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
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.
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
Maximize the magnitude of the force, F,
that you apply to the lever.
the distance, r, from the axis of
rotation of the point on the lever to which you apply the force.
the force in a direction perpendicular to the lever.
We can apply these three requirements to an equation for
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:
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
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
Let’s plug these values into the first equation we saw
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
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.
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
which represents an object’s resistance to being rotated. Using
the three variables,
, 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
is directed out of the page. From the linear version of Newton’s
Second Law, we know that F = ma
If we multiply both sides of this equation by r,
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.
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
||Make F and M as
large as possible and R as small as possible
||Make M as large as possible and F and R as
small as possible.
||Make F as large
as possible and M and R as small
||Make R as large as possible and F and M as
small as possible.
||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
, and so
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
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.
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
For Mass 2:
. Substitute into the
is positive, we know
that the pulley will spin in the counterclockwise direction and
block will drop.