Centripetal Force and Acceleration - AP Physics 1
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A 
boy is riding a merry-go-round with a radius of 
. What is the centripetal force on the boy if his velocity is 
?
A
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For this problem, we use the centripetal force equation:
We are given the mass, radius or rotation, and the linear velocity. Using these values, we can find the centripetal force.
For this problem, we use the centripetal force equation:
We are given the mass, radius or rotation, and the linear velocity. Using these values, we can find the centripetal force.
A ball is attached to a string that is 1.5m long. It is spun so that it completes two full rotations every second. What is the centripetal acceleration felt by the ball?
A ball is attached to a string that is 1.5m long. It is spun so that it completes two full rotations every second. What is the centripetal acceleration felt by the ball?
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We are simply asked to find the centripetal acceleration, which is given by:
We were given 
in the problem statement (radius will be equal to the length of the string), so we only need to find the velocity of the ball.
We are told that it travels in a circle with radius 1.5m and completes two full rotations per second. The length of each rotation is just the circumference of the circle:
The velocity can be found by multiplying that distance by the frequency:
Now we have all of our variables and can plug into our first equation:
We are simply asked to find the centripetal acceleration, which is given by:
We were given
We are told that it travels in a circle with radius 1.5m and completes two full rotations per second. The length of each rotation is just the circumference of the circle:
The velocity can be found by multiplying that distance by the frequency:
Now we have all of our variables and can plug into our first equation:
An amusement ride is used to teach students about centripetal force. The ride is a circlular wall that you place your back on. The wall and floor then begin to spin. Once it reaches a certain rotational velocity, the floor drops, and the students are pinned to the wall as a result of centripital force.
A student of mass 50kg decides to go on the ride. The coefficient of static friction between the student and wall is 0.8. If the diameter of the ride is 10m, what is the maximum period of the ride's rotation that will keep the student pinned to the wall once the floor drops?
An amusement ride is used to teach students about centripetal force. The ride is a circlular wall that you place your back on. The wall and floor then begin to spin. Once it reaches a certain rotational velocity, the floor drops, and the students are pinned to the wall as a result of centripital force.
A student of mass 50kg decides to go on the ride. The coefficient of static friction between the student and wall is 0.8. If the diameter of the ride is 10m, what is the maximum period of the ride's rotation that will keep the student pinned to the wall once the floor drops?
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There is a lot going on in this problem and it will take several steps to get to the answer. However, when you boil the question down, we are pretty much asked how fast must the ride spin so that the centripetal force on the student provides enough static friction to keep the student from falling. Let's work through this problem one step at a time.
First, let's figure out what minimal static frictional force is required. This force will be equal to the weight of the student; the student's weight will pull downward, while the friction of the wall pushes upward.
Now we can calculate the normal force required to reach that magnitude of frictional force. Note that the vector for the normal force will be perpendicular to the wal, directed toward the center of the circlel.
This normal force is the minimum centripital force required to keep the student pinned to the wall. We can convert this to centripital acceleration:
We can now convert centripital acceleration to a translational velocity using the equation:
Rearranging for velocity, we get:
This is the velocity that the outer wall of the ride must be spinning at. Since we know the radius of the ride, we can convert this velocity into a maximum period, the final answer:
If you weren't sure how to come about this equation, just think about your units. You know you need to get to units of seconds, and you have a value with units of m/s. Therefore, you need to cancel out the meter and get the second on top.
There is a lot going on in this problem and it will take several steps to get to the answer. However, when you boil the question down, we are pretty much asked how fast must the ride spin so that the centripetal force on the student provides enough static friction to keep the student from falling. Let's work through this problem one step at a time.
First, let's figure out what minimal static frictional force is required. This force will be equal to the weight of the student; the student's weight will pull downward, while the friction of the wall pushes upward.
Now we can calculate the normal force required to reach that magnitude of frictional force. Note that the vector for the normal force will be perpendicular to the wal, directed toward the center of the circlel.
