Ladybugs Problem 4bm3j
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Ladybugs on a Rotating Disk Two ladybugs sit on a rotating disk (the ladybugs are at rest with respect to the surface of the disk and
do not slip). ladybug 2 and the axis of rotation. A. What is the angular speed of ladybug 1?
Ladybug 1 is halfway between
onehalf the angular speed of ladybug 2 the same as the angular speed of ladybug 2 two times the angular speed of ladybug 2 onequarter the angular speed of ladybug 2 B. What is the ratio of the linear speed of ladybug 2 to that of ladybug 1? Answer numerically. 2 C. At the instant shown in the figure, what is the direction of the the radial component of the acceleration of ladybug 2? D. What is the ratio of the radial acceleration of ladybug 2 to that of ladybug 1? Answer numerically. 2 E. What is the direction of the vector representing the angular velocity of ladybug 2?
F.
In general, velocity is represented by a vector. Let represent the velocity of ladybug 1. Angular velocity is also often represented by a vector, so let that of ladybug 1 be given by Take to be the vector from the axis of rotation to ladybug 1. Which of the following equations correctly describes the relationship between the velocity and position of ladybug 1?
, angular velocity
,
.
G. Now assume that at the moment pictured in the figure, the disk is rotating but slowing down. What is the direction of the tangential component of the acceleration (i.e. acceleration tangent to the motion) of ladybug 2?
[ Print ]
Moment of Inertia and Center of Mass for Point Particles Ball A, of mass
, is connected to ball B, of mass
, by a massless rod of length .
The two vertical dashed lines in the figure, one through each ball, represent two different axes of rotation, axes a and b. These axes are parallel to each other and perpendicular to the rod. The moment of inertia of the twomass system about axis a is , and the moment of inertia of the system about axis b is . It is observed that the ratio of equal to 3:
to is
Assume that both balls are pointlike; that is, neither has any moment of inertia about its own center of mass. A. Find the ratio of the masses of the two balls. Express your answer numerically. =
B. Find
1/3 , the distance from ball A to the system's center of mass.
Express your answer in of , the length of the rod.
= 3*L/4 [ Print ]
Scaling of Moments of Inertia Learning Goal: To understand the concept of moment of inertia and how it depends on mass, radius, and mass distribution. In rigidbody rotational dynamics, the role analogous to the mass of a body (when one is considering translational motion) is played by the body's moment of inertia. For this reason, conceptual understanding of the motion of a rigid body requires some understanding of moments of inertia. This problem should help you develop such an understanding.
The moment of inertia of a body about some specified axis is , where is a dimensionless constant, is the mass of the body, and is the perpendicular distance from the axis of rotation. Therefore, if you have two similarly shaped objects of the same size but with one twice as massive as the other, the more massive object should have a moment of inertia twice that of the less massive one. Furthermore, if you have two similarly shaped objects of the same mass, but one has twice the size of the other, the larger object should have a moment of inertia that is four times that of the smaller one.
A. Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 meters and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass of the 1meter shell be for both shells to have the same moment of inertia about their centers of mass? = 4 kg It is important to understand how the dimensionless constant in the moment of inertia formula given in the problem introduction is determined. Consider a disk and a thin ring, both having the same outer radius and mass
. The moment of inertia of the disk is
, while the moment of inertia of the
ring is . (For each object, the axis is perpendicular to the plane of the object and es through the object's center of mass.) The factor of for the disk gives an indication of how the mass is distributed in that object. If both the disk and ring are spinning with the same angular velocity , the ring should have a greater kinetic energy, because all of the mass of the ring has linear speed , whereas the linear speeds of the different parts of the disk vary, depending on how far the part is from the center, and these speeds vary from zero to . In general, the value of reflects the distribution of mass within the object. A number close to indicates that most of the mass is located at a distance from the center of mass close to , while a number much less than indicates that most of the mass is located near the center of mass. A. Consider the moments of inertia of a solid uniform disk, a ring, a hollow spherical shell, and a solid sphere about their respective centers of mass. Assume that they all have the same mass and outer radius, that they have uniform mass distributions, and that the ring and disk are rotated about an axis perpendicular to their faces. List the objects (disk, ring, shell, sphere) in order of decreasing moment of inertia.
