Motion - AP Physics C: Electricity and Magnetism
Card 1 of 880
A guillotine blade weighing 
is accelerated upward into position at a rate of 
.
What is the the approximate mass of the guillotine blade?
A guillotine blade weighing
What is the the approximate mass of the guillotine blade?
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The force of gravity on the blade is 
, which is the same as 
This unit relationship comes from Newton's second law.
is the mathematical expression of Newton's second law. The units for force must be a product of the units for mass and the units for acceleration.
Solve the expression by plugging in known values.
The force of gravity on the blade is
This unit relationship comes from Newton's second law.
Solve the expression by plugging in known values.
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A guillotine blade weighing is accelerated upward into position at a rate of 
.
What is the tension on the rope pulling the blade, while it is accelerating into position?
A guillotine blade weighing
What is the tension on the rope pulling the blade, while it is accelerating into position?
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The tension in the rope is the sum of the forces acting on it. If one considers that the net force on an object must equal the mass of the object times the acceleration of the object, the net force on the object must be the force due to tension from the rope minus the force due to gravity.
Rearrange the equation.
Plug in known values.
The tension in the rope is the sum of the forces acting on it. If one considers that the net force on an object must equal the mass of the object times the acceleration of the object, the net force on the object must be the force due to tension from the rope minus the force due to gravity.
Rearrange the equation.
Plug in known values.
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An object is moving in two dimensions. Its vertical motion relative to the horizontal motion is described by the equation 
. Its motion in the horizontal direction is described by the equation 
. What is the object's velocity is the 
direction in terms of its horizontal position 
?
An object is moving in two dimensions. Its vertical motion relative to the horizontal motion is described by the equation
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The y velocity is thetime derivative of the 
position, and not the 
derivative. In order to find it, use the chain rule:
Of course, 
The y velocity is thetime derivative of the
Of course,
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Two objects moving in one dimension created the following velocity vs. time graph:
From the graph above, what is true about the two objects at time 
?
Two objects moving in one dimension created the following velocity vs. time graph:
From the graph above, what is true about the two objects at time
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Since this is a graph of velocity and not position, the curves intersect where the velocities match. Since we do not know the starting position, we do not know where the objects are relative to one another.
Since this is a graph of velocity and not position, the curves intersect where the velocities match. Since we do not know the starting position, we do not know where the objects are relative to one another.
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Two objects moving in one dimension created the following velocity vs. time graph:
From the graph above, which object has traveled a greater distance from its starting position when 
?
Two objects moving in one dimension created the following velocity vs. time graph:
From the graph above, which object has traveled a greater distance from its starting position when
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Since this is a graph of velocity vs. time, its integral is distance travelled. We can estimate the integral by looking at the area under the curves. Since Bbject 1 has a greater area under its velocity curve, it has covered a greater distance. Its velocity is greater that Object 2's for the entire time, so it makes sense that it will travel farther.
Since this is a graph of velocity vs. time, its integral is distance travelled. We can estimate the integral by looking at the area under the curves. Since Bbject 1 has a greater area under its velocity curve, it has covered a greater distance. Its velocity is greater that Object 2's for the entire time, so it makes sense that it will travel farther.
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Atwood's machine consists of two blocks connected by a string connected over a
pulley as shown. What is the acceleration of the blocks if their masses are 
and 
.
Assume the pulley has negligible mass and friction.
Atwood's machine consists of two blocks connected by a string connected over a
pulley as shown. What is the acceleration of the blocks if their masses are
Assume the pulley has negligible mass and friction.
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From the force diagram above, we can see that tension 
is pulling up on both sides of the string and gravity is pulling down on both blocks. With this information we can write 2 force equations:
If we add the two equations together, we get:
where 
Solving for 
, we get
From the force diagram above, we can see that tension
If we add the two equations together, we get:
where
Solving for
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An object starts from rest and accelerates to an angular velocity of 
in three seconds under a constant torque of 
. How many revolutions has the object made in this time?