This normal force is the minimum centripital force required to keep the student pinned to the wall. We can convert this to centripital acceleration:
We can now convert centripital acceleration to a translational velocity using the equation:
Rearranging for velocity, we get:
This is the velocity that the outer wall of the ride must be spinning at. Since we know the radius of the ride, we can convert this velocity into a maximum period, the final answer:
If you weren't sure how to come about this equation, just think about your units. You know you need to get to units of seconds, and you have a value with units of m/s. Therefore, you need to cancel out the meter and get the second on top.
Candy companies have long strived to catch the attention of children. One item that does this particularly well is the gumball machine. A certain gumball machine has a column that is 
tall with a spiral track of radius 
on which the gumball travels. The slope of the track is 
and the average frictional force exerted on the gumball as it travels down the track is 
. What is the the centripetal force on a gumball of mass 
as it reaches the end of the track?
Candy companies have long strived to catch the attention of children. One item that does this particularly well is the gumball machine. A certain gumball machine has a column that is
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We can use the expression for conservation of energy to solve this problem:
Substituting in our expressions for each variable and removing initial kinetic energy and final potential energy (which will each be zero), we get:
Rearranging for final velocity:
The only variable we don't have at this point is the distance the gumball travels. However, we can calculate it knowing the height of the track and its slope. We can imagine that the spiral track is unwound, creating a right triangle with an angle of 10 degrees and a height of 1.5m. For this triangle, the hypotenuse will be the total distance of the track.
Now that we have all of our variables, we can solve for the final velocity:
We can then use this to calculate the centripetal force on the gumball:
We can use the expression for conservation of energy to solve this problem:
Substituting in our expressions for each variable and removing initial kinetic energy and final potential energy (which will each be zero), we get:
Rearranging for final velocity:
The only variable we don't have at this point is the distance the gumball travels. However, we can calculate it knowing the height of the track and its slope. We can imagine that the spiral track is unwound, creating a right triangle with an angle of 10 degrees and a height of 1.5m. For this triangle, the hypotenuse will be the total distance of the track.
Now that we have all of our variables, we can solve for the final velocity:
We can then use this to calculate the centripetal force on the gumball:
A ball of mass 
is on a string of length 
. If the ball is being spun in vertical circles at a constant velocity and with a period of 
, what is the maxmium tension in the string?
A ball of mass
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First, we need to identify at which point in the circle the string is experiencing the most tension. There are two total forces in the system: gravity and tension. It is important to note that the tension isn't only resulting from gravity; it also includes the centripetal force required to keep the ball in circular motion. Thinking practically, we can say that the greatest tension will be when the ball is at its lowest point (gravity and tension are in opposite directions). At this point we can write:
Expand our terms for force:
We know the acceleration due to gravity, but we need to determine the centripetal accerleration. The formula for that is:
We know the radius (length of the string), so we need to develop an expression for velocity. We can use the period and circumference of circle:
Here, we use 
to denote period.
Substituting this into the expression for centripetal acceleration:
Substituting this back into the equation for tension, we get:
We have all of these values, allowing us to solve:
First, we need to identify at which point in the circle the string is experiencing the most tension. There are two total forces in the system: gravity and tension. It is important to note that the tension isn't only resulting from gravity; it also includes the centripetal force required to keep the ball in circular motion. Thinking practically, we can say that the greatest tension will be when the ball is at its lowest point (gravity and tension are in opposite directions). At this point we can write:
Expand our terms for force:
We know the acceleration due to gravity, but we need to determine the centripetal accerleration. The formula for that is:
We know the radius (length of the string), so we need to develop an expression for velocity. We can use the period and circumference of circle:
Here, we use
Substituting this into the expression for centripetal acceleration:
Substituting this back into the equation for tension, we get:
We have all of these values, allowing us to solve:
A car of mass 
is driving around a circular track of radius 
at a constant velocity of 
. The centripetal force acting on the car is 
. If the car's velocity is doubled, what is the new centripetal force required for the car to drive on the circular track?
A car of mass
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The equation for centripetal force is:
.
If 
is doubled and becomes 
, 
is quadrupled. Centripetal force is proportional to the square of velocity.