Give your answer in the form (for instance) "disk, ring, shell, sphere". ring , shell , disk , sphere [ Print ]
The ParallelAxis Theorem Learning Goal: To understand the parallelaxis theorem and its applications To solve many problems about rotational motion, it is important to know the moment of inertia of each object involved. Calculating the moments of inertia of various objects, even highly symmetrical ones, may be a lengthy and tedious process. While it is important to be able to calculate moments of inertia from the definition ( ), in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of frequently occurring shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily available from any mechanics text, including your textbook. However, one must take into that an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of tranlsational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moments of inertia that you can find in the textbooks are usually calculated with respect to an axis ing through the center of mass of the object. However, in many problems the axis of rotation does not through the center of mass. Does that mean that one has to go through the lengthy process of finding the moment of inertia from scratch? It turns out that in many cases, calculating the moment of inertia can be done rather easily if one uses the parallelaxis theorem. Mathematically, it can be expressed as about an axis ing through the center of mass,
, where
is the moment of inertia
is the total mass of the object, and is the moment
of inertia about another axis, parallel to the one for which is calculated and located a distance from the center of mass. In this problem you will show that the theorem does indeed work for at least one object: a dumbbell of length made of two small spheres of mass each connected by a light rod (see the figure). NOTE: Unless otherwise noted, all axes considered are perpendicular to the plane of the page.
A. Using the definition of moment of inertia, calculate
, the moment of inertia about the
center of mass, for this object. Express your answer in of
and .
= 2*m*r^2
B. Using the definition of moment of inertia, calculate , the moment of inertia about an axis
through point B, for this object. Point B coincides with (the center of) one of the spheres (see the figure). Express your answer in of
and .
= 4*m*r^2
C. Now calculate for this object using the parallelaxis theorem. Express your answer in of
,
, and .
= I_cm+(2*m)*r^2
D. Using the definition of moment of inertia, calculate , the moment of inertia about an axis through point C, for this object. Point C is located a distance from the center of mass (see the figure). Express your answer in of
and .
= 4*m*r^2
E. Now calculate for this object using the parallelaxis theorem. Express your answer in of
,
, and .
= I_cm+(2*m)*r^2 Consider an irregular object of mass . Its moment of inertia measured with respect to axis A (parallel to the plane of the page), which es through the center of mass (see the second diagram), is given by . Axes B, C, D, and E are parallel to axis A; their separations from axis A are shown in the diagram.
In the subsequent questions, the subscript indicates the axis with respect to which the moment of inertia is measured: for instance, is the moment of inertia about axis C. A. Which moment of inertia is the smallest?
B. Which moment of inertia is the largest?
C. Which moments of inertia are equal? and and and No two moments of inertia are equal.
D. Which moment of inertia equals
?
E. Axis X, not shown in the diagram, is parallel to the axes shown. It is known that
.
Which of the following is a possible location for axis X? between axes A and C between axes C and D between axes D and E to the right of axis E [ Print ]
Loop the Loop with a Twist In this problem you will consider the motion of a cylinder of radius
that is rolled from a certain
height so that it "loops the loop," that is, rolls around the track with a loop of radius the figure without losing with the track.
shown in
Unless otherwise stated, assume that friction is sufficient that the cylinder rolls without slipping. The radius of the cylinder is much smaller than the radius of the loop. A. Compared to an object that does not roll, but instead slides without friction, should a rolling object be released from the same,a greater, or a lesser height in order just barely to complete the loop the loop? The rolling object should be released from a greater height. The rolling object should be released from a lesser height. The rolling object should be released from exactly the same height. The answer depends on the moment of inertia of the rolling object.
B. Find the minimum height that will allow a solid cylinder of mass and radius the loop of radius
.
Express in of the radius
of the loop.
to loop
= (11/4)*r_loop [ Print ]
Top Spin A string is wrapped around a uniform solid cylinder of radius . The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass .
A. Find the magnitude of the angular acceleration of the cylinder as the block descends. Express your answer in of the cylinder's radius and the magnitude of the acceleration due to gravity . = (2/3)*g/r
Torques on a Seesaw: A Tutorial Learning Goal: To make the connection between your intuitive understanding of a seesaw and the standard formalism for torque.