An object starts from rest and accelerates to an angular velocity of
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Since it is experiencing a constant torque and constant angular acceleration, the angular displacement can be calculated using:
The angular acceleration is easily calculated using the angular velocity and the time:
Using this value, we can find the angular displacement:
Convert the angular displacement to revolutions by diving by 
:
Since it is experiencing a constant torque and constant angular acceleration, the angular displacement can be calculated using:
The angular acceleration is easily calculated using the angular velocity and the time:
Using this value, we can find the angular displacement:
Convert the angular displacement to revolutions by diving by
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A circular disk of radius 0.5m and mass 3kg has a force of 25N exerted perpendicular to its edge, causing it to spin. What is the angular acceleration of the disk?
A circular disk of radius 0.5m and mass 3kg has a force of 25N exerted perpendicular to its edge, causing it to spin. What is the angular acceleration of the disk?
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We can find the angular acceleration using the rotaional motion equivalent of Newton's second law. In rotational motion, torque is the product of moment of inertia and angular acceleration:
The moment of inertia for a circular disk is:
The tourque is the product of force and distance (in this case, the radius):
We can plug these into our first equation:
Simplify and rearrange to derive an equation for angular acceleration:
Use our given values to solve:
We can find the angular acceleration using the rotaional motion equivalent of Newton's second law. In rotational motion, torque is the product of moment of inertia and angular acceleration:
The moment of inertia for a circular disk is:
The tourque is the product of force and distance (in this case, the radius):
We can plug these into our first equation:
Simplify and rearrange to derive an equation for angular acceleration:
Use our given values to solve:
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A 0.18 m long wrench is used to turn the nut on the end of a bolt. A force of 85 N is applied downward to the end of the wrench, as shown in the figure. The angle between the force and the handle of the wrench is 65 degrees.
What is the magnitude and direction of the torque (around the center of the bolt) due to this force?
A 0.18 m long wrench is used to turn the nut on the end of a bolt. A force of 85 N is applied downward to the end of the wrench, as shown in the figure. The angle between the force and the handle of the wrench is 65 degrees.
What is the magnitude and direction of the torque (around the center of the bolt) due to this force?
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To calculate the magnitude of the torque,
where the radius 
is the distance between the center of rotation and the location of the force 
and the angle 
is between the radius 
and the force 
.
The magnitude is thus,
The direction of torque is perpendicular to the plane of the radius 
and the force 
, and is given by Right Hand Rule by crossing the radius vector into the force vector. For this situation, the radius vector is left and downward and the force is downward, resulting in the direction of the torque out of the page.
To calculate the magnitude of the torque,
where the radius
The magnitude is thus,
The direction of torque is perpendicular to the plane of the radius
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A meter stick is nailed to a table at one end and is free to rotate in a horizontal plane parallel to the top of the table. Four forces of equal magnitude are applied to the meter stick at different locations. The figure below shows the view of the meter stick from above.
You may assume the forces 
and 
are applied at the center of the meter stick, and the forces 
and 
are applied at the end opposite the nail.
What is the relationship among the magnitudes of the torques on the meter stick caused by the four different forces?
A meter stick is nailed to a table at one end and is free to rotate in a horizontal plane parallel to the top of the table. Four forces of equal magnitude are applied to the meter stick at different locations. The figure below shows the view of the meter stick from above.
You may assume the forces
What is the relationship among the magnitudes of the torques on the meter stick caused by the four different forces?
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Torque is given by,
Since all of the forces are equal in magnitude, the magnitude of the torque is then influenced by the radius r and the angle theta between the radius and the force.
For 
,
For 
,
For 
,
For 
,
Combining this information yields the relationship,
Torque is given by,
Since all of the forces are equal in magnitude, the magnitude of the torque is then influenced by the radius r and the angle theta between the radius and the force.
For
For
For
For
Combining this information yields the relationship,
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A man sits on the end of a long uniform metal beam of length 
. The man has a mass of 
and the beam has a mass of 
.
What is the magnitude of the net torque on the plank about the secured end of the beam? Use gravity 
.
