review: uniform circular motion in part 2 we looked at uniform circular motion: r = constant (def....

Review: Uniform Circular Motion

In Part 2 we looked at uniform circular motion:

r = constant (def. of circular motion)

= t ( changes with time at uniform rate where = constant)

vr = 0 (radius doesn’t change)

v = r (where t)

ar = 2r (v changes direction toward center)

a = 0 (angular speed doesn’t change)

Angular Acceleration

Here we generalize circular motion to include the case where the angular speed can change. We define angular acceleration as: t .

[Recall arclength = s, and =s/r, so s=r, v=r]

Since a = v/t = r)/t = r(t) = r.

We still have all the equations we had before, except a is no longer zero but is instead r.

NON-uniform Circular MotionRectangular viewpoint

Circular motion is defined by: r = constant.

We still have d/dt = , but is no longer held constant. This means that d/dt = rather than zero.

Converting polar to rectangular, we have:

x = r cos() = r cos[(t)]

y = r sin() = r sin[(t)] .

Non-uniform Circular MotionRectangular viewpoint

x = r cos[(t)]

y = r sin[(t)]

Now we use: vx = dx/dt and vy = dy/dt:

vx = dx(t)/dt = dxd * d/dt =

r (- sin() ) * = -r sin()

vy = dy(t)/dt = dyd * d/dt =

r (+ cos() ) * = r cos()

Non-uniform Circular MotionRectangular viewpoint

x = r cos()

y = r sin()

vx = -r sin() This v is the same as for

vy = r cos() uniform circular motion

Now we use: ax = dvx/dt and ay = dvy/dt. Since is

not constant, we need to use the product rule.

ax = - r sin() + -r cos()

ay = + r cos() + -r sin() .

Acceleration

vx = -r sin() vy = r cos()Now we use: ax = dvx/dt and ay = dvy/dt. Since is

not constant, we need to use the product rule:

ax = dvx/dt = d[-r sin() ]/dt = r d[-/dt sin() + -r d[sin()]/dt = - r sin() + -r cos() .

In a similar way, we get:

ay = + r cos() + -r sin() .

Non-uniform Circular MotionBack to Polar

To convert back to polar for position, we use the inverse transformation equations:

x = r cos() y = r sin()

r = [x2 + y2]1/2 = [r2 cos2() + r2 sin2()]1/2

= r[cos2() + sin2()]1/2 = r

= inv tan[y/x] = inv tan[ r sin() / r cos()]

= inv tan[tan()] = .

Non-uniform Circular MotionBack to Polar

To convert back to polar for velocity, we use the inverse transformation equations:

vx = -r sin() vy = r cos()

v = [vx2 + vy

2]1/2 = [2r2 sin2() + 2r2 cos2()]1/2

= r[sin2() + cos2()]1/2 = r

v = inv tan[vy/vx] = inv tan[r cos() / -r sin()]

Note that cos() = sin(+90o) and –sin() = cos(+90o) so

v = . Note that the direction of the velocity, v, is perpendicular to the direction of the position (the radius), which means the velocity is tangent to the circle.

Non-uniform Circular MotionBack to Polar

v =

Note that the direction of the velocity, v, is perpendicular to the direction of the position (the radius), which means the velocity is tangent to the circle.

Non-uniform Circular MotionBack to Polar

To convert acceleration:

ax = - r sin() + -r cos()

ay = + r cos() + -r sin() .

a = [ax2+ay

2]1/2 = [2r2 sin2() + 22r2sin()cos()+ 4r2 cos2() + 2r2 cos2() - 22r2sin()cos() +

4r2 sin2()]1/2 = [(r)2 + (r)2]1/2 = [at2+ar

2]1/2.

We can identify (2r) as the radial acceleration - due to the turning of the velocity, and we now identify (r) as the tangential acceleration - due to the changing speed of the velocity.

Non-uniform Circular MotionBack to Polar

ax = - r sin() + -r cos()

ay = + r cos() + -r sin() .

a = [at2+ar

2]1/2 where at = r and ar = 2r.