This problem deals with the concept of torque, the "twist" that an offcenter force applies to a body that
tends to make it rotate.
Try to use your intuition to answer the following question. If your intuition fails, work the rest of the problem and return here when you feel that you are more comfortable with torques.
A. A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground. Given that her heavier child of weight
is sitting a distance to the left of the pivot, at what
distance must she place her second child of weight on the right side of the pivot to balance the seesaw? Express your answer in of ,
, and .
= L*W/w Now consider this problem as a more formal introduction to torque. The torque of each child about the pivot point is the product of the child's weight and the distance of the child (strictly speaking, the child's center of mass) from the pivot. The sign of the torque is positive if it would cause a counterclockwise rotation of the seesaw. The distance is measured perpendicular to the line of force and is called the moment arm. The concept of torque requires both a force and a specification of the pivot point, emphasized by the first subscript on the torque. A. Find , the torque about the pivot due to the weight of the smaller child on the seesaw. Express your answer in of
and .
= L_1*w The children's mother wants the seesaw to balance, which means that there can be no angular acceleration about the pivot. The balanced seesaw will then be in equilibrium since it has neither linear acceleration nor rotational acceleration. For the linear acceleration to be zero, the vector sum of forces acting on the seesaw and children must equal zero. For the angular acceleration to be zero, the sum of the torques about the pivot must equal zero. This can be written ,
where
is the torque about the pivot due to the th force.
A. Determine
by the children.
, the sum of the torques on the seesaw. Consider only the torques exerted
Express your answer in of
=
, , , and
.
W*L w*L_1
The child with weight has an identical twin also of weight . The two twins now sit on the same side of the seesaw, with one twin a distance
from the pivot and the other a distance
.
A. Where should the mother position the child of weight
Express your answer in of
,
,
to balance the seesaw now?
, and .
= w*(L_3 + L_2)/W Bad news! When the mother finds the distance from the previous part it turns out to be greater than , the distance from the pivot to the end of the seesaw. Hence, even with the child of weight at the very end of the seesaw the twins exert more torque than the heavier child does. The mother now
elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is a height
above the pivot.
A. With what force in the rightwards direction,
, should the mother push? Note that if you
think the force exerted by the mother should be toward the left, your final answer negative. Express your answer in of
,
, ,
,
should be
, and .
= (W*L_endw*(L_2+L_3))/h [ Print ]
Pulling a String Adds Energy to a Wheel
A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel has moment of inertia , where is its mass, is its radius, and is a dimensionless number between zero and one. The wheel is rotating counterclockwise with angular velocity , when
at time
someone starts pulling the string with a force of magnitude
.
Assume that the string does not slip on the wheel.
A. Suppose that after a certain time , the string has been pulled through a distance . What is the final rotational speed of the wheel? Express
in of , , , ,
, and
.
= (2*F*L/(k*m*r^2) + omega_0^2)^.5
B. What is the instantaneous power delivered to the wheel via the force at time
?
Express the power in of the variables given in the problem introduction. = F*r*omega_0 [ Print ]
Hoop on a Ramp A circular hoop of mass
, radius , and infinitesimal thickness rolls without slipping down a ramp
inclined at an angle with the horizontal.
A. What is the acceleration of the center of the hoop? Express the acceleration in of physical constants and all or some of the given quantities:
, , and .
= (g/2)*sin(theta) B. What is the minimum coefficient of friction
needed for the hoop to roll without slipping?
Express the minimum coefficient of friction in of all or some of the given quantities:
, , and .
= 1/2*tan(theta)
[ Print ]
Record and Turntable Learning Goal: To understand how to use conservation of angular momentum to solve problems involving collisions of rotating bodies. Consider a turntable to be a circular disk of moment of inertia rotating at a constant angular velocity around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is ed by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis. Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is . The initial angular velocity of the second disk is zero. There is friction between the two disks. After this "rotational collision," the disks will eventually rotate with the same angular velocity. A. What is the final angular velocity, , of the two disks? Express
in of , , and
.
= (I_t/(I_t+I_r))*omega_i
B. Because of friction, kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final kinetic energy,
, of the two spinning disks?