The equation for centripetal force is:
If
Troy spins a ball attached to a string around in a vertical circle. Assume the ball travels at constant speed during its circular path. Which of the following is not constant throughout the ball's path?
Troy spins a ball attached to a string around in a vertical circle. Assume the ball travels at constant speed during its circular path. Which of the following is not constant throughout the ball's path?
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Let's go through each answer choice one by one. The force of gravity on the ball is always its mass times gravity, which is constant. The string's pull on the ball is always perpendicular to the ball's motion, so it isn't doing any work, and thus has a constant rate of work 
. The ball's angular momentum is always constant since its mass, the radius of the circle it's traveling in, and its speed are constant. The only the changing is the tension in the string. This is true because the ball needs the same center pulling force (centripetal force) throughout its path, but this centripetal force is constantly changing direction. For example, at the top of its path, the centripetal force must be down because that's the direction of the center of its circle of motion, but at the bottom, the centripetal force must be up. There are two forces on the ball, gravity and tension. The force of gravity is always down and constant, so the tension needs to change in order to keep the center pulling force the same. For example, when the ball is at the bottom of the circle, the force of gravity is away from the center of the circle, so tension must be higher to keep the center pull constant.
Let's go through each answer choice one by one. The force of gravity on the ball is always its mass times gravity, which is constant. The string's pull on the ball is always perpendicular to the ball's motion, so it isn't doing any work, and thus has a constant rate of work
At the 2015 NCAA Track Championships, a competitor in the hammer is in the midst of his throw. During a period of his pre-throw spinning, his hammer is moving at a constant angular velocity of 
in its circular orbit. Being a regulation hammer, it weighs 
, and the length of its rope is 
. What is the magnitude of the force exerted towards the competitor along the hammer's rope?
At the 2015 NCAA Track Championships, a competitor in the hammer is in the midst of his throw. During a period of his pre-throw spinning, his hammer is moving at a constant angular velocity of
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Because the hammer is moving at a constant angular velocity, the centripetal force along the hammer rope can be expressed as 
. By plugging in the numbers given by the problem we get 
which, when simplified gives us approximately 
or 
Because the hammer is moving at a constant angular velocity, the centripetal force along the hammer rope can be expressed as
Consider a situation in which a 2kg block slides down a ramp and then around a circular loop with a radius of 12m. If friction between the surfaces is negligible, what is the minimum height that the block can start off at so that it will go all the way around the loop without falling off?
Consider a situation in which a 2kg block slides down a ramp and then around a circular loop with a radius of 12m. If friction between the surfaces is negligible, what is the minimum height that the block can start off at so that it will go all the way around the loop without falling off?
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To start off with, it is useful to consider the energy of the system. Initially, the block is at a certain unknown height, and will thus have gravitational potential energy.
Since we are assuming that there is no friction in this case, then we know that total mechanical energy will be conserved. Therefore, all of the gravitational potential energy contained in the block will become kinetic energy when it slides down to the bottom of the loop. But, once the block begins to slide up the loop, it will lose kinetic energy and will regain some gravitational potential energy. Therefore, at the top of the loop, the block will have a combination of kinetic and gravitational potential energy whose sum is equal to the initial energy of the system.
Due to conservation of energy, we can equate the two.
In addition to considering energy, it is also necessary to consider the forces acting on the block in this scenario. When at the top of the loop, the block will experience a downward force due to its weight, and another downward force due to the normal force of the loop on the block. Furthermore, because the block is traveling along a circular path while in the loop, it will experience a centripetal force. At the top of the loop, the centripetal force will be due to a combination of the weight of the block as well as the normal force.
We're looking for a starting height that will just allow the block to travel around the loop. The minimum height will be the height such that the block will just start to fall off. When falling off, the block will no longer be touching the loop and therefore, the normal force will be equal to zero. This is the situation we are looking for, and since the normal force is zero, only the block's weight will contribute to the centripetal force at the top of the loop.
The preceding expression gives us the value of velocity that will allow the block to have enough kinetic energy while at the top of the loop to not fall off. We can plug this expression into the previous energy equation.
