Rotational Dynamics

While there are some analogs between linear and rotational dynamics, the rotational problems are generally much more troublesome (to the beginning student.) Still, recognizing similarities between the linear problems and rotational problems is a good start!

In the linear world of dynamics, we were concerned with forces that caused linear accelerations of objects. (Or, with balances of forces that resulted in equilibrium or zero acceleration.)

In the rotational world of dynamics, we are concerned with torques that cause angular accelerations of objects. (Or, with balances of torques that result in rotational equilibrium or zero angular acceleration.)

Torque – Is not a force, but, it will work, conceptually, if you think of torque as any force that causes rotation (rather than translation) or would tend to cause rotation (given the chance) of the object (about some axis of rotation or pivot point.)

Any force whose line of action is not directed towards the axis of rotation (or the center of mass) for an object) will provide a torque on that object.

Torque

t = F l

where t is torque, F is force, and l is the length of the torque arm or lever arm.

Some very important points:

- The force F must be the component of the applied force

that is perpendicular to the lever arm.

- The length of the lever arm is that distance between the

pivot point and the point of application of the force.

- “Pivot point”, “axis of rotation”, “fulcrum” all mean the

same thing. (There will be slightly different contexts.)

- Units would be newton meter (N m) in SI, foot pound

(ft lb) in English, or any appropriate combination of force

and distance. (Note: in “real” life, there are other options

such as kg cm which you might find in a automotive shop

manual but it is technically incorrect. (Why?!))

- Torque is a vector quantity. Direction matters!

- Torque is positive for a force that causes or tends to

cause counterclockwise rotation.

- Torque is negative for a force that causes or tends to

cause clockwise rotation.

Note that this analysis is different than that presented in the text (which defines the lever arm to be the distance from the pivot point to a point perpendicular to the force’s “line of action” and uses the applied force rather than its component. This is different conceptually but identical mathematically.)

Torque (continued)

Torque, by itself, isn’t especially interesting. But, this simple little torque equation pops up in wide variety of different problems of varying complexity.

An object is in translational (straight line) equilibrium when the sum of all forces acting = 0 (and, thus, its acceleration = 0.)

SF = 0

An object is in rotational equilibrium when the sum of all torques acting = 0 (and, thus, its angular acceleration = 0.)

St = 0

An object is in both translational and rotational equilibrium when both conditions are met.

Equilibrium

Solving Equilibrium Problems

1) Clearly identify the object in question. Make a sketch.

2) Identify all external forces acting on the object. Locate

clearly where these forces are being applied.

3) Choose an appropriate coordinate system. Resolve all

forces into x and y components.

4) Set up equations for SF_x = 0 and SF_y = 0.

5) Select an appropriate axis of rotation (pivot point.) This is the

coolest thing – ANY POINT IN THE UNIVERSE CAN BE

YOUR SELECTED PIVOT POINT! BUT, pick a point that

is “obvious”, or, just as likely, a point that makes the

problem easier!

- “Obvious” might be the fulcrum of a lever. Or the

axle of a wheel. Or the hinge on a door.

- “Easier” means that you pick the pivot point at the

point of application of an unknown force, thus

eliminating it from the torque problem. (Since l=0,

t=0.)

6) Identify all forces and their lever arms for chosen pivot point.

7) Avoid positive and negative signs by setting up as

St_c.c.w. = St_c.w.

8) Solve all equations!

Note, in that step 7, you are summing torques. Not forces. Not lever arms. But torques.

Center of Mass/Gravity

What matters to us, solving torque problems, is the following:

- The center of mass of an object is the point at which all

mass of the object can be considered to be concentrated.

It’s not really all at that point, but it ACTS like it!

- The center of gravity is the point at which all weight can

be considered to be concentrated. When computing torque

due to gravitational force, the object acts as if all the

gravitational force on the object acts at only that point.

- For all common examples we will use, the center of mass

and center of gravity are the same point.

- For any uniform object, with uniform mass distribution,

the center of gravity is the geometric center of the object.

- It does not matter if some parts of the object are on one

side of the chosen pivot point while other parts of the

object are on the other side of the pivot point - the

gravitational force acts only at the center of gravity when

figuring its contribution to the total torque on object.

- For “compound” objects, one can find the centers of

gravity of each subsection and proceed.

Inertia versus Moment of Inertia

Inertia is defined as that property of matter that resists changes in linear motion (acceleration.) It is dependent on (and proportional) to the mass of the object.

Moment of Inertia is defined as that property of matter that resists changes in rotational motion (angular acceleration.) It is dependent not only on the mass of the object, but how that mass is distributed about the rotational axis.

“I” is the variable used for moment of inertia.

For a particle of mass m, at distance R from the rotational axis:

I = m R²

For a distribution of “n” particles, it is found by

I = S m_n R_n²

For a continuous distribution of mass, it is found by

I = ò R² dm

Note, I is proportional to mass but is also proportional to the distance squared. The distribution (R) of the mass can thus be a much bigger factor than the mass itself.

Moment of Inertia (cont)

Note that the units for moment of inertia will be kg m² (SI).

For many real objects, I can get to be complicated.

- Use calculus to compute I for extended objects.

- Tables of equations for I for common shapes.

- For objects that are combinations of simpler shapes,

the total moment of inertia is just the sum of the moments

of inertia of the component parts!

Newton’s 2^nd Law for Rotation

Newton’s Second Law of Motion (F_net = ma) must be true even for something that is rotating. By defining the F in terms of some applied torque, and the acceleration in terms of resulting angular acceleration, and doing the appropriate substitutions, we come up with Newton’s 2^nd Law for Rotating Things:

t_net = I a

where t_net is the vector sum of all acting external torques (in N m), I is the moment of inertia of the object (in kg m²), and a is the resulting angular acceleration of the object (in rad/s²).

Note that our table of analogs between the linear world and the rotational world is growing:

Concept Linear Rotational

displacement x q

velocity v w

acceleration a a

“force” F t

inertia m I

2^nd Law SF = ma S t = Ia

work F s t q

kinetic energy ½ m v² ½ I w²

momentum p = m v L = I w

Kinetic Energy

Conservation of Energy

Angular Momentum

Conservation of Angular Momentum

Rotational KE = ½ I w²

Conservation of Energy concept is the same as before but now the total energy, E, picture is enlarged to include rotational effects!

E = translational KE + rotational KE + gravitational PE

E = ½ m v² + ½ I w² + mgh

Angular Momentum L = I w

Conservation of Angular Momentum – If the net average external torque is zero, the total angular momentum of the system is constant (is conserved.)

Σ L_before = Σ L_after

Dynamics Summary

Concept Linear Rotational

displacement x q

velocity v w

acceleration a a

force F t (= r X F )

inertia m I

2^nd Law SF = ma S t = Ia

work F s t q

kinetic energy ½ m v² ½ I w²

momentum p = m v L = I w