Express the final kinetic energy in of , , and the initial kinetic energy twodisk system. No angular velocities should appear in your answer.
of the
= K_i*I_t/(I_t+I_r)
C. Assume that the turntable deccelerated during time before reaching the final angular velocity ( is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, , acting on the bottom disk due to friction with the record? Express the torque in of ,
,
, and
.
= I_t*(omega_fomega_i)/(deltat)
Spinning Situations Suppose you are standing on the center of a merrygoround that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis is pointing upward. The wheel is rotating counterclockwise when observed from above.
For this problem, neglect any air resistance or friction between the merrygoround and its foundation. A. Suppose you now grab the edge of the wheel with your hand, stopping it from spinning. What happens to the merrygoround? It remains at rest. It begins to rotate counterclockwise (as observed from above). It begins to rotate clockwise (as observed from above). [ Print ]
Precessing Tilted Gyroscope
A gyroscope consists of a flywheel of mass , which has a moment of inertia for rotation about its axis. It is mounted on a rod of negligible mass, which is ed at one end by a frictionless pivot attached to a vertical post, as shown in the diagram. The distance between the center of the wheel and the pivot is . The wheel rotates about its axis with angular velocity , where positive refers to counterclockwise rotation as seen by an observer looking at the face of the wheel that is opposite the pivot. The rod is tilted upward, making an angle with respect to the horizontal. Gravity acts downward with a force of magnitude . Adopt a coordinate system with the z axis pointing upward and the x and y axes in the horizontal plane. The gyroscope is moving, but at
, the rod is in the yz plane.
A. Assuming that the only significant contribution to the angular momentum comes from the spinning of the flywheel about its center, what is the angular momentum vector about the pivot at
?
Specify the components of with respect to the axes shown in the diagram. Write the components in order
,
,
separated by commas.
,
B. At
,
= 0 , I*omega*cos(theta) , I*omega*sin(theta) , what is the torque acting on the wheel about the pivot?
Express your answer in of components
, , , separated by commas.
, , = m*g*d*cos(theta) , 0 , 0
C. The gyroscope is observed to precess about the vertical axis, with an angular velocity of precession , defined as positive for counterclockwise precession as seen from above. Find in of the given quantities. = (m*g*d)/(I*omega)
do not slip). ladybug 2 and the axis of rotation. A. What is the angular speed of ladybug 1?
Ladybug 1 is halfway between
onehalf the angular speed of ladybug 2 the same as the angular speed of ladybug 2 two times the angular speed of ladybug 2 onequarter the angular speed of ladybug 2 B. What is the ratio of the linear speed of ladybug 2 to that of ladybug 1? Answer numerically. 2 C. At the instant shown in the figure, what is the direction of the the radial component of the acceleration of ladybug 2? D. What is the ratio of the radial acceleration of ladybug 2 to that of ladybug 1? Answer numerically. 2 E. What is the direction of the vector representing the angular velocity of ladybug 2?
F.
In general, velocity is represented by a vector. Let represent the velocity of ladybug 1. Angular velocity is also often represented by a vector, so let that of ladybug 1 be given by Take to be the vector from the axis of rotation to ladybug 1. Which of the following equations correctly describes the relationship between the velocity and position of ladybug 1?
, angular velocity
,
.
G. Now assume that at the moment pictured in the figure, the disk is rotating but slowing down. What is the direction of the tangential component of the acceleration (i.e. acceleration tangent to the motion) of ladybug 2?
[ Print ]
Moment of Inertia and Center of Mass for Point Particles Ball A, of mass
, is connected to ball B, of mass
, by a massless rod of length .
The two vertical dashed lines in the figure, one through each ball, represent two different axes of rotation, axes a and b. These axes are parallel to each other and perpendicular to the rod. The moment of inertia of the twomass system about axis a is , and the moment of inertia of the system about axis b is . It is observed that the ratio of equal to 3:
to is
Assume that both balls are pointlike; that is, neither has any moment of inertia about its own center of mass. A. Find the ratio of the masses of the two balls. Express your answer numerically. =
B. Find
1/3 , the distance from ball A to the system's center of mass.
Express your answer in of , the length of the rod.
