ladybugs problem

Ladybugs on a Rotating Disk

Ladybugs on a Rotating DiskTwo ladybugs sit on a rotating disk (the ladybugs are at rest with respect to the surface of the disk and do not slip). A. What is the angular speed of ladybug 1?

one-half the angular speed of ladybug 2the same as the angular speed of ladybug 2two times the angular speed of ladybug 2one-quarter 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?

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/h/%2C5Chat%2C7Bx%2C7D.gif" \* MERGEFORMATINET

HTMLCONTROL Forms.HTML:Option.1

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/h/%2C2D%2C5Chat%2C7Bx%2C7D.gif" \* MERGEFORMATINET


INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/h/%2C5Chat%2C7By%2C7D.gif" \* MERGEFORMATINET


INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/h/%2C7B%2C7D%2C2D%2C5Chat%2C7By%2C7D.gif" \* MERGEFORMATINET


INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/h/%2C5Chat%2C7Bz%2C7D.gif" \* MERGEFORMATINET


INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/h/%2C7B%2C7D%2C2D%2C5Chat%2C7Bz%2C7D.gif" \* MERGEFORMATINET

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?



INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/h/%2C7B%2C7D%2C2D%2C5Chat%2C7Bx%2C7D.gif" \* MERGEFORMATINET









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 , angular velocity , and position of ladybug 1?

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/v/%2C5Cvec%2C7Bv_1%2C-z4hMxmPyVwMAddSHJrDoiw.gif" \* MERGEFORMATINET

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/v/%2C5Cvec%2C7Bv_1%2C-bHt1sfmKIXdXL_V5CFJv4w.gif" \* MERGEFORMATINET

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/v/%2C5Cvec%2C7Bv_1%2C-R6CVtW5CVUmWWsiwYxUqUQ.gif" \* MERGEFORMATINET

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/v/%2C5Cvec%2C7Bv_1%2C-X140lhAoo6MkYYGfmBycXg.gif" \* MERGEFORMATINET

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/v/%2C5Cvec%2C7Bv_1%2C-RYkvPCr8jNCK8Pxvp4OApg.gif" \* MERGEFORMATINET

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?



INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/h/%2C7B%2C7D%2C2D%2C5Chat%2C7Bx%2C7D.gif" \* MERGEFORMATINET









[Print]

Moment of Inertia and Center of Mass for Point ParticlesBall 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 two-mass system about axis a is , and the moment of inertia of the system about axis b is . It is observed that the ratio of to is equal to 3:

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.

=1/3

B. Find , the distance from ball A to the system's center of mass.

Express your answer in terms of , the length of the rod.

=3*L/4

[Print]

Scaling of Moments of InertiaLearning Goal: To understand the concept of moment of inertia and how it depends on mass, radius, and mass distribution.In rigid-body 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 1-meter 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 passes 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 Parallel-Axis TheoremLearning Goal: To understand the parallel-axis theorem and its applicationsTo 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 account 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 passing through the center of mass of the object. However, in many problems the axis of rotation does not pass 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 parallel-axis theorem. Mathematically, it can be expressed as , where is the moment of inertia about an axis passing through the center of mass, 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 terms 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).


=4*m*r^2

C. Now calculate for this object using the parallel-axis theorem.

Express your answer in terms 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).


=4*m*r^2

E. Now calculate for this object using the parallel-axis theorem.


=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 passes 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?

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20A%2C7D.gif" \* MERGEFORMATINET

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20B%2C7D.gif" \* MERGEFORMATINET

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20C%2C7D.gif" \* MERGEFORMATINET

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20D%2C7D.gif" \* MERGEFORMATINET

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20E%2C7D.gif" \* MERGEFORMATINET

B. Which moment of inertia is the largest?

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20A%2C7D.gif" \* MERGEFORMATINET





C. Which moments of inertia are equal?

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20A%2C7D.gif" \* MERGEFORMATINET and

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20B%2C7D.gif" \* MERGEFORMATINET and

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/I/I_%2C7B%2C5Crm%2C20C%2C7D.gif" \* MERGEFORMATINET 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 Cbetween axes C and Dbetween axes D and Eto the right of axis E

[Print]

Loop the Loop with a TwistIn 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 shown in the figure without losing contact with the track. 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 to loop the loop of radius .

Express in terms of the radius of the loop.

=(11/4)*r_loop

[Print]

Top SpinA 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 terms of the cylinder's radius and the magnitude of the acceleration due to gravity .

=(2/3)*g/r

Torques on a Seesaw: A TutorialLearning 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 off-center 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?


=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.


=-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 , the sum of the torques on the seesaw. Consider only the torques exerted by the children.

Express your answer in terms 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 to balance the seesaw now?

Express your answer in terms of , , , 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 should be negative.

Express your answer in terms of , , , , , and .

=(W*L_end-w*(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 terms 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 terms 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 terms 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 terms of all or some of the given quantities: , , and . =1/2*tan(theta)

[Print]

Record and TurntableLearning 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 supported 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 terms 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 terms of , , and the initial kinetic energy of the two-disk system. No angular velocities should appear in your answer.

=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 terms of , , , and .

=I_t*(omega_f-omega_i)/(deltat)

Spinning SituationsSuppose you are standing on the center of a merry-go-round 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 merry-go-round and its foundation. A. Suppose you now grab the edge of the wheel with your hand, stopping it from spinning.

What happens to the merry-go-round?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 supported 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

INCLUDEPICTURE "http://session.masteringphysics.com/latexgif/g/g.gif" \* MERGEFORMATINET .

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.

, , =0,I*omega*cos(theta),I*omega*sin(theta)

B. At , what is the torque acting on the wheel about the pivot?

Express your answer in terms 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 terms of the given quantities.

=(m*g*d)/(I*omega)

_1160386176.unknown

_1160386337.unknown

_1160386343.unknown

_1160386396.unknown

_1160387095.unknown

_1160387097.unknown

_1160386397.unknown

_1160387094.unknown

_1160386345.unknown

_1160386393.unknown

_1160386394.unknown

_1160386346.unknown

_1160386344.unknown

_1160386340.unknown

_1160386341.unknown

_1160386339.unknown

_1160386327.unknown

_1160386332.unknown

_1160386335.unknown

_1160386336.unknown

_1160386334.unknown

_1160386330.unknown

_1160386331.unknown

_1160386328.unknown

_1160386181.unknown

_1160386322.unknown

_1160386324.unknown

_1160386326.unknown

_1160386323.unknown

_1160386184.unknown

_1160386319.unknown

_1160386321.unknown

_1160386185.unknown

_1160386183.unknown

_1160386179.unknown

_1160386180.unknown

_1160386177.unknown

_1160386166.unknown

_1160386171.unknown

_1160386174.unknown

_1160386175.unknown

_1160386172.unknown

_1160386168.unknown

_1160386170.unknown

_1160386167.unknown

_1160386161.unknown

_1160386163.unknown

_1160386164.unknown

_1160386162.unknown

_1160386158.unknown

_1160386159.unknown

_1160386155.unknown

_1160386157.unknown

_1160386153.unknown

_1160386154.unknown

_1160386150.unknown

ladybugs problem

Documents

mergeformatinet htmlcontrol

angular speed of ladybug

angular velocity of

position of ladybug

linear speed of ladybug

radial acceleration

rotating disk ladybugs

rotating disktwo ladybugs