If we look at just the r terms, then the direction of these terms is: at = inv tan[ cos() / -sin() ] = +/- 90o (just like for v which is tangent to the circle),

And if we look just at the 2r terms, then the direction of these terms is: ar = inv tan[ -sin() / -cos() ] +/- 180o (which is towards the center).

Constant Angular Acceleration

In the special case of constant angular acceleration ( = constant), we have equations analogous to those we had for constant (regular) acceleration:

d/dt = = constant, becomes o d = t=0t dt, or

t like vx = vxo + axt . d/dt =

becomes o d = t=0t dt = t=0t (ot)dt , or

t + (1/2)t2 like x =

xo+vxot+(1/2)axt2.

Constant Angular Acceleration

If a car rolls on wheels that do not slip, we can relate the motion of the car to the spinning of the wheels:

s = distance of car = arclength of wheels,

v = speed of car = v of wheels,

a = acceleration of car = a of wheels;

= angle of wheel, = angular speed of wheel, and

= angular acceleration of wheel.

Constant Angular AccelerationFor circular motion:

s = r, r = constant (circular)

v = rv vr=0

a = r = a ar = 2r (due to turning)

For constant acceleration:

s = so + vot + (1/2)at2 v = vo + at

t + (1/2)t2 t .

This gives many relations so that, by knowing just a few things, we can solve for many other things about the motion.

ExampleConstant Angular Acceleration

Consider a car that accelerates from rest to a speed of 20 m/s (45 mph) in 10 seconds on wheels (tires) of outside diameter 66 cm. Assume that the tires roll nicely and do not slip on the road. Through what angle (θ) do the tires rotate during the 10 seconds? What is the final angular speed (ω) at the 10 second mark? What is the angular acceleration (α) of the wheels during the 10 seconds?

Example

We can use the equations for constant angular acceleration, or we can use the relations between distance and angle.

xo= 0 m vo = 0 m/s t = 10 seconds x = ? v = 20 m/s

θo = 0 rad ωo = 0 rad/sec θ = ? ω = ?

α = ?

d = 66 cm, r = d/2 = 33 cm

a = ?

t = 0

ExampleSince the car is rolling (and not slipping), we

recognize that x = s = rθ, and that

v = v= r

Using the equations for constant angular acceleration along with rv :

t + (1/2)t2 t

v = 20 m/s, t = 10 sec., 0rad, r = .33 m,

vo = 0 m/s, so = 0 rad/sec.

Example(1) rv becomes: .33 m * 20 m/s

t + (1/2)t2 becomes = 0 + 0 + (1/2)**(10 sec)2

t becomes: = 0 + *(10 sec)

From Eq. 1: = (20 m/s) / .33 m = 60.6 rad/sFrom Eq. 3: = /t = (60.6 rad/s) / 10 sec = 6.06 rad/s2.From Eq. 2: = (1/2)**(10 sec)2

= (1/2)*rad/s2)*(10 sec)2 = 303 rad * (1 cycle/2 rad) = 48.2 revolutions

Example – another wayInstead of using the constant angular

acceleration formulas, we could solve the regular motion and use the relations between regular quantities and angular quantities:

s = r, r = constant (circular)

v = rv vr = 0

a = r = a ar = 2r (due to turning)

Example another ways = r, r = constant (circular)

v = rv vr = 0

a = r = a ar = 2r (due to turning)

s = so + vot + (1/2)at2 v = vo + at20 m/s = 0 + a*(10 sec), so a = (20 m/s) / (10 s) = 2 m/s2 , soa = r, or = a/r = (2 m/s2) / (.33 m) = 6.06 rad/s2 .s = 0 + 0 + (1/2)*a*(10 sec)2 = (1/2)*(6.06 rad/s2)*(10 sec)2 = 303 radians.v = r, or = v/r = (20 m/s) / (.33 m) = 60.6 rad/s.

Rotational Kinematics

The computer homework program, Rotational Kinematics, Volume 2 #7, contains explanations and problems involving Rotational Kinematics using the equations in the previous slide.

Rotational Force (Torque)

Forces cause change in the motion, but in Part 2 we only considered motion that changes the position of the object.