= 3*L/4 [ Print ]
Scaling of Moments of Inertia Learning Goal: To understand the concept of moment of inertia and how it depends on mass, radius, and mass distribution. In rigidbody rotational dynamics, the role analogous to the mass of a body (when one is considering translational motion) is played by the body's moment of inertia. For this reason, conceptual understanding of the motion of a rigid body requires some understanding of moments of inertia. This problem should help you develop such an understanding.
The moment of inertia of a body about some specified axis is , where is a dimensionless constant, is the mass of the body, and is the perpendicular distance from the axis of rotation. Therefore, if you have two similarly shaped objects of the same size but with one twice as massive as the other, the more massive object should have a moment of inertia twice that of the less massive one. Furthermore, if you have two similarly shaped objects of the same mass, but one has twice the size of the other, the larger object should have a moment of inertia that is four times that of the smaller one.
A. Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 meters and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass of the 1meter shell be for both shells to have the same moment of inertia about their centers of mass? = 4 kg It is important to understand how the dimensionless constant in the moment of inertia formula given in the problem introduction is determined. Consider a disk and a thin ring, both having the same outer radius and mass
. The moment of inertia of the disk is
, while the moment of inertia of the
ring is . (For each object, the axis is perpendicular to the plane of the object and es through the object's center of mass.) The factor of for the disk gives an indication of how the mass is distributed in that object. If both the disk and ring are spinning with the same angular velocity , the ring should have a greater kinetic energy, because all of the mass of the ring has linear speed , whereas the linear speeds of the different parts of the disk vary, depending on how far the part is from the center, and these speeds vary from zero to . In general, the value of reflects the distribution of mass within the object. A number close to indicates that most of the mass is located at a distance from the center of mass close to , while a number much less than indicates that most of the mass is located near the center of mass. A. Consider the moments of inertia of a solid uniform disk, a ring, a hollow spherical shell, and a solid sphere about their respective centers of mass. Assume that they all have the same mass and outer radius, that they have uniform mass distributions, and that the ring and disk are rotated about an axis perpendicular to their faces. List the objects (disk, ring, shell, sphere) in order of decreasing moment of inertia.
Give your answer in the form (for instance) "disk, ring, shell, sphere". ring , shell , disk , sphere [ Print ]
The ParallelAxis Theorem Learning Goal: To understand the parallelaxis theorem and its applications To solve many problems about rotational motion, it is important to know the moment of inertia of each object involved. Calculating the moments of inertia of various objects, even highly symmetrical ones, may be a lengthy and tedious process. While it is important to be able to calculate moments of inertia from the definition ( ), in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of frequently occurring shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily available from any mechanics text, including your textbook. However, one must take into that an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of tranlsational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moments of inertia that you can find in the textbooks are usually calculated with respect to an axis ing through the center of mass of the object. However, in many problems the axis of rotation does not through the center of mass. Does that mean that one has to go through the lengthy process of finding the moment of inertia from scratch? It turns out that in many cases, calculating the moment of inertia can be done rather easily if one uses the parallelaxis theorem. Mathematically, it can be expressed as about an axis ing through the center of mass,
, where
is the moment of inertia
is the total mass of the object, and is the moment
of inertia about another axis, parallel to the one for which is calculated and located a distance from the center of mass. In this problem you will show that the theorem does indeed work for at least one object: a dumbbell of length made of two small spheres of mass each connected by a light rod (see the figure). NOTE: Unless otherwise noted, all axes considered are perpendicular to the plane of the page.
A. Using the definition of moment of inertia, calculate
, the moment of inertia about the
center of mass, for this object. Express your answer in of
and .
= 2*m*r^2
B. Using the definition of moment of inertia, calculate , the moment of inertia about an axis
through point B, for this object. Point B coincides with (the center of) one of the spheres (see the figure). Express your answer in of
and .
= 4*m*r^2
C. Now calculate for this object using the parallelaxis theorem. Express your answer in of
,
, and .
= I_cm+(2*m)*r^2
D. Using the definition of moment of inertia, calculate , the moment of inertia about an axis through point C, for this object. Point C is located a distance from the center of mass (see the figure). Express your answer in of
and .
= 4*m*r^2
E. Now calculate for this object using the parallelaxis theorem. Express your answer in of
,
, and .
