SAT II Physics (SN) Episode 1 Part 7 ppsx

The First Law states: If the net torque acting on a rigid object is zero, it will rotate with a constant angular velocity.. Newton’s Second Law We have seen that acceleration has a rotat

A student exerts a force of 50 N on a lever at a distance 0.4 m from its axis of rotation The student pulls at an angle that is 60º above the lever arm What is the torque experienced by the lever arm?

Let’s plug these values into the first equation we saw for torque:

This vector has its tail at the axis of rotation, and, according to the right-hand rule, points out of the page

Newton’s First Law and Equilibrium

Newton’s Laws apply to torque just as they apply to force You will find that solving problems involving torque is made a great deal easier if you’re familiar with how to apply Newton’s Laws

to them The First Law states:

If the net torque acting on a rigid object is zero, it will rotate with a constant angular velocity.

The most significant application of Newton’s First Law in this context is with regard to the

concept of equilibrium When the net torque acting on a rigid object is zero, and that object is not

already rotating, it will not begin to rotate

When SAT II Physics tests you on equilibrium, it will usually present you with a system where more than one torque is acting upon an object, and will tell you that the object is not rotating That means that the net torque acting on the object is zero, so that the sum of all torques acting in the clockwise direction is equal to the sum of all torques acting in the counterclockwise direction A typical SAT II Physics question will ask you to determine the magnitude of one or more forces acting on a given object that is in equilibrium

EXAMPLE

Two masses are balanced on the scale pictured above If the bar connecting the two masses is

horizontal and massless, what is the weight of mass m in terms of M?

Since the scale is not rotating, it is in equilibrium, and the net torque acting upon it must be zero

In other words, the torque exerted by mass M must be equal and opposite to the torque exerted by mass m Mathematically,

Because m is three times as far from the axis of rotation as M, it applies three times as much torque per mass If the two masses are to balance one another out, then M must be three times as heavy as m.

Newton’s Second Law

We have seen that acceleration has a rotational equivalent in angular acceleration, , and that force has a rotational equivalent in torque, Just as the familiar version of Newton’s Second Law tells us that the acceleration of a body is proportional to the force applied to it, the rotational version of Newton’s Second Law tells us that the angular acceleration of a body is proportional to the torque applied to it

Of course, force is also proportional to mass, and there is also a rotational equivalent for mass: the

moment of inertia, I, which represents an object’s resistance to being rotated Using the three

variables, , I, and , we can arrive at a rotational equivalent for Newton’s Second Law:

As you might have guessed, the real challenge involved in the rotational version of Newton’s Second Law is sorting out the correct value for the moment of inertia

Moment of Inertia

What might make a body more difficult to rotate? First of all, it will be difficult to set in a spin if it has a great mass: spinning a coin is a lot easier than spinning a lead block Second, experience shows that the distribution of a body’s mass has a great effect on its potential for rotation In general, a body will rotate more easily if its mass is concentrated near the axis of rotation, but the calculations that go into determining the precise moment of inertia for different bodies is quite complex

MOMENT OF INERTIA FOR A SINGLE PARTICLE

Consider a particle of mass m that is tethered by a massless string of length r to point O, as

pictured below:

The torque that produces the angular acceleration of the particle is = rF, and is directed out of the page From the linear version of Newton’s Second Law, we know that F = ma or F = m r If

we multiply both sides of this equation by r, we find:

If we compare this equation to the rotational version of Newton’s Second Law, we see that the

moment of inertia of our particle must be mr2

MOMENT OF INERTIA FOR RIGID BODIES

Consider a wheel, where every particle in the wheel moves around the axis of rotation The net torque on the wheel is the sum of the torques exerted on each particle in the wheel In its most general form, the rotational version of Newton’s Second Law takes into account the moment of inertia of each individual particle in a rotating system:

Of course, adding up the radius and mass of every particle in a system is very tiresome unless the system consists of only two or three particles The moment of inertia for more complex systems can only be determined using calculus SAT II Physics doesn’t expect you to know calculus, so it will give you the moment of inertia for a complex body whenever the need arises For your own reference, however, here is the moment of inertia for a few common shapes

In these figures, M is the mass of the rigid body, R is the radius of round bodies, and L is the

distance on a rod between the axis of rotation and the end of the rod Note that the moment of inertia depends on the shape and mass of the rigid body, as well as on its axis of rotation, and that

for most objects, the moment of inertia is a multiple of MR2

EXAMPLE 1

A record of mass M and radius R is free to rotate around an axis through its center, O A tangential

force F is applied to the record What must one do to maximize the angular acceleration?