What about changing the “spin” or rotation of an object?

To get a nice introduction to the idea of torque, see the computer homework program on Introduction to torque (Vol.2 #5), and continue with the next slide.

Torque

There are two important quantities in torque: Force and where you apply the force (called radius).

= r x F = r F sin(rF)

To get a large torque, we need to use a large radius, a large force, and apply the force perpendicular to the radius!

Statics and Torque

Just as F = 0 when the object is static, so also = 0 when the object is not spinning (or spinning at a constant rate).

In static cases, there is no obvious center to measure the radius from, so we are free to choose any point. However, some points may be simpler to use than others.

Your elbow

Let’s consider as an example of torque how your muscles, bones and joints work.

Consider holding up a ball of weight 5 lb.

How does this work? (We will choose to measure distances from the elbow.)

First we draw a diagram: biceps

triceps weight

= elbow

rwrb

Your elbow

In terms of forces and distances, the diagram looks like this:

Estimate the distance

from your elbow joint Fc Fb

to your biceps connect

point, rb; and to

your hand, rw. rc=0 rb rwW

Your elbow

If the ball weights 5 lb, how much force does your biceps pull up with? How much force of contact does your upper arm push down with on your lower arm at the elbow?

What is the basic principle to use? Statics:

F = 0 and .

Your elbow

From F = 0 we have:

-Fc + Fb - W = 0

And from and measuring from the elbow gives: Fc*rc + Fb*rb - W*rw = 0 .

We have two equations and we have two unknowns (Fc and Fb).

Your elbow

We can use the torque equation first, since rc=0 eliminates one of the unknowns, Fc.

Fc*rc + Fb*rb - W*rw = 0 or Fb = W*rw/rb .

Then we can use the force equation to find Fc :-Fc + Fb - W = 0, or Fc = Fb - W.

Your elbow

By putting in reasonable values for rb and rw, you can see that the biceps have to exert a large force to hold up a relatively light weight!

What advantage does this give? Note how far the biceps have to contract in order to move the weight! This is the advantage of the elbow set-up!

In practice, we use clubs and rackets to make this difference even greater!

Car going around a turn:Sliding off the road or flipping over

For a car going around a turn, if the car goes too fast the car will either slide off the road or flip over. What determines which will occur, and how fast can the car go around the turn without either happening?

Since we now introduce “flipping over”, we have to consider rotational motion, and hence consider torque. This means we need to consider not only what the forces are, but where they act.

Sliding or Flipping

We’ll consider a car making a right turn on an unbanked (horizontal) turn.

We have the weight of the car (W=mg), which acts at the center of gravity.

We have the contact force, which acts on each of the left and right wheels at the road: FcL and FcR.

We have the friction force, which also acts on each of the left and right wheels at the road: FfL and FfR, where Ff Fc

We also have an acceleration to the right: ac = v2/r .

ac = v2/r

Sliding or Flipping

h

The weight, W=mg, acts at the center of gravity, which is at a height, h, above the road.

The contact forces, Fc act at the tires which are on the road.

The friction forces, Ff, alsoact at the road.

The tires are a distance, w, apart.w

Sliding or Flipping

h

Fx = FfL + FfR = m v2/r

Fy = FcL + FcR - mg = 0

Let’s choose to consider the center of gravity as the point for rotations. The distances for the torques need to be distances that are to the direction of the forces. We will also choose clockwise as positive for torques.

w

ac = v2/r

w

Sliding or Flipping

h

Fx = FfL + FfR = m v2/r

Fy = FcL + FcR - mg = 0

= -FfL*h - FfR*h + FcL*(w/2) - FcR*(w/2) + mg*0 = 0.

Since the FcL is the only force causing a + torque, it must increase and FcR must decrease

if the = 0. This can happenuntil FcR = 0 and FcL = mg. When this

happens, FfR= 0 .

ac = v2/r

Sliding or FlippingFx = FfL + 0 = m v2/r, or FfLm v2/r

Fy = FcL + 0 - mg = 0, or FcL = mg

= -FfL*h - 0*h + FcL*(w/2) - 0*(w/2) + mg*0 = 0, or FfL*h = FcL*(w/2) .