= I_cm+(2*m)*r^2 Consider an irregular object of mass . Its moment of inertia measured with respect to axis A (parallel to the plane of the page), which es through the center of mass (see the second diagram), is given by . Axes B, C, D, and E are parallel to axis A; their separations from axis A are shown in the diagram.
In the subsequent questions, the subscript indicates the axis with respect to which the moment of inertia is measured: for instance, is the moment of inertia about axis C. A. Which moment of inertia is the smallest?
B. Which moment of inertia is the largest?
C. Which moments of inertia are equal? and and and No two moments of inertia are equal.
D. Which moment of inertia equals
?
E. Axis X, not shown in the diagram, is parallel to the axes shown. It is known that
.
Which of the following is a possible location for axis X? between axes A and C between axes C and D between axes D and E to the right of axis E [ Print ]
Loop the Loop with a Twist In this problem you will consider the motion of a cylinder of radius
that is rolled from a certain
height so that it "loops the loop," that is, rolls around the track with a loop of radius the figure without losing with the track.
shown in
Unless otherwise stated, assume that friction is sufficient that the cylinder rolls without slipping. The radius of the cylinder is much smaller than the radius of the loop. A. Compared to an object that does not roll, but instead slides without friction, should a rolling object be released from the same,a greater, or a lesser height in order just barely to complete the loop the loop? The rolling object should be released from a greater height. The rolling object should be released from a lesser height. The rolling object should be released from exactly the same height. The answer depends on the moment of inertia of the rolling object.
B. Find the minimum height that will allow a solid cylinder of mass and radius the loop of radius
.
Express in of the radius
of the loop.
to loop
= (11/4)*r_loop [ Print ]
Top Spin A string is wrapped around a uniform solid cylinder of radius . The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass .
A. Find the magnitude of the angular acceleration of the cylinder as the block descends. Express your answer in of the cylinder's radius and the magnitude of the acceleration due to gravity . = (2/3)*g/r
Torques on a Seesaw: A Tutorial Learning Goal: To make the connection between your intuitive understanding of a seesaw and the standard formalism for torque.
This problem deals with the concept of torque, the "twist" that an offcenter force applies to a body that
tends to make it rotate.
Try to use your intuition to answer the following question. If your intuition fails, work the rest of the problem and return here when you feel that you are more comfortable with torques.
A. A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground. Given that her heavier child of weight
is sitting a distance to the left of the pivot, at what
distance must she place her second child of weight on the right side of the pivot to balance the seesaw? Express your answer in of ,
, and .
= L*W/w Now consider this problem as a more formal introduction to torque. The torque of each child about the pivot point is the product of the child's weight and the distance of the child (strictly speaking, the child's center of mass) from the pivot. The sign of the torque is positive if it would cause a counterclockwise rotation of the seesaw. The distance is measured perpendicular to the line of force and is called the moment arm. The concept of torque requires both a force and a specification of the pivot point, emphasized by the first subscript on the torque. A. Find , the torque about the pivot due to the weight of the smaller child on the seesaw. Express your answer in of
and .
= L_1*w The children's mother wants the seesaw to balance, which means that there can be no angular acceleration about the pivot. The balanced seesaw will then be in equilibrium since it has neither linear acceleration nor rotational acceleration. For the linear acceleration to be zero, the vector sum of forces acting on the seesaw and children must equal zero. For the angular acceleration to be zero, the sum of the torques about the pivot must equal zero. This can be written ,
where
is the torque about the pivot due to the th force.
A. Determine
by the children.
, the sum of the torques on the seesaw. Consider only the torques exerted
Express your answer in of
=
, , , and
.
W*L w*L_1
The child with weight has an identical twin also of weight . The two twins now sit on the same side of the seesaw, with one twin a distance
from the pivot and the other a distance
.
A. Where should the mother position the child of weight
Express your answer in of
,
,
to balance the seesaw now?
, and .
= w*(L_3 + L_2)/W Bad news! When the mother finds the distance from the previous part it turns out to be greater than , the distance from the pivot to the end of the seesaw. Hence, even with the child of weight at the very end of the seesaw the twins exert more torque than the heavier child does. The mother now
elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is a height
above the pivot.
