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# What is meant by the torque produced by a force. - PowerPoint PPT Presentation

We know that a net force applied to a body gives that body an acceleration. But what does it take to give a body an angular acceleration?

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Presentation Transcript

• We know that a net force applied to a body gives that body an acceleration. But what does it take to give a body an angular acceleration?

• The physical quantity that make things turn is called torque. The net torque acting on a rigid body determines its angular acceleration, in the same way that the net force on a body determines its linear acceleration.

Ch 10 – learning goals an acceleration. But what does it take to give a body an

• What is meant by the torque produced by a force.

• How the net torque on a body affects the rotational motion of the body.

• How to analyze the motion of a body that both rotates and moves as a whole through space.

• How to solve problems that involve work and power for rotating bodies.

• What is meant by the angular momentum of a particle or of a rigid body.

• How does the angular momentum of a system changes with time.

The quantitative measure of the tendency of a an acceleration. But what does it take to give a body an force to change a body’s rotational motion is called torque;

Fa applies a torque about point O to the wrench.

Fb applies a greater torque about O,

and Fc applies zero torque about O.

• The tendency of a force an acceleration. But what does it take to give a body an F to cause a rotation about O depends on

• its magnitude F

• the perpendicular distancel1 between point O and the line of action of the force. We call the distance l1 the lever arm of force F1about O.

• We define the torque of the force F1 with respect to O as the product F1l1. we use the Greek letterτ (tau) for torque.

• τ = Fl

F: force

l: the perpendicular distancel1 between point O and the line of action of the force

CAUTION: Torque is always measured about a point an acceleration. But what does it take to give a body an

The direction of torque an acceleration. But what does it take to give a body an

• counterclockwise torques are positive and clockwise torques are negative.

The units of torque an acceleration. But what does it take to give a body an

• The SI unit of torque is the Newton-meter.

• Torque is not work or energy, and torque should be expressed in Newton-meters, notjoules.

If an acceleration. But what does it take to give a body an φ is angle between force F and distance r

τ = F∙(r∙sin)

r∙sin - perpendicular distance

τ = r∙(F∙sin)

F∙sin - perpendicular force

Magnitude an acceleration. But what does it take to give a body an :

When a force F acts at a point having a position vector r with respect to an origin O, the torque of the force with respect to O is the vector quantity

The direction of torque is perpendicular to both r and F. The torque vector is directed along the axisof rotation, with a sense given by the right-hand rule.

A dot ● means pointing an acceleration. But what does it take to give a body an out of the screen

A cross × means pointing into the screen

45 an acceleration. But what does it take to give a body an o

60o

sign

sign

example

• Rank the design scenarios (A through C) on the basis of the tension in the supporting cable from largest to smallest. In scenarios A, and C, the cable is attached halfway between the midpoint and end of the pole. In B, the cable is attached to the end of the pole.

30o

sign

A

B

C

A, C, B

example an acceleration. But what does it take to give a body an

• If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what does she have to do with the mass of the smiley face to keep the mobile in perfect balance? Note that she may have to change masses of other objects to keep the entire structure balanced.

• make it eight times more massive

• make it four times more massive

• make it twice as massive

• Nothing

• impossible to tell

since an acceleration. But what does it take to give a body an

• Newton’s second law for the tangential component is:

We write an equation like this for every particle in the body and then add all these equations:

Just as Newton’s second law says that the net force on a particle equals the particle’s mass times its acceleration, the equation says that the net torque on a rigid body equals the body’s moment of inertia about the rotation axis times its angular acceleration.

4 body and then add all these equations: things to note in Eq.

• The equation is valid only for rigid bodies.

• Since we used atan = r∙αz, αz must be measured in rad/s2.

• Since all the internal torques add to zero, so the sum ∑τ in Eq. ∑τ = Iα includes only the torques of the external forces.

• Often, an important external force acting on body is its weight. We assume that all the weight is concentrated at the center of mass of the body to get the correct torque (about any specified axis).

• The figure shows a glider of mass m of the block of mass m.1 that can slide without friction on horizontal air tract. It is attached to an object of mass m2 by a massless string. The pulley has radius R and moment of inertia I about it axis of rotation. When released, the hanging object accelerates downward, the glider accelerates to the right, and the string turns the pulley without slipping or stretching. Rank the magnitudes of the following forces that acting during the motion, in order from largest to smallest magnitude.