To start off with, it is useful to consider the energy of the system. Initially, the block is at a certain unknown height, and will thus have gravitational potential energy.
Since we are assuming that there is no friction in this case, then we know that total mechanical energy will be conserved. Therefore, all of the gravitational potential energy contained in the block will become kinetic energy when it slides down to the bottom of the loop. But, once the block begins to slide up the loop, it will lose kinetic energy and will regain some gravitational potential energy. Therefore, at the top of the loop, the block will have a combination of kinetic and gravitational potential energy whose sum is equal to the initial energy of the system.
Due to conservation of energy, we can equate the two.
In addition to considering energy, it is also necessary to consider the forces acting on the block in this scenario. When at the top of the loop, the block will experience a downward force due to its weight, and another downward force due to the normal force of the loop on the block. Furthermore, because the block is traveling along a circular path while in the loop, it will experience a centripetal force. At the top of the loop, the centripetal force will be due to a combination of the weight of the block as well as the normal force.
We're looking for a starting height that will just allow the block to travel around the loop. The minimum height will be the height such that the block will just start to fall off. When falling off, the block will no longer be touching the loop and therefore, the normal force will be equal to zero. This is the situation we are looking for, and since the normal force is zero, only the block's weight will contribute to the centripetal force at the top of the loop.
The preceding expression gives us the value of velocity that will allow the block to have enough kinetic energy while at the top of the loop to not fall off. We can plug this expression into the previous energy equation.
A person is crossing a canyon by swinging on a vine, as shown in the given figure. The person has a mass of 90kg and the vine he is using is 15m long from where it connects to the tree to the person's center of mass. At the instant he is at the lowest point of the swing, so the vine is going straight up from his hands, the person is moving at 
. At that instant, what is the tension force in the vine?
A person is crossing a canyon by swinging on a vine, as shown in the given figure. The person has a mass of 90kg and the vine he is using is 15m long from where it connects to the tree to the person's center of mass. At the instant he is at the lowest point of the swing, so the vine is going straight up from his hands, the person is moving at
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Like all force problems, this one starts with a clear free body diagram:
The tension 
points along the vine (tensions can only pull), so it goes straight up. The force of gravity 
points straight down, as it always does. The two do not add to zero, however, since the person is undergoing circular motion. Instead, they add to a net force pointing towards the center of the circle that the person is making, which is up at the place where the vine is attached to the tree. Draw a vector diagram:
Then write the equation about the lengths of the vectors: the length of the gravity vector plus the length of the net force vector equals the length of the tension vector:
The net force for an object undergoing circular motion is mass times speed squared divided by the radius of the circle. 
is the gravity force constant. Some use 
, but the AP physics 1 test allows you to use 
, which makes it a lot easier. Plug in the numbers, and solve for the tension:
This answer is reasonable since the vine has to both hold the person up and provide a centripetal force; that is why the tension is is greater than his weight alone.
Like all force problems, this one starts with a clear free body diagram:
The tension
Then write the equation about the lengths of the vectors: the length of the gravity vector plus the length of the net force vector equals the length of the tension vector:
The net force for an object undergoing circular motion is mass times speed squared divided by the radius of the circle.
This answer is reasonable since the vine has to both hold the person up and provide a centripetal force; that is why the tension is is greater than his weight alone.
You tie a 
rock to a rope and swing it above your head with a frequency of 
. The distance from your hand to the rope is 
. Determine the centripital force experience by the rock.
You tie a
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Centripital acceleration is
Force is defined as
Using subsitution
The rock has a frequency of 
.
Each circle it makes has a circumference of 
, which is equivalent to 
.
That means it travel 
in 
, which is
Plugging in our values:
Centripital acceleration is
Force is defined as
Using subsitution
The rock has a frequency of
Each circle it makes has a circumference of
That means it travel
Plugging in our values:
Imagine a car driving over a hill at a constant speed. Once the car has reached the apex of the hill, what is the direction of the acceleration?
Imagine a car driving over a hill at a constant speed. Once the car has reached the apex of the hill, what is the direction of the acceleration?