A. With what force in the rightwards direction,
, should the mother push? Note that if you
think the force exerted by the mother should be toward the left, your final answer negative. Express your answer in of
,
, ,
,
should be
, and .
= (W*L_endw*(L_2+L_3))/h [ Print ]
Pulling a String Adds Energy to a Wheel
A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel has moment of inertia , where is its mass, is its radius, and is a dimensionless number between zero and one. The wheel is rotating counterclockwise with angular velocity , when
at time
someone starts pulling the string with a force of magnitude
.
Assume that the string does not slip on the wheel.
A. Suppose that after a certain time , the string has been pulled through a distance . What is the final rotational speed of the wheel? Express
in of , , , ,
, and
.
= (2*F*L/(k*m*r^2) + omega_0^2)^.5
B. What is the instantaneous power delivered to the wheel via the force at time
?
Express the power in of the variables given in the problem introduction. = F*r*omega_0 [ Print ]
Hoop on a Ramp A circular hoop of mass
, radius , and infinitesimal thickness rolls without slipping down a ramp
inclined at an angle with the horizontal.
A. What is the acceleration of the center of the hoop? Express the acceleration in of physical constants and all or some of the given quantities:
, , and .
= (g/2)*sin(theta) B. What is the minimum coefficient of friction
needed for the hoop to roll without slipping?
Express the minimum coefficient of friction in of all or some of the given quantities:
, , and .
= 1/2*tan(theta)
[ Print ]
Record and Turntable Learning Goal: To understand how to use conservation of angular momentum to solve problems involving collisions of rotating bodies. Consider a turntable to be a circular disk of moment of inertia rotating at a constant angular velocity around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is ed by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis. Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is . The initial angular velocity of the second disk is zero. There is friction between the two disks. After this "rotational collision," the disks will eventually rotate with the same angular velocity. A. What is the final angular velocity, , of the two disks? Express
in of , , and
.
= (I_t/(I_t+I_r))*omega_i
B. Because of friction, kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final kinetic energy,
, of the two spinning disks?
Express the final kinetic energy in of , , and the initial kinetic energy twodisk system. No angular velocities should appear in your answer.
of the
= K_i*I_t/(I_t+I_r)
C. Assume that the turntable deccelerated during time before reaching the final angular velocity ( is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, , acting on the bottom disk due to friction with the record? Express the torque in of ,
,
, and
.
= I_t*(omega_fomega_i)/(deltat)
Spinning Situations Suppose you are standing on the center of a merrygoround that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis is pointing upward. The wheel is rotating counterclockwise when observed from above.
For this problem, neglect any air resistance or friction between the merrygoround and its foundation. A. Suppose you now grab the edge of the wheel with your hand, stopping it from spinning. What happens to the merrygoround? It remains at rest. It begins to rotate counterclockwise (as observed from above). It begins to rotate clockwise (as observed from above). [ Print ]
Precessing Tilted Gyroscope
A gyroscope consists of a flywheel of mass , which has a moment of inertia for rotation about its axis. It is mounted on a rod of negligible mass, which is ed at one end by a frictionless pivot attached to a vertical post, as shown in the diagram. The distance between the center of the wheel and the pivot is . The wheel rotates about its axis with angular velocity , where positive refers to counterclockwise rotation as seen by an observer looking at the face of the wheel that is opposite the pivot. The rod is tilted upward, making an angle with respect to the horizontal. Gravity acts downward with a force of magnitude . Adopt a coordinate system with the z axis pointing upward and the x and y axes in the horizontal plane. The gyroscope is moving, but at
, the rod is in the yz plane.
A. Assuming that the only significant contribution to the angular momentum comes from the spinning of the flywheel about its center, what is the angular momentum vector about the pivot at
?
Specify the components of with respect to the axes shown in the diagram. Write the components in order
,
,
separated by commas.
,
B. At
,
= 0 , I*omega*cos(theta) , I*omega*sin(theta) , what is the torque acting on the wheel about the pivot?
Express your answer in of components
, , , separated by commas.
, , = m*g*d*cos(theta) , 0 , 0
C. The gyroscope is observed to precess about the vertical axis, with an angular velocity of precession , defined as positive for counterclockwise precession as seen from above. Find in of the given quantities. = (m*g*d)/(I*omega)
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