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Chapter 7 Review Questions
Chapter 13 contains answers and explanations.
There is an object of mass 2 kg with a linear momentum of 6 km/s.
Find the average strength of the force applied to the box.
One of the objects is moving with a speed of 2 m/s and the other is moving with a speed of 5 km/s.
Find the speed of the objects after the collision if the collision is inelastic.
The mass of Object 2 is twice that of Object 1 and it is initially at rest.
Two objects collide and separate.
Two carts are sitting at rest on a table.
One cart is pushed off the other by the teacher.
The mass of Object 1 is half that of Object 2.
The collision is strong.
The wooden block is suspended from a horizontal support by cords.
The block swings upward as a result of the perfectly inelastic impact of a bullet hitting it.
There is a change in momentum.
A system with no external forces has a conserved quantity of momentum.
A coordinate system can be created.
Sometimes you have to rebuild in the end.
Since man has looked up to the stars, we have always tried to understand why they move the way they do.
The stars and planets make elliptical circles, which is against the First Law in which objects will continue to move straight.
The answers explained motion in a linear fashion.
Our motion became parabolic when we added a second dimension.
When objects begin to move in a circular motion will be explored.
We will have a better understanding of the Moon's position around the Earth.
The object's speed around its path should be constant.
Although the speed may be constant, the direction of the velocity is always changing.
There must be acceleration since the velocity is changing.
This acceleration does not change the speed of the object, it only changes the direction of the velocity to keep it on its path.
The object would move off in a straight line if there wasn't a force.
The figures are below.
There is a figure on the left that shows an object moving along a circular trajectory.
The object's path is always tangential to the velocity vector.
The magnitudes are the same.
Most objects don't undergo uniform circular motion.
They follow ellipticals with different speeds.
This won't be tested for the AP physics 1 exam.
An object moving at constant speed in a circular path is undergoing uniform circular motion.
The magnitude of the force is given by this equation.
The force that produces centripetal acceleration points to the center of the path.
Centrifugal force is not a real force.
The net force from the physical forces on the object is called capillary force.
Identifying what forces produce the centripetal acceleration is the first thing to do in a problem like this.
This is a horizontal circle.
The unit has changed from 80 cm to 0.80 m.
The centripetal force is provided by static friction.
The centripetal force needed to keep him running in a circle would increase if the radius of the arcs were smaller.
He would slip if the centripetal force increased enough.
The speed of the car is 15 m/s at the very top of the circle, where the people are upside down.
If the diameter of the loop is 40 m and the total mass of the car is 1,200 kg, find the magnitude of the normal force by the track on the car at this point.
There are two forces acting on the car at its topmost point, both of which point downward.
The normal force is directed downward by the surface of the track.
Force tells an object how to move.
The car would fall straight down if it had zero velocity at this point.
The normal force pushes ninety degrees to the surface, even though the gravity still points downward.
The forces are against one another.
The centripetal force is still provided by the combination of these two forces.
We will make anything that points toward the center of the circle positive and anything that points away from the circle negative, because the centripetal acceleration points inward.
You would feel little force between you and the seat at the top of the loop, but you would feel a big slam at the bottom of the loop.
The strength is proportional to the product of the objects' mass and the distance between them as measured from center to center.
The pulling force is gravity.
2-on-1 act along the line that joins the bodies and form an action/reaction pair.
The law was published more than a hundred years ago.
The mass of the Earth and the radius of the Earth can be used to calculate gravity.
The mass of the Earth is determined by the radius of the Earth.
From above the North Pole, you can see the Earth.
It is possible to recognize the relationship between variables in formulas.
The centripetal force is provided by N.
The difference is so small that it can usually be ignored.
Satellites are often parked above Earth's surface.
The satellites have the same position on Earth's surface because they have the same amount of time in the air.
Determine the altitude that a satellite has to be above the equator.
The answer is that the Earth pulls.
Any object at the same distance from the Earth as the moon must move at the same speed.
There are questions about banking on the AP physics 1 exam.
Engineers often use banked curves to design and build roads.
Banking allows for cars to travel around a curve at or below the posted speed limit, without relying on the tires and road.
The curve is banked at 11.8 degrees if the radius of curvature is 60 m.
The centripetal force that the car experiences as it rounds the curve is produced by the horizontal component of the normal force.
The recommended speed is 40 km/hour.
We covered objects that move in a circular motion.
Taking those objects and spinning them is the next part of this chapter.
There were previous equations where objects were moved in a linear orientation.
We need a new set of equations that are similar to the physics of linear motion.
An object's mass is its resistance to acceleration.
The harder it is to change an object's speed, the more inertia it has.