A man sits on the end of a long uniform metal beam of length
What is the magnitude of the net torque on the plank about the secured end of the beam? Use gravity
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The net torque on the beam is given by addition of the torques caused by the weight of the man and the weight of the beam itself, each at its respective distance from the end of the beam:
Let's assign the direction of positive torque in the direction of the torques of the man's and the beam's weights, noting that they will add together since they both point in the same direction.
We can further simplify by combining like terms:
Using the given numerical values,
The net torque on the beam is given by addition of the torques caused by the weight of the man and the weight of the beam itself, each at its respective distance from the end of the beam:
Let's assign the direction of positive torque in the direction of the torques of the man's and the beam's weights, noting that they will add together since they both point in the same direction.
We can further simplify by combining like terms:
Using the given numerical values,
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Two children sit on the opposite sides of a seasaw at a playground, doing so in a way that causes the seasaw to balance perfectly horizontal. The 
child on the left is 
from the pivot.
What is the mass of the second child if she sits 
from the pivot?
Two children sit on the opposite sides of a seasaw at a playground, doing so in a way that causes the seasaw to balance perfectly horizontal. The
What is the mass of the second child if she sits
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A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,
This is a static situation. There are two torques about the pivot caused by the weights of two children. We will note that these weights cause torques in opposite directions about the pivot, such that
Consequently,
Or more simply,
Solving for 
,
A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,
This is a static situation. There are two torques about the pivot caused by the weights of two children. We will note that these weights cause torques in opposite directions about the pivot, such that
Consequently,
Or more simply,
Solving for
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A gymnast is practicing her balances on a long narrow plank supported at both ends. Her mass is 
. The 
long plank has a mass of 
.
Calculate the force the right support provides upward if she stands 
from the right end. Use gravity 
.
A gymnast is practicing her balances on a long narrow plank supported at both ends. Her mass is
Calculate the force the right support provides upward if she stands
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A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,
This is a static situation. As such, any pivot point can be chosen about which to do a torque analysis. The quickest way to the unknown force asked for in the question is to do the torque analysis about the left end of the plank. There are three torques about this pivot: two clockwise caused by the weights of the gymnast and the plank itself, and one counterclockwise caused by the force from the right support. Designating clockwise as positive,
Consequently,
This simplifies to
Solving for 
,
Solving with numerical values,
A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,
This is a static situation. As such, any pivot point can be chosen about which to do a torque analysis. The quickest way to the unknown force asked for in the question is to do the torque analysis about the left end of the plank. There are three torques about this pivot: two clockwise caused by the weights of the gymnast and the plank itself, and one counterclockwise caused by the force from the right support. Designating clockwise as positive,
Consequently,
This simplifies to
Solving for
Solving with numerical values,
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A square of side lengths 
and mass 
is shown with 
possible axes of rotation.
Which statement of relationships among the moments of inertia is correct?
A square of side lengths
Which statement of relationships among the moments of inertia is correct?
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The moments of inertia for both axes 
and 
are equal because both of these axes are equivalently passing through the center of the mass.
By the parallel axis theorem for moments of inertia (
), the moment of inertia for axis 
is larger than 
or 
because it is located a distance
away from the center of mass.
The moments of inertia for both axes
By the parallel axis theorem for moments of inertia (
because it is located a distance
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A wind catcher is created by attaching four plastic bowls of mass 
each to the ends of four lightweight rods, which are then secured to a central rod that is free to rotate in the wind. The four lightweight rods are of lengths 
, 
, 
, and 
.
Calculate the moment of inertia of the four bowls about the central rod. You may assume to bowls to be point masses.
A wind catcher is created by attaching four plastic bowls of mass
Calculate the moment of inertia of the four bowls about the central rod. You may assume to bowls to be point masses.
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The moment of inertia for a point mass is 
.
