th and 11th second, its speed goes from to , a much bigger increase in the same time.
Finally, for the sake of argument, let's say the cliff is mount Everest and the ball keeps falling for 100 seconds. Now, in one second, its speed goes from to .
In other words, it is much easier for a force to change the speed of an object when it points along its velocity vector. Forces are capable of changing an object's velocity --- speed and direction: the more parallel to motion the force, the more it changes speed; the more perpendicular, direction (if we had analyzed the effect on the angle of the velocity vector above instead of speed, the results would have been reversed).
Circular Motion
Knowing this, we can understand why, when force is always perpendicular to the velocity vector, an object's speed never changes, while its direction changes continuously. If the force is constant and magnitude, the direction of the object's velocity must change at a constant rate --- otherwise the situation would be asymmetrical. In other words, the object will travel in a circle, with instantaneous velocity tangent to it and instantaneous force pointing toward the center. At any given time, the relationship between force, acceleration, and velocity is illustrated here:
Circular motion is kind of a limiting case of the 'first second' scenario above --- if the force had always been perpendicular to the ball's velocity, it wouldn't have accelerated downward for an entire second. No matter how weak a centripetal force, it will in principle always cause a moving object to travel in a circle. This may seem counterintuitive, but is actually a direct result of the arguments above.
Characterizing The Force and Motion
If a mass is traveling with velocity and experiences a centripetal ---always perpendicular --- force , it will travel in
a circle of radius
Gravity as a Centripetal Force
When can Gravity Act as a Centripetal Force?
We saw last chapter that the force of Gravity causes an attraction between two objects of mass and at a distance of By Newton's Third Law, both objects experience the force: equal in magnitude and opposite in direction, and both will move as an effect of it. If one of the objects is much lighter than the other (like the earth is to the sun, or a satellite is to earth) we can approximate the situation by saying that the heavier mass (the sun) does not move, since its acceleration will be far smaller due to its large mass. Then, if the lighter mass remains at a relatively constant absolute distance from the heavier one (remember, centripetal force needs to be constant in magnitude), we can say that the lighter mass experiences an effectively centripetal force.
Math of Centripetal Gravity
Gravity is not always a centripetal force. This is a really important point. It only acts as a centripetal force when conditions approximate those listed above --- very much like it isn't constant near the surface of the earth, but very close to it.
If gravity provides centripetal force and acceleration, we can set [2] equal to [4]. It's important to remember that in [2] refers to the lighter mass, since that is the one traveling. Then,
Key Concepts
For something in orbit, an orbital period, , is the time it takes to make one complete rotation.
If a particle travels a full circle or orbit --- a distance of --- in an amount of time , then its speed is distance over time or .
An object moving in a circle has an instantaneous velocity vector tangential to the circle of its path. The force and acceleration vectors point to the center of the circle.
Net force and acceleration always have the same direction.
Centripetal acceleration is just the acceleration provided by centripetal forces.
A geosynchronous orbit occurs when a satellite completes one orbit of the Earth every 24 hours. Since it revolves at the same rate as the earth, it will stay above the same location.
Key Applications
To find the maximum speed that a car can take a corner on a flat road without skidding out, set the force of friction equal to the centripetal force.
To find the tension in the rope of a swinging pendulum, remember that it is the sum of the tension and gravity that produces a net upward centripetal force. A common mistake is just setting the centripetal force equal to the tension.
To find the speed of a planet or satellite in an orbit, follow the example above.
Examples
Example 1
Question: You buy new tires for your car in order to take turns a little faster (uh, not advised---always drive slowly). The new tires double your coefficient of friction with the road. With the old tires you could take a particular turn at a speed of . What is the maximum speed you can now take the turn without skidding out.
Solution: To find the maximum speed a car can take a corner on a flat road without skidding out, set the force of friction equal to the centripetal force. This is because the centripetal force pushes the car off the road and the frictional force keeps the car on the road. Therefore, if the centripetal force and the frictional force were equal, the car would be going the maximum speed it could go on that turn without sliding off the road. Then we solve for v.
If we replace the frictional force with twice the frictional force we get the new speed. The new speed is greater than the original speed.