The greater the inertia, the greater the force that is required in order for an object to be moved.
If the same force is applied on both objects, Object 1 will experience a smaller acceleration.
The force, mass, acceleration, and velocity are put in the linear model.
The relationship between the three rotational parameters and the linear parameters will be explored.
There are some basic definitions.
A toy car is going around a circle.
If you follow the path of the car, you will find that your fingers are counterclockwise.
The direction of your thumb is determined by your thumb.
It points out something on the page.
Many of the equations reflect linear equations.
Four children are on a carousel.
The objects were treated as a single particle in the preceding chapters.
The force is being delivered at a single point on the object.
Imagine a bunch of experiments.
We walk into a large room with a hammer and a small light.
The light will be attached to the end of the hammer in the first experiment.
We throw the hammer across the room after turning off the light.
We repeat the experiment again.
The small light should be attached to the head of the hammer.
We turn off the light, throw the hammer across the room, and trace the path of the hammer.
There was something important about that point.
The one point that gave a smooth path was the only one that gave spiraled trajectory.
If we place that point on our fingers, we can see that the hammer is horizontal with the floor.
The center of mass is this point.
It is possible to say that the center of mass is the point at which all the mass of the object can be concentrated.
The center of mass is the geometric center of the object.
The center of mass is motionless as every other point moves around it.
Pick a location that is convenient.
The formula above can be used to calculate the center of mass.
The stick matters.
If the stick has mass, it must be taken into account to determine its center position.
We could either palm the ball or put our hands on the opposite side of the ball and push one hand forward and the other backward.
In both cases, we need to exert force to make the object's center of mass accelerate.
We need to exert a Torque in order to make an object spin.
The measure of a force's effectiveness is called Torque.
Something must have a Torque if it starts to spin after being at rest.
If an object is spinning, something has to exert a Torque to stop it.
The systems that can spin have a "center" of turning.
While the rest of the object is rotating, the point that does not move becomes the center of the circle.
There are many different terms used to describe this point.
Students have difficulty understanding the topic of Torque.
There is a door with a pivot point on the left side of the drawing.
Some examples can be tried at home on a door to get a better understanding.
The door will close the fastest in the first situation.
In both situations, the door will close the fastest.
The door will not close in situation 3.
If you try this at home, you will see that it will be easier to close the door if you push it like Situation 2.
There were a few points that mattered when trying to close the door.
The amount of force used to close the door mattered.
The angle in which we pushed the door mattered.
In scenario 3, the place in which we pushed mattered.
Our force's effectiveness at rotating was determined by a few factors.
This is a cross product between your force and your radius.
The unit is called a newton-meter.
Torque is not a force because it is not in newtons.
There was a force being applied to the door straight into the pivot point.
The force was not enough to close the door.
It's the equivalent of force in trying to accelerate something.
A newton-meter became a joule in the previous chapter.
This isn't the case with Torque.
Torque problems can involve putting systems in equilibrium.
A student pulls down a rope with a force of 40 N and a pulley of 5 cm.
The two forces produce a Torque, but they don't like each other.
The Torque of F1 is counterclockwise and the Torque of F2 is counterclockwise.
Imagine the effect of each force if the other wasn't there.
It's important to balance one force's effectiveness at turning something clockwise with another force's effectiveness at turning it counter-clockwise.
The wall is connected to the bar's center by a wire.
The wall on the bar exerts contact force.
The components are sin 55deg.
When using center-of- mass, some problems are easier than others.
You can choose which one is easier.
This system can't be solved as is.
The second condition for equilibrium requires that the sum of the Torques about any point is zero.
cot 55o is what we use here.
We can put together the pieces of making an object spin now that we've studied Torque and rotation.
The moment of inertia is the tendency of an object in motion to rotate until acted upon by an outside force.
We are taking a ball at rest and speeding it up, which is rotational acceleration.
A force is needed to achieve this.
In terms of inertia, Torque is required.
We need to apply a force that works.
Some key relationships are made by this equation.
The larger the inertia, the smaller the value for an object.
It will be more difficult to rotation Object 1 than it will be to rotation Object 2.
If the same Torque is applied to both objects, Object 1 will undergo a smaller rotational acceleration.
There is more to it than the object's mass.
There are two objects that have the same mass.
How the mass is distributed in an object affects rotational inertia.
The greater the rotation inertia, the farther away the mass is from it.
Imagine a barbell with weights near each end and an identical barbell with weights near the middle of the bar.
The barbells have the same mass, but their inertias are different.
The first barbell has its attached mass farther away from the axis of rotation than the second barbell.
It was more difficult to rotation the first barbell than the second one.
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