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If we imagine the hill as a semi-circle, it appears that the car is moving along a circle. At the apex of the hill, the car's acceleration points downwards as this points towards the center of the circle. If an object travels in a circular fashion, at a constant speed, the direction of acceleration is always towards the center of the circle. This type of acceleration arises do to the change in velocity. Although the speed is constant, the direction changes.
If we imagine the hill as a semi-circle, it appears that the car is moving along a circle. At the apex of the hill, the car's acceleration points downwards as this points towards the center of the circle. If an object travels in a circular fashion, at a constant speed, the direction of acceleration is always towards the center of the circle. This type of acceleration arises do to the change in velocity. Although the speed is constant, the direction changes.
A 
ball is being swung faster and faster in a horizontal circle. When the speed reaches 
, the tension (
) is 
and the string breaks. What is the radius 
of the circle?
A
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In order to find the radius of the circle that corresponds to the maximum tension the string can experience without breaking, we must rearrange the centripetal force equation for objects moving in a circle:
In order to find the radius of the circle that corresponds to the maximum tension the string can experience without breaking, we must rearrange the centripetal force equation for objects moving in a circle:
The passengers in a car driving over a hill at a speed of 
feel weightless.
Calculate the radius of curvature of the hill.
The passengers in a car driving over a hill at a speed of
Calculate the radius of curvature of the hill.
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We can solve this problem by looking at all of the forces acting on the car at the top of the circular path.
The weight is positive, because it points towards the center of the circle, while the normal force 
is negative because it points away from the circle.
Since we are asked to look at the special case where the occupants of the car feel weightless, this tells us that the normal force experienced by the occupants must be zero.
We can solve this problem by looking at all of the forces acting on the car at the top of the circular path.
The weight is positive, because it points towards the center of the circle, while the normal force
Since we are asked to look at the special case where the occupants of the car feel weightless, this tells us that the normal force experienced by the occupants must be zero.
You are riding a rollercoaster going around a vertical loop, on the inside of the loop. If the loop has a radius of 
, how fast must the cart be moving in order for you to feel three times as heavy at the top of the loop?
You are riding a rollercoaster going around a vertical loop, on the inside of the loop. If the loop has a radius of
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To solve this problem, it is important to remember the signs of each force acting on the coaster. Since we are inside the loop at the top, both the weight and the normal force point toward to center of the circular path, and are thus positive.
Next, we note that the normal force is what a rider "feels" during their ride on the rollercoaster. We want the speed at which a rider would feel three times as heavy, or three times their weight.
To solve this problem, it is important to remember the signs of each force acting on the coaster. Since we are inside the loop at the top, both the weight and the normal force point toward to center of the circular path, and are thus positive.
Next, we note that the normal force is what a rider "feels" during their ride on the rollercoaster. We want the speed at which a rider would feel three times as heavy, or three times their weight.
Imagine a table that has a hole drilled through the center. Through this hole is a string that connects one mass that hangs under the table, to a mass that is on top of the table and spinning. mass M weight 10 kg, mass m weighs 5 kg and the radius, R, of the circle mass M traces out is 50 cm. At what speed, v, does the mass M need to rotate at in order to keep mass m suspended?
Imagine a table that has a hole drilled through the center. Through this hole is a string that connects one mass that hangs under the table, to a mass that is on top of the table and spinning. mass M weight 10 kg, mass m weighs 5 kg and the radius, R, of the circle mass M traces out is 50 cm. At what speed, v, does the mass M need to rotate at in order to keep mass m suspended?
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For this problem, we recognize that the force keeping the mass m suspended is the tension in the string. This tension is balanced by the gravitational force on m such that m does not accelerate. So how do we relate the speed of M to the tension? We must ask ourselves why is there tension in the string at all? Any object that travels in a circle has some centripital acceleration acting on it that point towards the center of the circle (for constant speed). This force is equal to the mass time the centripial acceleration.