If friction is maximum (limit for sliding) we have: FfLm v2/r with Ff = Fc = mg, so mg = m v2/r, or vmax = [gr]1/2 .

If flipping is to be prevented we have FfL = mv2/r, FcL=mg, or mv2/r * h = mg*(w/2), or vmax = [grw/2h]1/2 .

Sliding or Flipping

vmax = [gr]1/2 applies when sliding is a concern;

vmax = [grw/2h]1/2 applies when flipping over is a concern.

Which maximum speed is lower - that tells us the true maximum speed to continue around the turn:

If < w/2h, then sliding is the main concern;

if > w/2h, then flipping is the main concern.

For big trucks, h is large, and unless the street is very slick, flipping over will be bigger; for sports cars, h is small and sliding is more of a concern.

Statics and Torque

The computer homework program, Statics and Torque, Volume 2 #6, contain explanations and problems dealing with this topic.

Torque and Rotations

Forces (when not balanced) cause changes in motion. Torques (when not balanced) cause changes in rotational motion. Forces are related to acceleration by Newton’s Second Law: F = ma. How are and related?

Rotational Dynamics

Consider: F = ma and = r F sin(rF) .

If we multiply both sides of F = ma by

r sin(rF), we get: = m a r sin(rF) ;

but a sin(ra) = ar, so = mr2 .

If the mass is distributed at various radii, then the mr2 becomes miri

2.

We define the moment of inertia be be

I = miri2 so that we have = I .

Moment of Inertia

Note that the moment of inertia relates the torque to the angular acceleration just like the mass relates the force to the regular acceleration.

Note that the moment of inertia depends on the mass of the object and also on the shape of the object.

Moment of InertiaObjects that have most of their mass in the center

have smaller moments of inertia than objects that have most of their mass on the outside.

Using the calculus, we can replace the summation of small mass pieces with an integral: I = miri

2 becomes I = r2 dm.

Each small dm has its own radius, or distance from the axis of rotation, so r is really a function of which dm we are using: r(m).

Moment of Inertia

I = r2 dm

It is awkward to express r(m). We can, however, use density: = mass/Volume = dm/dV to convert dm into dV. Then the integral is really a Volume integral (triple integral).

I = r2 dVUsing this, we can derive nice formulas for the moment

of inertias of certain shapes: Iring = MR2

Isphere = (2/5)MR2; Icylinder = (1/2)MR2 .

Sphere versus CylinderThe blue object in the figures is theside view of a cylinder with thesame radius and volume (mass)as the orange object (a sphere). [The top view would like the same: circles.]

Note that most of the mass is in the same place, but the topand bottom of the sphere mustbe moved away to provide thecorners for the cylinder. Hence, the Moment ofInertia of the cylinder is slightly larger than that of

the sphere: Isphere = (2/5)MR2 ; Icylinder = (1/2)MR2 .

axis

R

Rotational Energy

Now that we have the equivalent of Newton’s Second Law for rotations: = I , let’s see if we can get a rotational energy relation.

Recall that Work = F ds. If we exert a force to spin an object, the ds becomes a small arc length that is perpendicular to the radius. Recall the definition of an angle in radians: = s/r, so that ds = r d. But F times r is torque, so we get: Work = d .

Rotational Power

Work = d

The definition of power is: P = dWork/dt, and so for rotations we have:

P = dW/dt = d[ ddt = This formula for rotational power is similar to

that for regular power: P = F v

P = .

Rotational Kinetic Energy

If something is spinning, it can still not be moving so that it’s regular kinetic energy is zero. However, parts of the object are moving, so there is kinetic energy associated with its spinning. We use three relations we have already developed:

KE = ½mv2, v = r, and I = miri2

KEtotal = ½mivi2 = ½mi(ri)2 = ½2 miri

2 = ½I2 .

This energy is used in the Conservation of Energy relation just as any other energy term is used.

Angular MomentumAnother important tool in solving problems is

Conservation of Momentum. Is there a similar tool for rotations?