(A) Make F and M as large as possible and R as small as possible

(B) Make M as large as possible and F and R as small as possible.

(C) Make F as large as possible and M and R as small as possible.

(D) Make R as large as possible and F and M as small as possible.

(E) Make F , M, and R as large as possible.

To answer this question, you don’t need to know exactly what a disc’s moment of inertia is—you

just need to be familiar with the general principle that it will be some multiple of MR2

The rotational version of Newton’s Second Law tells us that = I , and so = FR/I Suppose we

don’t know what I is, but we know that it is some multiple of MR2 That’s enough to formulate an equation telling us all we need to know:

As we can see, the angular acceleration increases with greater force, and with less mass and

radius; therefore C is the correct answer.

Alternately, you could have answered this question by physical intuition You know that the more force you exert on a record, the greater its acceleration Additionally, if you exert a force on a small, light record, it will accelerate faster than a large, massive record

EXAMPLE 2

The masses in the figure above are initially held at rest and are then released If the mass of the

pulley is M, what is the angular acceleration of the pulley? The moment of inertia of a disk spinning around its center is MR2

This is the only situation on SAT II Physics where you may encounter a pulley that is not

considered massless Usually you can ignore the mass of the pulley block, but it matters when your knowledge of rotational motion is being tested

In order to solve this problem, we first need to determine the net torque acting on the pulley, and then use Newton’s Second Law to determine the pulley’s angular acceleration The weight of each mass is transferred to the tension in the rope, and the two forces of tension on the pulley block exert torques in opposite directions as illustrated below:

To calculate the torque one must take into account the tension in the ropes, the inertial resistance

to motion of the hanging masses, and the inertial resistence of the pulley itself The sum of the torques is given by:

Solve for the tensions using Newton’s second law For Mass 1:

For Mass 2:

Remember that Substitute into the first equation:

Because is positive, we know that the pulley will spin in the counterclockwise direction and the

3m block will drop.

Kinetic Energy

There is a certain amount of energy associated with the rotational motion of a body, so that a ball rolling down a hill does not accelerate in quite the same way as a block sliding down a frictionless

slope Fortunately, the formula for rotational kinetic energy, much like the formula for translational kinetic energy, can be a valuable problem-solving tool.

The kinetic energy of a rotating rigid body is:

Considering that I is the rotational equivalent for mass and is the rotational equivalent for

velocity, this equation should come as no surprise

An object, such as a pool ball, that is spinning as it travels through space, will have both rotational and translational kinetic energy:

In this formula, M is the total mass of the rigid body and is the velocity of its center of mass

This equation comes up most frequently in problems involving a rigid body that is rolling along a surface without sliding Unlike a body sliding along a surface, there is no kinetic friction to slow the body’s motion Rather, there is static friction as each point of the rolling body makes contact with the surface, but this static friction does no work on the rolling object and dissipates no energy

EXAMPLE

A wheel of mass M and radius R is released from rest and rolls to the bottom of an inclined plane of height h without slipping What is its velocity at the bottom of the incline? The moment of inertia of a wheel of mass M and radius R rotating about an axis through its center of mass is 1 / 2 MR2

Because the wheel loses no energy to friction, we can apply the law of conservation of mechanical

energy The change in the wheel’s potential energy is –mgh The change in the wheel’s kinetic

energy is Applying conservation of mechanical energy:

It’s worth remembering that an object rolling down an incline will pick up speed more slowly than

an object sliding down a frictionless incline Rolling objects pick up speed more slowly because only some of the kinetic energy they gain is converted into translational motion, while the rest is converted into rotational motion.