I have stated previously that this force must be equal to the gravititational force acting on mass m. So,
Solving for v,
Plugging in values yields a speed of 1.6 m/s
For this problem, we recognize that the force keeping the mass m suspended is the tension in the string. This tension is balanced by the gravitational force on m such that m does not accelerate. So how do we relate the speed of M to the tension? We must ask ourselves why is there tension in the string at all? Any object that travels in a circle has some centripital acceleration acting on it that point towards the center of the circle (for constant speed). This force is equal to the mass time the centripial acceleration.
I have stated previously that this force must be equal to the gravititational force acting on mass m. So,
Solving for v,
Plugging in values yields a speed of 1.6 m/s
On a wet road, the coefficient of static friction between a car's tires and the flat road is 
. What is the maximum speed a car can safely navigate a turn with a 
radius of curvature?
On a wet road, the coefficient of static friction between a car's tires and the flat road is
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When a car is going around a flat turn, the only force keeping the car going along a circular path is the force of friction, which is equal to the centripetal force. We do not need the mass of the car to calculate the maximum speed, as we will show below.
Rearrange these equations and solve for the velocity.
When a car is going around a flat turn, the only force keeping the car going along a circular path is the force of friction, which is equal to the centripetal force. We do not need the mass of the car to calculate the maximum speed, as we will show below.
Rearrange these equations and solve for the velocity.
We can determine the centripetal force exerted by the nucleus on an electron. The diameter of an atom is 
, and an electron is moving at
in a circular motion around the nucleus. The mass of an electron is roughly 
.
What is the centripetal force exerted by the nucleus on an electron?
We can determine the centripetal force exerted by the nucleus on an electron. The diameter of an atom is
What is the centripetal force exerted by the nucleus on an electron?
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The formula for centripetal force is:
Where 
is the mass of the object, 
is its velocity, and 
is the radius of the circle made by the motion of the object around the center. We can model an atom like a circle, with its nucleus being its center. Since we are given the diameter of an atom, its radius will be
The formula for centripetal force is:
Where
A 
object moves in a circular motion with a diameter of 
. What is the magnitude of this object's velocity?
A
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Let's start with what we know, and what we want to know. We are given the mass of an object, and we're told that it's traveling in a circular orbit. Also, we have been given the diameter of the orbit. What we are trying to find is the velocity of the object.
In this problem, we'll need to consider the centripetal force of the object as it travels in a circular orbit. The force that is contributing to the centripetal force is the weight of the object. Therefore, we can write the following expression:
The above expression tells us that the object's velocity is dependent on the acceleration due to gravity, as well as the radius of the orbit. Since we're told that the diameter of the orbit is 
, we can find the radius by taking half of this value, which is 
. Moreover, the plus-or-minus sign indicates that the object can either be traveling in a clockwise or counter-clockwise orbit. In either case, the object will have the same magnitude of velocity.
Solving for velocity, we can plug in values to obtain:
Let's start with what we know, and what we want to know. We are given the mass of an object, and we're told that it's traveling in a circular orbit. Also, we have been given the diameter of the orbit. What we are trying to find is the velocity of the object.
In this problem, we'll need to consider the centripetal force of the object as it travels in a circular orbit. The force that is contributing to the centripetal force is the weight of the object. Therefore, we can write the following expression:
The above expression tells us that the object's velocity is dependent on the acceleration due to gravity, as well as the radius of the orbit. Since we're told that the diameter of the orbit is
Solving for velocity, we can plug in values to obtain:
A ball of mass 
is being swung in a vertical circle on a string, as shown in the given figure. At the instant it is directly above the center of the circle it is making, its speed is 
. The string has negligible mass, and its length is 
. What is the tension, 
, in the string at the instant the ball is at the top of its circle?
A ball of mass
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The tension and gravity both point down toward the center of the circle. The net force on the ball is the vector sum of these two:
For an object in circular motion, the net force must be equal to:
In this case, the force of gravity is helping the tension provide this force, so the tension is not as great as it would be if it were supplying the entire centripetal force.
Substitute and solve.
The tension and gravity both point down toward the center of the circle. The net force on the ball is the vector sum of these two:
For an object in circular motion, the net force must be equal to:
In this case, the force of gravity is helping the tension provide this force, so the tension is not as great as it would be if it were supplying the entire centripetal force.
Substitute and solve.