We already have: p = mv and F = dp/dt.

If we multiply both sides of Newton’s Second law by the radius we have (with v = r):

= r F = r dp/dt = d(r p)/dt = dL/dt(where we have used the fact that dr/dt p = v mv = 0)

where we have defined L = r p, where L is called the Angular Momentum.

Conservation of Angular Momentum

= dL/dt

Just like F = dp/dt leads to Conservation of momentum if no external forces are present, so = dL/dt leads to Conservation of Angular momentum if no external torques are present.

Note: p = mv, and L = r p =

r mv = r m v = r m r = mr2 = I .

Angular Momentum

In class we will observe several demonstrations about Angular Momentum and talk about its uses in controlling rockets and spinning athletes (divers and gymnasts in particular). We will also show how angular momentum relates to riding bicycles, and how this relates to the spinning earth!

Review of Rotational Equations

Basically we replace F with , m with I, v with , a with andp with L (where L is the angular momentum):

F = ma = IWork = = F ds Work = d

Power = F v Power =

KE = (1/2)mv2 KErotation = (1/2)I2

p = mv L = I F = p/t = L/t .

Rotational Motion

The computer homework program, Volume 2 #8 on Rotational Dynamics gives an overview and several problems involving rotational motion.

Example: rolling ball

How fast will a ball be going at the base of a ramp if it is released at the top of the ramp, where the length of the ramp is 2 meters and the high end is 50 cm above the floor (base of the ramp)? We assume the ball rolls without slipping. Does the mass or radius of the ball affect the speed? If a cylinder of the same mass and radius were also released, would the ball or cylinder win the race down the ramp?

Rolling ball

Ball: vi = 0, i=0

mass = m

radiius = r

hi = 50 cm s = 2 meters

vf = ? f = ?

Rolling ball

We could use Newton’s Laws of Motion:

F = ma, and = I along with a=r and the equations for constant acceleration, or

we could use Conservation of Energy

(KEregular + KErotational + PEgravity)initial =

(KEregular + KErotational + PEgravity)initial + Elost

along with v=r.

Rolling ball

Since we’re not interested in direction or time, we’ll use Conservation of Energy.

We know that for a ball (sphere), I = (2/5)mr2.

We know that KEregular = (1/2)mv2 .

We know that KErotational = (1/2)I2.

We know that PEgravity = mgh .

We know that without slipping, even though friction causes the ball to roll, there is no Elost because the ball does not slide.

Rolling ball

Conservation of Energy

(KEregular + KErotational + PEgravity)initial =

(KEregular + KErotational + PEgravity)initial + Elost

0 + 0 + mgh = (1/2)mv2 + (1/2)I2 + 0 + 0.

Substituting I=(2/5)mr2 and =v/r gives:

mgh = (1/2)mv2 + (1/2)[(2/5)mr2][v2/r2] , or

mgh = (1/2)mv2 + (1/5)mv2 = (7/10)mv2 .

Rolling ball

mgh = (7/10)mv2 which becomes when solving for v: v = [(10/7)gh]1/2 = [(10/7)*9.8m/s2 * .50m] = 2.645 m/s.

Note: both the mass and the radius cancel out of the solution, and so the mass and radius don’t matter.

Rolling ball

Note: the angle of the ramp doesn’t make any difference either, since only the height and not the length is involved. The steeper the ramp, the faster the ball will reach its final speed, but the final speed will not be affected!

Sphere versus Cylinder

Would a cylinder or a sphere win a race down the ramp?

What difference does the shape make in our equation? The difference comes in the KE for rotations in the moment of inertia, I. For a sphere (ball), I = (2/5)mr2, while for a cylinder it would be I = (1/2)mr2. This ultimately changes the (10/7) factor into (4/3), the value from 2.65 m/s to 2.55 m/s.

Sphere versus Cylinder

We can understand the slightly lower speed for the cylinder by considering that the moment of inertia of the cylinder is larger. This will make the kinetic energy larger for the same speed. Hence there will be less kinetic energy for the regular speed if we have the same potential energy to begin with!