1206 - Concepts of Physics

Wednesday, October 7th

Rotational dynamics

• Dynamics deals with the action of forces

• We will discuss torques and Newton’s second law for rotational motion

• We will introduce rotational Work and Energy

• We will discuss angular momentum

TorqueThe mass of most rigid objects, such as a propeller of wheel, is spread out and not concentrated in

one point. These objects can move in a number of ways. Lets look at gymnastics as an example

This first picture shows translation. The body doesn’t change during the movement. In pure translation there is not rotation of any line in the body. It can still occur on a

curved line, like here - a person jumping.

This second picture shows a combination of translation and rotation. The gymnast moves forward while rotating.

We know that for linear motions the net force causes an object to accelerate. But what causes rotational motion to change? We will see that the external net torque causes angular acceleration.

To create a picture for torque, imagine a door and somebody pushing to open it. The larger the force used to push, the faster the door open, the greater the torque.

Door example - top view

The hinges of the door appear as black dot ( ) and defines the axis of rotation. The line of action and lever arm l are illustrated for a force applied to the door perpendicularly and in an angle. The

last one has no lever arm, sine the line of action passes through the axis of rotation.

F

lever arm l

line of action

axis of rotationF

line of action

line of action

Definition of TorqueTorque = (Magnitude of the force) x (Lever arm)

τ = Fl

The torque is positive when the force tends to produce a counterclockwise rotation about the axis, and negative when the force tends to produce a clockwise rotation.

The SI unit of torque is Nm

Example in numbersLet’s assume the magnitude of the force we apply to the door is 55 N. However the lever arms are

different in the three parts of the drawing: (a) l = 0.80 m, (b) l = 0.60 m and (c) l = 0 m. Find the magnitude of the torque in each case.

In each case the lever arm is the perpendicular distance between the axis of rotation and the line of action of the force. In part (a) this perpendicular distance is equal to the width of the door. In parts (b) and (c), however, the lever arm is less than the width. Since the lever arm is different in

each case, the torque is different, even though the magnitude of the applied force is the same.

(a) τ = Fl = (55N)(0.80 m) = 44 Nm(b) τ = Fl = (55N)(0.60 m) = 33 Nm

(c) τ = Fl = (55N)(0 m) = 0 Nm

Example:

The Achilles Tendon: The achilles tendon is attached to the heel at point P. The tendon exerts a force of magnitude F = 720 N. Determine the torque (magnitude and direction) of this force about the ankle joint, which is located 3.6 x 10-2 m away from point P.

Pankle joint

F

55°

To calculate the torque we need to force (given) and the lever arm. The lever arm not given, we have to calculate it. The lever arm is the perpendicular distance between the axis of rotation at the ankle joint and the the line of action of the force F (dashed red line).

l = (3.6 x 10-2 m) cos55°

τ = Fl = (720 N) (3.6 x 10 -2 m) cos55° = 15 Nm

To produce a clockwise rotation about the ankle joint, the torque needs to be negative.

Rigid Objects in EquilibriumIf a rigid body is in equilibrium, neither its linear motion nor its rotational motion changes. In this situation certain equations apply. For example a rigid body who’s linear motion doesn’t change has

no acceleration a. Therefore the net force ∑F must be zero. (Newton’s second law: F = ma)For two-dimensional motions, we have learned, that the horizontal and vertical components can be treated separately, therefore the x and y components of the net force have to sum up to zero.

∑Fx = 0 and ∑Fy = 0.

Also the rotational motion doesn’t change in equilibrium. this means that the net external torque action of the object must be zero. ∑τ = 0.

Equilibrium of a rigid body: A rigid body is in equilibrium fit has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the

sum of the externally applied torques is zero:

∑Fx = 0 and ∑Fy = 0 and ∑τ = 0

ExampleA woman whose weight is 530 N is poised at the right end of a diving board with a length of 3.90 m. The board has negligible weight and is bolted down at the left end, while being

supported 1.40 m away. Find the forces F1 and F2 exerted on the board.

l2 = 1.40 m

lW = 3.90 m

F2

F1W

Shown here is the free body diagram for the situation described above.

The board is in equilibrium, therefore the sum of the vertical forces must be zero.

∑Fy = -F1 + F2 - W = 0

With the same argument (the board is in equilibrium), we can say that the sum of the torques must be zero. ∑τ = F2l2 - WlW = 0

Solving the equation for F2 yields: F2 = WlW/l2 = (530 N)(3.90 m)/(1.40 m) = 1480 N

Solving first equation for F1 and using result for F2 we obtain: F1 = F2 - W = 1480 N - 530 N = 950 N

Fighting a Fire - exampleAn 8.00 m ladder of weight WL = 355 N leans against a smooth vertical wall. The term “smooth” means that the wall can exert only a normal force directed perpendicular to the wall and cannot

exert a frictional force parallel to it. A firefighter, whose weight is WF = 875 N stands 6.30 m from the bottom of the ladder. Assume that the ladder’s weight acts at the ladder’s center. Find the forces

that the wall and the ground exert on the ladder.