Angular Momentum

The rotational analogue of linear momentum is angular momentum, L After torque and

equilibrium, angular momentum is the aspect of rotational motion most likely to be tested on SAT

II Physics For the test, you will probably have to deal only with the angular momentum of a particle or body moving in a circular trajectory In such a case, we can define angular momentum

in terms of moment of inertia and angular velocity, just as we can define linear momentum in terms of mass and velocity:

The angular momentum vector always points in the same direction as the angular velocity vector

Angular Momentum of a Single Particle

Let’s take the example of a tetherball of mass m swinging about on a rope of length r:

The tetherball has a moment of inertia of I = mr2 and an angular velocity of = v/r Substituting

these values into the formula for linear momentum we get:

This is the value we would expect from the cross product definition we saw earlier of angular

momentum The momentum, p = mv of a particle moving in a circle is always tangent to the circle

and perpendicular to the radius Therefore, when a particle is moving in a circle,

Newton’s Second Law and Conservation of Angular Momentum

In the previous chapter, we saw that the net force acting on an object is equal to the rate of change

of the object’s momentum with time Similarly, the net torque acting on an object is equal to the rate of change of the object’s angular momentum with time:

If the net torque action on a rigid body is zero, then the angular momentum of the body is constant

or conserved The law of conservation of angular momentum is another one of nature’s

beautiful properties, as well as a very useful means of solving problems It is likely that angular

momentum will be tested in a conceptual manner on SAT II Physics.

EXAMPLE

One of Brian Boitano’s crowd-pleasing skating moves involves initiating a spin with his arms extended and then moving his arms closer to his body As he does so, he spins at a faster and faster rate Which

of the following laws best explains this phenomenon?

(A) Conservation of Mechanical Energy

(B) Conservation of Angular Momentum

(C) Conservation of Linear Momentum

(D) Newton’s First Law

(E) Newton’s Second Law

Given the context, the answer to this question is no secret: it’s B, the conservation of angular

momentum Explaining why is the interesting part

As Brian spins on the ice, the net torque acting on him is zero, so angular momentum is

conserved That means that I is a conserved quantity I is proportional to R2, the distance of the parts of Brian’s body from his axis of rotation As he draws his arms in toward his body, his mass

is more closely concentrated about his axis of rotation, so I decreases Because I must remain constant, must increase as I decreases As a result, Brian’s angular velocity increases as he

draws his arms in toward his body

2 A washing machine, starting from rest, accelerates within 3.14 s to a point where it is revolving at

a frequency of 2.00 Hz Its angular acceleration is most nearly:

3 What is the direction of the angular velocity vector for the second hand of a clock going from 0 to

30 seconds?

(A) Outward from the clock face

(B) Inward toward the clock face

(C) Upward

(D) Downward

(E) To the right

4 Which of the following are means of maximizing the torque of a force applied to a rotating object?

II Apply the force as close as possible to the axis of rotation III Apply the force perpendicular to the displacement vector between the axis of rotation and the point of applied force

(A) I only

(B) II only

(C) I and II only

(D) I and III only

(E) I, II, and III

5 What is the torque on the pivot of a pendulum of length R and mass m, when the mass is at an

7 What is the angular acceleration of a 0.1 kg record with a radius of 0.1 m to which a torque of 0.05

N · m is applied? The moment of inertia of a disk spinning about its center is 1/2MR2

8 A disk of mass m and radius R rolls down an inclined plane of height h without slipping What is the

velocity of the disk at the bottom of the incline? The moment of inertia for a disk is 1 / 2 mR2 (A)

(A) The mass should be concentrated at the outer edge of the body

(B) The mass should be evenly distributed throughout the body

(C) The mass should be concentrated at the axis of rotation

(D) The mass should be concentrated at a point midway between the axis of rotation and the outer edge of the body

(E) Mass distribution has no impact on angular velocity

Explanations

1 D

An object that experiences 120 revolutions per minute experiences 2 revolutions per second; in other words, it rotates with a frequency of 2 Hz We have formulas relating frequency to angular velocity and angular velocity to linear velocity, so solving this problem is simply a matter of finding an expression for linear velocity in terms of frequency Angular and linear velocity are related by the formula , so we need to plug this formula into the formula relating frequency and angular velocity:

4 D

The torque on an object is given by the formula , where F is the applied force and r is the distance of the applied force from the axis of rotation In order to maximize this cross product, we need to maximize the two quantities and insure that they are perpendicular to one another Statement I maximizes

F and statement III demands that F and r be perpendicular, but statement II minimizes r rather than maximizes it, so statement II is false.

5 C