To calculate the total moment of inertia, we add the moment of inertia for each part of the object, such that
The masses of the bowls are all equal in this problem, so this simplifies to
Plugging in and solving with numerical values,
The moment of inertia for a point mass is
To calculate the total moment of inertia, we add the moment of inertia for each part of the object, such that
The masses of the bowls are all equal in this problem, so this simplifies to
Plugging in and solving with numerical values,
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A long uniform thin rod of length 
has a mass of 
.
Calculate the moment of rotational inertia about an axis perpendicular to its length passing through a point 
from one of its ends.
A long uniform thin rod of length
Calculate the moment of rotational inertia about an axis perpendicular to its length passing through a point
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For a long thin rod about its center of mass,
According to the parallel axis theorem,
where 
is defined to be the distance between the center of mass of the object and the location of the axis parallel to one through the center of mass. For this problem, 
is the difference between the given distance and half the length of the rod.
Combining the above,
Inserting numerical values,
For a long thin rod about its center of mass,
According to the parallel axis theorem,
where
Combining the above,
Inserting numerical values,
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The moment of inertia of a long thin rod about its end is determined to be 
.
What is the new value if the mass and length of the rod are both reduced by a factor of 
?
The moment of inertia of a long thin rod about its end is determined to be
What is the new value if the mass and length of the rod are both reduced by a factor of
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The moment of inertia for a long uniform thin rod about its end is given by
Reducing the mass and the length by a factor of four introduces the following factors into the equation,
Simplifying,
The moment of inertia for a long uniform thin rod about its end is given by
Reducing the mass and the length by a factor of four introduces the following factors into the equation,
Simplifying,
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An art sculpture comprised of a long hollow painted cylinder is designed to pivot about an axis located along its long edge. The cylinder has mass 
, radius 
, and length 
.
Calculate the moment of inertia of the sculpture about this axis.
An art sculpture comprised of a long hollow painted cylinder is designed to pivot about an axis located along its long edge. The cylinder has mass
Calculate the moment of inertia of the sculpture about this axis.
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The moment of inertia for a hollow cylinder about its center is given by
The parallel axis is applied here because the pivot axis is located on the outside edge of the cylinder, a distance 
from the center of mass:
Incidentally, for this problem, the length does not contribute to the moment of inertia.
The moment of inertia for a hollow cylinder about its center is given by
The parallel axis is applied here because the pivot axis is located on the outside edge of the cylinder, a distance
Incidentally, for this problem, the length does not contribute to the moment of inertia.
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A
meter stick is nailed to a table at one end and is free to rotate in a horizontal plane parallel to the top of the table. A force of 
is applied perpendicularly to its length at a distance 
from the nailed end.
Calculate the resulting angular acceleration experienced by the meter stick.
A
Calculate the resulting angular acceleration experienced by the meter stick.
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For a net torque applied to an object free to rotate,
The net torque for this problem is supplied by the 
force applied at a distance 
from the pivot. The torque of a force is calculated by
The moment of inertia for a long uniform thin rod about its end is
Combining above,
Solving for the angular acceleration,
For a net torque applied to an object free to rotate,
The net torque for this problem is supplied by the
The moment of inertia for a long uniform thin rod about its end is
Combining above,
Solving for the angular acceleration,
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A 
meter stick is secured at one end so that it is free to rotate in a vertical circle. It is held perfectly horizontally, and released.
Calculate the instantaneous angular acceleration the moment it is released.
A
Calculate the instantaneous angular acceleration the moment it is released.
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For a net torque applied to an object free to rotate,
The net torque for this problem is supplied by the 
force applied at a distance 
from the pivot. The torque of a force is calculated by
The moment of inertia for a long uniform thin rod about its end is
Combining above,
where the force causing the torque is the meter stick's weight, which is applied at the center of mass of the meter stick - a distance of half the length away from its end.
Solving for the angular acceleration,
Solving numerically,
For a net torque applied to an object free to rotate,
The net torque for this problem is supplied by the
The moment of inertia for a long uniform thin rod about its end is
Combining above,
where the force causing the torque is the meter stick's weight, which is applied at the center of mass of the meter stick - a distance of half the length away from its end.
Solving for the angular acceleration,
Solving numerically,
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