Gy

Gx

WL

WF

P

2.30 m

4.00 m

50°

⎠⎞

x

y

+τ

50°

WL

WF

P

lL

lWAxis

lP

The figure shows the free body diagram for a firefighteron a ladder. The following forces act on the ladder:

1.) The ladders weight WL

2.) The firefighters weight WF

3.) From wall, perpendicular P 4.) From ground Gy and Gx

The ground is not smooth, so Gx is produced by friction andprevents the ladder from slipping. Gy is the normal force applied

to the ladder by the ground. The ladder is in equilibrium, so the sum of these forces and the sum of the torques produced by them

must be zero. Since the net force acting on the ladder is zero, we have:

(a) ∑Fx = Gx - P = 0(b) ∑Fy = Gy - WL - WF = 0

or Gy = WL + WF = 355 N + 875 N = 1230 N

Equation (a) contains two unknowns, so it cannot be solved directly. However, another equation can be obtained from the fact

that the net torque acting is zero as well (equilibrium).

The lever arm for both components of the force exerted by the ground are zero, so we won’t have to take them into account. The other lever

arms are indicated in the drawing on the left.

(a) ∑Fx = Gx + P = 0

50°

WL

WF

P

lL

lWAxis

lP

Force Lever Arm Torque

WL = 355 N lL = (4.00 m) cos 50.0° -WLlLWF = 875 N lF = (6.30 m) cos 50.0° -WFlF

P = ? lP = (8.00 m) sin 50.0° +PlP

Setting the sum equals zero gives: ∑τ = -WLlL -WFlF +PlP = 0

Solving this equation for P gives: P = (WLlL + WFlF)/lP = 727 N

Now we can go back into (a): P = Gx = 727 N

Center of GravityOften it is important to know the torque produced by the weight of an extended body. This is

similar to using the center of mass in collision of extended objects - like cars. In the previous ladder example we have somewhat already used this principle. The weight was considered to act at a

definite point for the purpose of calculating the torque. This point is called the center of gravity cg.

Definition of the center of gravity: The center of gravity of a rigid body is the point at which its weight can be considered to act

when the torque due to the weight is being calculated.

When an object is symmetrical in its shape and its weight is distributed uniformly, the center of gravity lies at its geometrical center. For example take a plank made out of wood that is 10 m

long and 50 cm wide. The center of gravity is located 5 m in and 25 cm in. The lever arm for the weight W is L/2, and the magnitude of the torque is τ = W(L/2).

When only one dimension matters, we can write (similar to the center of mass):

xcg = W1x1 + W2x2 + ...

W1 + W2 + ...

Example: Loading a Cargo Plane

Shown is a cargo plane with its front landing up in the air. This accident occurred because the plane was overloaded toward the rear. How did a shift in the center of gravity of

the loaded plane cause the accident?

FN1FN2

cg cg

FN2

WW

assume front lifted up

The left show’s a correctly loaded plane - with the center of gravity between the front and rear landing gears. The weight acts downward at the center of gravity, and the normal forces act upward

at the front and at the back landing gear. The counterclockwise torque due to FN1 balances the clockwise torque due to W and the plane stays in equilibrium.

The right picture shows the plane with too much weight loaded towards the rear, just after the plane has begun to rotate counterclockwise. Because of the overloading, the center of gravity

has shifted behind the rear landing gear. The torque due to W is now counterclockwise and not balanced.

Newton’s second Law for rotational motion about fixed axisLets begin by assuming a circular motion - imagine the model plane on a guide wire. The plane’s

engine produces a net external tangential force FT that gives the plane a tangential acceleration aT. Using Newton’s law, we can write FT = m aT

The torque τ = FT r, where the radius r of the circular path is also the lever arm. As a result, the torque is τ = m aT r. But the tangential acceleration is related to the angular acceleration α

according to aT = rα, where α must be expressed in rad/s2. With this substitution for aT, the torque becomes

τ = (mr2)α

This is the form of Newton’s law we were looking for. The net external torque τ is directly proportional to the angular acceleration α. This can be applied to any rigid body rotating.

The expression mr2 is called the moment of inertia of the particle i. τ = i α

Rotational analog of Newton’s second law for a rigid body rotating about a fixed axis.

Net external torque = moment of inertia * angular acceleration

Στ = i α

α must be expressed in rad/s2

ExampleThe moment of inertia depends on where the axis is. Two particles each have a mass m and are

fixed to the ends of a thin rigid rod, whose mass can be ignored. The length of the rod is L. Find the moment of inertia when this object rotated relative to an axis that is perpendicular to the rod at

one end (a) and the center (b).

m1m2

axis

(a)

axisWhen the axis of rotation changes, the distance r between the axis and each particle changes. In determining the moment of inertia, using

i = Σmr2, we must be careful to use the distances that apply for each axis.

(b)

(a) Particle 1 lies on the axis, as part (a) indicates, and has a zero radial distance r1 = 0. In contrast, particle 2 moves on a circle whose radius is r2 = L. Noting that m1 = m2 = m, we find that the

moment of inertia is: i = Σmr2 = m1r12 + m2r22 = mL2

(b) Particle 1 lies no longer on the axis, as part (b) shows, and has radial distance r1 = L/2. Particle 2 moves on a circle whose radius is r2 = L/2 = r1. Therefore,

i = Σmr2 = m1r12 + m2r22 = 2m(L/2)2 = 1/2mL2

The results for (a) and (b) are different, because the axis of rotation is different.