• The tension force (magnitude T1) in the horizontal part of the string;

• The tension force (magnitude T2) in the vertical part of the string;

• The weight m2g of the hanging object.

example of the block of mass m.

• Find the magnitude of the angular acceleration α of the swing bar.

• The the body is kinetic energy of a rigid body that has both translational and rotational motions is the sum of a part ½ Mvcm2 associated with motion of the center of mass and a part ½ Icmω2 associated with rotation about an axis through the center of mass.

The point on the wheel that contacts the surface must be instantaneously at rest so that it does not slip. Hence the velocity v1’ of the point of contact relative to the center of mass must have the same magnitude but opposite direction as the center-of-mass velocity vcm. If the radius of the wheel is R and its angular speed about the center of mass is ω, then the magnitude of v1’ is R∙ω; hence we must have

vcm = Rω (condition for rolling without slipping)

• Note that the relationship instantaneously vcm = Rω holds only if there is rolling without slipping.

• When a drag racer first starts to move, the rear tires are spinning very fast even though the racer is hardly moving, so Rω is greater than vcm.

• If a driver applies the brakes too heavily so that the car skids, the tires will spin hardly at all and Rω is less than vcm.

• If a rigid body changes height as it moves, we must also consider gravitational potential energy. U = Mgycm

• The speed doesn’t depend on either the mass M of the body or its radius R. All uniform solid cylinders have the same speed at the bottom, even if their masses and radii are different, because they have the same c. All solid spheres also have the same speed, an so on.

• The smaller the value of c, the faster the body is moving at the bottom (and at any point on the way down).

• Small-c bodies always beat large-c bodies because they have less of their kinetic energy tied up in rotation and have more available for translation.

• Reading the values of c from the reference sheet, the order of finish is as follows:

• Solid sphere

• Solid cylinder,

• Thin-walled hollow sphere

• Thin-walled hollow cylinder

When a rigid body with total mass M moves, its motion can be described by combining translational motion and rotational motion

In translation, the acceleration acm of the center of mass is the same as that of a point mass M acted on by all the external forces on the actual body:

The rotational motion about the center of mass is described by the rotational analog of Newton’s 2nd law:

• Note: when we learned this equation, we assumed that the axis of rotation was stationary. But in fact, this equation is valid even when the axis of rotation moves, provided the following two conditions are met:

• The axis through the center of mass must be an axis of symmetry.

• The axis must not change direction.

When a perfectly rigid sphere is rolling down a perfectly rigid incline, there is no sliding at the point of contact, so friction does no work. However, in reality, when a not so perfectly rigid sphere rolling down a not so perfectly rigid incline, there are some deformations at the points of contact. As a result, there is rolling friction.

Often the rolling body and the surface are rigid enough that rolling friction can be ignored.

example is replaced by a hollow cylinder of the same mass and radius

• Two uniform identical solid spherical balls each of mass M, radius r and moment of inertial about its center 2/5MR2, are released from rest from the same height h above the horizontal ground Ball A falls straight down, while ball B rolls down the distance x along the inclined plane without slipping.

• If the velocity of ball A as it hits the ground is VA, what is the velocity VB of ball b as it reaches the ground?

• In terms of acceleration due to earth’s gravity g, the acceleration of ball B along the inclined plane would be

The work is replaced by a hollow cylinder of the same mass and radiusdW done by the force Ftan while a point on the rim moves a distance ds is dW = Ftan∙ds. If dθ is measured in radians, then ds = R∙dθ

The work done by a constant torque is the product of torque and the angular displacement. If torque is expressed in Newton-meters and angular displacement in radian, the work is in joules.

Only the tangent component of force does work, other components do no work.

When a torque does work on a rotating rigid body, the kinetic energy changes by an amount equal to the work done.

Wtot = ½ I∙ω22 – ½ I∙ω12

P = dW/dt = τz(dθ/dt) = τzω

When a torque acts on a body the rotates with angular velocity ωz, its power is the product of τz and ωz. This is the analog of the relationship P = F∙v

• You apply equal torques to two different cylinders, one of which has a moment of inertial twice as large as the other cylinder. Each cylinder is initially at rest after one complete rotation, which cylinder has the greater kinetic energy?