Centripetal Forces Problem Set
When you make a right turn at constant speed in your car what is the force that causes you (not the car) to change the direction of your velocity? Choose the best possible answer. Friction between your butt and the seat
Inertia
Air resistance
Tension
All of the above
None of the above
You buy new tires for your car in order to take turns a little faster (uh, not advised — always drive slowly). The new tires double your coefficient of friction with the road. With the old tires you could take a particular turn at a speed . What is the maximum speed you can now take the turn without skidding out?
Not enough information given
A pendulum consisting of a rope with a ball attached at the end is swinging back and forth. As it swings downward to the right the ball is released at its lowest point. Decide which way the ball attached at the end of the string will go at the moment it is released. Straight upwards
Straight downwards
Directly right
Directly left
It will stop
A ball is spiraling outward in the tube shown to the above. Which way will the ball go after it leaves the tube? Towards the top of the page
Towards the bottom of the page
Continue spiraling outward in the clockwise direction
Continue in a circle with the radius equal to that of the spiral as it leaves the tube
None of the above
An object of mass is in a circular orbit of radius at a velocity of . Calculate the centripetal force (in ) required to maintain this orbit.
What is the acceleration of this object?
Suppose you are spinning a child around in a circle by her arms. The radius of her orbit around you is meter. Her speed is . Her mass is . What is the tension in your arms?
In her arms?
A racecar is traveling at a speed of on a circular racetrack of radius . What is its centripetal acceleration in ?
What is the centripetal force on the racecar if its mass is ?
What provides the necessary centripetal force in this case?
The radius of the Earth is . Calculate the velocity of a person standing at the equator due to the Earth’s 24 hour rotation. Calculate the centripetal acceleration of this person and express it as a fraction of the acceleration g due to gravity. Is there any danger of “flying off”?
Neutron stars are the corpses of stars left over after supernova explosions. They are the size of a small city, but can spin several times per second. (Try to imagine this in your head.) Consider a neutron star of radius that spins with a period of seconds. Imagine a person is standing at the equator of this neutron star. Calculate the centripetal acceleration of this person and express it as a multiple of the acceleration due to gravity (on Earth).
Now, find the minimum acceleration due to gravity that the neutron star must have in order to keep the person from flying off.
Calculate the force of gravity between the Sun and the Earth. (The relevant data are included in Appendix .)
Calculate the force of gravity between two human beings, assuming that each has a mass of and that they are standing apart. Is this a large force?
Prove g is approximately on Earth by following these steps: Calculate the force of gravity between a falling object (for example an apple) and that of Earth. Use the symbol to represent the mass of the falling object.
Now divide that force by the object’s mass to find the acceleration of the object.
Our Milky Way galaxy is orbited by a few hundred “globular” clusters of stars, some of the most ancient objects in the universe. Globular cluster is orbiting at a distance of light-years (one light-year is ) and has an orbital period of million years. The mass of the cluster is times the mass of the Sun. What is the amount of centripetal force required to keep this cluster in orbit?
What is the source of this force?# Based on this information, what is the mass of our galaxy? If you assume that the galaxy contains nothing, but Solar-mass stars (each with an approximate mass of ), how many stars are in our galaxy?
Calculate the centripetal acceleration of the Earth around the Sun.
You are speeding around a turn of radius at a constant speed of What is the minimum coefficient of friction µ between your car tires and the road necessary for you to retain control?
Even if the road is terribly icy, you will still move in a circle because you are slamming into the walls.
What centripetal forces must the walls exert on you if you do not lose speed? Assume .
Calculate the gravitational force that your pencil or pen pulls on you. Use the center of your chest as the center of mass (and thus the mark for the distance measurement) and estimate all masses and distances. If there were no other forces present, what would your acceleration be towards your pencil? Is this a large or small acceleration?
Why, in fact, doesn’t your pencil accelerate towards you?
A digital TV satellite is placed in geosynchronous orbit around Earth, so it is always in the same spot in the sky. Using the fact that the satellite will have the same period of revolution as Earth, calculate the radius of its orbit.
What is the ratio of the radius of this orbit to the radius of the Earth?
Draw a sketch, to scale, of the Earth and the orbit of this digital TV satellite.
If the mass of the satellite were to double, would the radius of the satellite’s orbit be larger, smaller, or the same? Why?
CK-12 People's Physics Book Version 2 Page 6