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 10.1 Important Definitions 10.2 Rotational Kinematics 10.3 Frequency and Period 10.4 Rotational Dynamics 10.5 Kinetic Energy

 10.6 Angular Momentum 10.7 Key Formulas 10.8 Practice Questions 10.9 Explanations
Rotational Dynamics
Just as we have rotational counterparts for displacement, velocity, and acceleration, so do we have rotational counterparts for force, mass, and Newton’s Laws. As with angular kinematics, the key here is to recognize the striking similarity between rotational and linear dynamics, and to learn to move between the two quickly and easily.
Torque
If a net force is applied to an object’s center of mass, it will not cause the object to rotate. However, if a net force is applied to a point other than the center of mass, it will affect the object’s rotation. Physicists call the effect of force on rotational motion torque.
Torque Defined
Consider a lever mounted on a wall so that the lever is free to move around an axis of rotation O. In order to lift the lever, you apply a force F to point P, which is a distance r away from the axis of rotation, as illustrated below.
Suppose the lever is very heavy and resists your efforts to lift it. If you want to put all you can into lifting this lever, what should you do? Simple intuition would suggest, first of all, that you should lift with all your strength. Second, you should grab onto the end of the lever, and not a point near its axis of rotation. Third, you should lift in a direction that is perpendicular to the lever: if you pull very hard away from the wall or push very hard toward the wall, the lever won’t rotate at all.
Let’s summarize. In order to maximize torque, you need to:
1. Maximize the magnitude of the force, F, that you apply to the lever.
2. Maximize the distance, r, from the axis of rotation of the point on the lever to which you apply the force.
3. Apply the force in a direction perpendicular to the lever.
We can apply these three requirements to an equation for torque, :
In this equation, is the angle made between the vector for the applied force and the lever.
Torque Defined in Terms of Perpendicular Components
There’s another way of thinking about torque that may be a bit more intuitive than the definition provided above. Torque is the product of the distance of the applied force from the axis of rotation and the component of the applied force that is perpendicular to the lever arm. Or, alternatively, torque is the product of the applied force and the component of the length of the lever arm that runs perpendicular to the applied force.
We can express these relations mathematically as follows:
where and are defined below.
Torque Defined as a Vector Quantity
Torque, like angular velocity and angular acceleration, is a vector quantity. Most precisely, it is the cross product of the displacement vector, r, from the axis of rotation to the point where the force is applied, and the vector for the applied force, F.
To determine the direction of the torque vector, use the right-hand rule, curling your fingers around from the r vector over to the F vector. In the example of lifting the lever, the torque would be represented by a vector at O pointing out of the page.
Example
 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 60Âº 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 = mr. 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.
 Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II PhysicsStrategies for Taking SAT II PhysicsVectorsKinematicsDynamicsWork, Energy, and PowerSpecial Problems in MechanicsLinear MomentumRotational MotionCircular Motion and GravitationThermal PhysicsElectric Forces, Fields, and PotentialDC CircuitsMagnetismElectromagnetic InductionWavesOpticsModern PhysicsPhysics GlossaryPractice Tests Are Your Best Friends
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