• The cylinder with the larger moment of inertia;

• The cylinder with the smaller moment of inertia;

• Both cylinders have the same kinetic energy.

∑F = ma which has a moment of inertial twice as large as the other cylinder. Each cylinder is initially at rest after one complete rotation, which cylinder has the greater kinetic energy?

K = ½ mv2

Wtot = ½ mv22 - ½ mv12

P = Fv

• Every rotational quantity that we have encountered so far is the analog of some quantity in the translational motion of a particle.

• The analog of which has a moment of inertial twice as large as the other cylinder. Each cylinder is initially at rest after one complete rotation, which cylinder has the greater kinetic energy?momentum of a particle is angular momentum, a vector quantity denoted as L.

L = I ∙ ω

The value of L depends on the choice of origin O, since it involves the particle’s position vector relative to O. the units of angular momentum are kg∙m2/s.

If at moment t the position of the particle is r and its linear momentum is p = m∙v, then the magnitude of angular momentum is:

L = (mvsinΦ)r

or

L = mv(rsinΦ) = mvl

Where l is the perpendicular distance from the line of v to O. this distance plays the role of “lever arm” for the momentum vector.

= F the angular momentum vector net

• When a net force F acts on a particle, its velocity and momentum changes:

• The rate of change dp/dt of the linear momentum of a particle equals the net force acting on it.

• Similarly, when the torque of the net force acting on a particle, its angular velocity and angular momentum changes:

• The rate of change of angular momentum of a particle equals the torque of the net force acting on it.

The derivation of the angular momentum vector

If we take the time derivative of , using the rule for the derivative of a product:

Consider a rigid body rotating about the z-axis with angular speed ω. First let’s think about a thin slice of the body lying in the xy-plane.

Each particle moves in a circle centered at the origin, and at each instant its velocity vi is perpendicular to its position vector ri. Hence Φ = 90o for every particle. A particle with mass mi at a distance ri from O has a speed vi = riω. The magnitude of its angular momentum is:

Li = (mivisinΦ)ri = (miriω)ri = miri2ω

The direction of each particle's angular momentum, as given by the right-hand rule for the vector product, is along the +z-axis.

We can do this same calculation for the other slides of the body, all parallel to the xy-plane. Since the rigid body is rotating the axis of symmetry, their angular momentum vector sum L1 + L2 also lies along the symmetry, and its magnitude is L = Iω.

Since ω and L have the same direction, we have a vector relationship: L = Iω

• The total angular momentum of the slice of the body lying in the xy-plane is the sum ∑Li of the angular momenta Li of the particles:

Where I is the moment of inertia of the slice about the z-axis.

• For any system of particles (including both rigid and non-rigid bodies), the rate of change of the total angular momentum equals the sum of the torques of all forces acting on all the particles. The torques of the internal forces add to zero. So the sum of the torques includes only the torques of the external forces. If the total angular momentum of the system of particles is L and the sum of the external torques is ∑τ, then

If the system of particles is a rigid body rotating about a symmetry axis (z-axis), the Lz = Iωz and I is constant. If this axis has a fixed direction in space, then the vectors L and ω change only in magnitude, not in direction. dLz/dt = Idωz/dt = Iαz ∑τ = Iαz

practice the other end of the string and whirl the ball in a circle around your hand.

• A uniform rod of mass M and length L has a moment of inertia about one end I = ML2/3. It is released from rest in horizontal direction about the fixed axis perpendicular to the paper as shown below.

• What is the linear velocity of the center of mass, cm, when the rod is in the vertical position?

• What is the angular momentum of the rod about the axis of rotation at one end?

• Like conservation of energy and of linear momentum, the other end of the string and whirl the ball in a circle around your hand. the principle of conservation of angular momentum is a universal conservation law, valid at all scales from atomic and nuclear systems to the motions of galaxies.

• The principle follows directly from equation: ∑τ = dL/dt

• If ∑τ = 0, then dL/dt = 0, and L is constant.

When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved).