Rotational Work and EnergyWork and Energy are among the most fundamental and useful concepts in physics. We have discussed

them for linear motion (translation) before. They are equally useful for rotational motion, provided they are expressed in terms of angular variables. The work done by a constant force that points in the same direction as the displacement is W = Fd, where F and d are the magnitudes of the force and the

displacement, respectively. Rotational Work WR done by a constant torque τ in turning an object through an angle Θ is

WR = τ Θ (Θ must be expressed in radians, SI unit is Joule J)

Rotational kinetic energy KErot is the kinetic energy of the entire rotating body and is given:

KErot = Σ(1/2mr2ω2) = 1/2 i ω2

If a bicycle coasts down a hill, for instance, its tires are both translating and rotating. An object such as a rolling bicycle tire has both translational and rotational kinetic energies, so that the total

mechanical energy is

E = 1/2mv2 + 1/2iω2 + mghtotal mechanical energy = translational kinetic energy + rotational kinetic energy + gravitational potential energy

Rolling cylinders

A thin-walled hollow cylinder (mass = mh, radius = rh) and a solid cylinder (mass = ms, radius = rs) start from rest at the top of an incline. Both cylinders start at the same vertical height h0. All

heights are measured relative to an arbitrarily chosen zero level that passes through the center of mass of a cylinder when it is at the bottom of the incline. Ignoring energy losses due to retarding forces, determine which cylinder has the greatest translational speed upon reaching the bottom.

h0

zero level hf = 0

Only the conservative force of gravity does work on the cylinders, so the total mechanical energy is conserved as they

roll down. Which one do you think will be faster and why?

Rolling cylinders

h0

zero level hf = 0

Only the conservative force of gravity does work on the cylinders, so the total mechanical energy is conserved as they

roll down. Which one do you think will be faster and why?

We expect the solid cylinder to have the greater translational speed, because more of its mass is located near the rotational axis and, thus, possesses less rotational kinetic energy.

The total mechanical energy Ef at the bottom (hf = 0 m) is the same as the total mechanical energy E0 at the top (h0, v0=0 m/s, ω0=0 rad/s):

1/2mvf2 + 1/2iωf2 + mghf = 1/2mv02 + 1/2iω02 + mgh0

1/2mvf2 + 1/2iωf2 = mgh0

Each cylinder rolls without slipping, the final rotational speed ωf and the final translational speed vf

of its center of mass are related according to ωf = vf/r, where r is the radius of the cylinder:1/2mvf2 + 1/2ivf2/r2 = mgh0

vf2 (m + i/r2) = 2mgh0 --> vf = sqrt[(2mgh0)/(m + i/r2)]

hollow cylinder: (m = mh, r = rh, i = mrh2) vf = sqrt(gh0) solid cylinder: (m = ms, r = rs, i = 1/2mrs2) vf = sqrt(1/3*4gh0) = 1.15 sqrt(gh0)

vf

The solid cylinder arrives at the bottom first.

Angular momentum

The linear momentum p of an object is defined as the product of its mass m and linear velocity v. For rotational motion the analogous concept is called the angular momentum L. The

mathematical form of angular momentum is analogous to that of linear momentum, with the mass m and the linear velocity v being replaced with their rotational counterparts, the moment

of inertia i and the angular velocity ω.

L = i ω

When the sum of the average external torques is zero, then the final and initial angular momenta are the same: Lf = L0, which is the principle of angular momentum.

Analogies between rotational and translational concepts

Physical concept Rotational TranslationalDisplacement Θ d (or x or s)Velocity ω vAcceleration α aCause of acceleration Torque τ Force FInertia Moment of inertia i Mass mNewton’s second law Στ = iα ΣF = maWork τΘ FdKinetic energy 1/2iω2 1/2mv2

Momentum L=iω p = mv

Example

An ice skater is spinning with both arms and a leg outstretched (a) and then she pulls her arms and leg

inward (b). Does her angular velocity increase or decrease?

Example

An ice skater is spinning with both arms and a leg outstretched (a) and then she pulls her arms and leg

inward (b). Does her angular velocity increase or decrease?

Her spinning motion changes dramatically - use the principle of angular momentum and explain why it

changes.

Choosing the skater as the system, we can apply the conservation principle provided that the net external torque produced by air resistance and by friction between the skates and the ice is negligibly small. WE assume that it is. Then the skater would spin forever at the same angular

velocity, since her angular momentum is conserved. The inward movement of her arms and leg involves internal, not external, torques and, therefore, does not change her angular momentum.

But angular momentum is the product of the moment of inertia i and angular velocity ω. By moving the mass of her arms an leg inward, the skater decreases the distance r of the mass and

consquently her moment of inertia. If the product has to stay the same, her velocity must increase. Therefor she spins with a larger angular velocity.