I the other end of the string and whirl the ball in a circle around your hand. 1ω1z = I2ω2z

• A circus acrobat, a diver, and an ice skater pirouetting on the toe of one skate all take advantage of this principle.

When a system has several parts, the internal forces that the parts exert on each other cause changes in the angular momenta of the parts, but the total angular momentum doesn’t change. The total angular momentum of the system is constant.

• If the polar ice caps were to completely melt due to global warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• Increase

• Decrease

• Remain the same

(hint: use angular momentum ideas. Assume that the sun, moon, and planets exert negligibly small torques on the earth.)

L = Iω

I increases, ω decreases, days is longer.

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A uniform rod of mass M and length L has a moment of inertia about one end I = ML2/3. It is released from rest in horizontal direction about the fixed axis perpendicular to the paper as shown below.

• What is the linear velocity of the center of mass, cm, when the rod is in the vertical position?

• What is the angular momentum of the rod about the axis of rotation at one end?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• Two equal masses, each m, are resting at the ends of a uniform rod of length 2a and negligible mass. The system is in equilibrium about the center C of the rod. A piece of clay of mass m is dropped down on the mass at the right end, hits it with velocity v as shown below and sticks to it.

• What is the ratio of the kinetic energy Ef just after the collision to the kinetic energy Ei just before the collision, Ef/Ei, of the system?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A skater is spinning on ice with her arms outstretched about the vertical axis at an angular speed of ω. When she brings her arms close to her body, which of the following statements is correct?

• Her angular velocity and angular momentum remain constant.

• Her angular momentum is increased.

• Her kinetic energy is increased.

• Her kinetic energy is decreased.

• The net torque on her about the axis of rotation increases.

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A uniform diving board, 12 meters long and 20 kg in mass, is hinged at P, which is 5 meters from the edge of the platform. An 80 kg diver is standing at the other end of the board.

• What will be the force exerted by the hinge on the board?

• What will be the normal force on the board at the edge of the platform?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A mass M slides down a smooth surface from height h and collides inelastically with the lower end of a rod that is free to rotate about a fixed axis at P as shown below. The mass of the rod is also M, the length is l, and the moment of inertial about P is ml2/3.

• What is the angular velocity of the rod about the axis P jut after the mass sticks to it?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A square metal plate 0.180 m on each side is pivoted about an axis through point at its center and perpendicular to the plate. Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are  F1 = 21.0 N ,   F2 = 17.0 N, and   F3 = 14.9 N . The plate and all forces are in the plane of the page. Take positive torques to be counterclockwise.

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• Masses M1 and M2 are separated by a distance L. what is the distance of the center of mass of the system at P from M1 as shown above?

• What is the moment of inertial of the system about the center of mass at P?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A solid, uniform cylinder with mass 8.45 kg and diameter 11.0 cm is spinning with angular velocity 230 rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction-brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.334. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.15 revolutions?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A wheel with a weight of 395 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 26.8 rad/s. The radius of the wheel is 0.652 m and its moment of inertia about its rotation axis is 0.800 MR2 . Friction does work on the wheel as it rolls up the hill to a stop, at a height of above the bottom of the hill; this work has a magnitude of 3520 J. Calculate h. Use 9.81 m/s2  for the acceleration due to gravity.

r warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

M

m

h

example

• The radius of the pulley is r and mass M is initially at height h. The system is initially at rest and is then released at time t = 0. Assume M>>m.

• Assuming the pulley to be massless and frictionless, what is the angular acceleration of the pulley while M is falling?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• An object of moment of inertial I is initially at rest when torque T begins to act on it as shown below. After t seconds,

• what is the angular velocity of the object in terms of T, I and t?

• What is the kinetic energy of the object?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A solid, uniform cylinder with mass 8.45 kg and diameter 11.0 cm is spinning with angular velocity 230 rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction-brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.334. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.15 revolutions?

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• A hoop is rolling to the right without slipping on a horizontal floor at a steady 1.8 m/s (Vcm).

• Find the velocity vector of each of the following points, as viewed by a person at rest on the ground:

• The highest point on the hoop

• The lowest point on the hoop

• A point on the right side of the hoop, mideway abetween the top and the bottom

• find the velocity vector of each of the above points, as viewed by a person moving along with the same velocity as the hoop.

example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to

• Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end.