AP Physics Angular Motion

Lesson 1 – Rotational Motion Linear motion involves displacements, velocities and accelerations. We know that velocity is the time rate of change of displacement and is given by the equation

v =ΔdΔt

When objects move in a circular fashion, we instead speak of angular displacements.

Here, an object moves from θ1 to θ2 in a time t, and we define angular velocity:

ω =ΔθΔt

(rad/s)

Angular velocity is defined as the time rate of change of angular position.

Angular acceleration is defined in a similar way to linear acceleration – how velocity changes with time. We can define a new equation for this angular acceleration:

α =ΔωΔt

You can also draw a parallel between linear and angular quantities:

Example 1 A boy turns 45° in 2 seconds. What is his angular velocity?

AP Physics Angular Motion

Velocity Acceleration Make sure that you do not confuse accelerations associated with angular motion:

1) With uniform circular motion we have a =v2

r and it acts towards the centre.

2) When the angular velocity is changing we have α =ωt

.

Linking Angular Equations with Linear Equations There is a 1-to-1 correspondence between angular and linear equations of motion. This helps us to create equations of angular motion. The proof and derivation of the angular equations below is trivial and uses the fact that

d = rθ (arclength) v = rω a = rα

Example 2 At low speed, a fan blade is turning at 80 rad/s clockwise. The fan is turned up a notch to rotate at 125 rad/s. If the time to change the speed is 0.73 s, find the angular acceleration of the fan blades.

AP Physics Angular Motion

Here, we adhere to the standard definitions of our variables θ, ω and α: And as long as α is constant, we have:

Linear Angular

Our convention for angular direction is standard:

Example 3 A flywheel at rest undergoes an angular acceleration of 5 rad/s2 for 10 s. What is its final angular velocity?

AP Physics Angular Motion

Example 5 A disc, initially rotating at 100 rad/s is stopped by a force, acting for 25 s. Through what angle did the disc rotate during this time? What was the angular acceleration of the disc during this time?

Example 4 A disc initially rotating at 2 rpm accelerates at – 6 rad/s2 for 20 seconds. What is its final angular velocity?

AP Physics Angular Motion

Lesson 2 – Rotational Dynamics

Just as there is a correspondence between linear motion and angular motion,

there are also many similarities between forces that act linearly and those that

act to cause rotation.

Moment of Inertia

As it turns out, Newton’s laws also apply to rotating objects. The concept of

inertia however, becomes slight more involved. Mass, you will recall, is a

measure of an object’s inertia. If an object has more mass, it becomes much

more difficult to accelerate. For a rotating object though, not only is mass an

important consideration, but also its mass distribution about its axis of

revolution. Consider the following rotating rods:

The moment of inertia is the angular equivalent to mass. It depends on the

objects mass, but also the object’s rotation axis. Different objects, spinning along

different axes produce different moments of inertia. For a ring of mass, rotating

about a central axis:

I =mr 2

Here I is the moment of inertia in units of kg ⋅ m2, m is the mass of the ring in kg

and r is the radius of the ring in m.

The following chart provides many different moments of inertia, each for a given

object rotating about a specific axis.

AP Physics Angular Motion

Object Description Equation

Hoop about central axis I =mr 2

Hoop about any diameter I = 12mr 2

Hollow cylinder about central axis

I = 12m(r

12 + r

22)

Thin hollow cylinder I =mr 2

Solid cylinder or disk I = 12mr 2

Thin rod about axis perpendicular to length

I = 112ml 2

This rod about end I = 13ml 2

Solid sphere about any diameter

I = 25mr 2

Thin spherical shell about any diameter

I = 23mr 2

Thin sheet, axis parallel to one edge and passing through

centre of other edge I = 1

12ml 2

Thin sheet axis along edge I = 13ml 2

Thin sheet axis through centre I = 112m(a2 +b2)

AP Physics Angular Motion

Torque The rotational effect caused by a force is known as torque. You apply a torque when you open a bottle of pop, tighten a screw, or turn on a tap. There are two main influences on torque: applied force and radius. Consider the following diagram of a rod with a pivot at one end. The applied force can be at different angles, denoted by θ. Using right triangles, we can resolve the applied force. And we can find a formula for torque:

τ = r F sinθ This expression implies that we know the direction of rotation, and so more formally, torque is a vector which is perpendicular to the plane created by

r and

F . Mathematically, a vector perpendicular to both r and

F can be found using

the cross product.

τ =r ×F

We use the right hand rule to determine the direction of torque.

Example 1 To open a door, you provide a 45.0 N force at an angle of 5° from the perpendicular a distance of 0.6 m from the hinges. What torque is acting on the door?

AP Physics Angular Motion

Newton’s 2nd Law for Rotation We know Newton’s second law well. But using our new found expressions for angular motion we can derive this well known law in terms of rotations. The angular motion equivalent for Newton’s 2nd Law is

τ = Iα

where τ represents the applied torque. Just as a force causes an acceleration, torque causes an angular acceleration.

Example 1 With a force of 30N, a rider pushes down on the rim of her bicycle wheel, which has a diameter of 0.60 m. Find the applied torque and the mass of the wheel if the wheel experiences an angular acceleration of 26.5 rad/s2.

AP Physics Angular Motion

Angular Motion Problems

1. An airliner arrives at the terminal and its engines are shut off. The rotor

of one of its engines has an initial clockwise angular speed of 2000 rad/s.

The engine’s rotation slows with an angular acceleration of magnitude 80.0

rad/s2.

(a) Determine the angular speed of the rotor after 10.0 s. 1200 rad/s (b) How long does it take for the rotor to come to rest? 25.0 s

2. A rotating wheel requires 3.00 s to complete 37.0 revolutions. Its angular

speed at the end of the 3.00 s interval is 98.0 rad/s. What is the constant

angular acceleration of the wheel? 13.7 rad/s2

3. A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in

9.00 s. If the diameter of the tire is 58.0 cm, find (a) the number of

revolutions the tire makes during this motion, assuming that no slipping

occurs. 54.3 rev (b) What is the final angular speed of the tire in

revolutions/s? 12.1 rev/s

4. Find the net torque on the wheel shown

about the axle O if a = 10.0 cm and b =

25.0 cm. -3.55 N⋅m

5. A block of mass m1 = 2.00 kg and a block of mass m

2 = 6.00 kg are

connected by a massless string over a pulley in the shape of a disk having

a radius R = 0.250 m and mass M = 10.0 kg. These blocks are released

and allowed to move as shown. The coefficient for both blocks is 0.360

and the ramp is at an angle of 30.0°.

Determine (a) the

acceleration of the

blocks and (b) the

tension in the string

on both sides of the

pulley. 0.309 m/s2; 7.67 N, 9.22 N

b

a

O

10.0 N

9.00 N12.0 N

30.0°

m1

m2

4"3,,,JL,. Motion Prob(ervrS

St>l,,.tra,n:

dt=o<=

b) -b' (Ar=()

N2 = ,^:, -t- "'tQ = -2Ooo a2ooo - L

r_go

-2@@sSO rai?

a) t= los , 6>= ?

dz= dr 't- o.tuz ? -zoooryl-r 8o(r")

@z= *lzoo@

80t l+ lalrs 25osstDp

{zr tl,e nr{o. ta

rev9=&r<v^t = 3.O sdr=q8o@

s

c(' ?

7Ir vd = z72.t8 rad.

49= drt - *nt'272.r{8 = (rr)13) _ **(=),

g2

a|+"n tos, -the a"SuQar sgeed is\zoO ro.d f cloctwEe]

s

;. The ovrqu0ar qsgQevat ano€ {t\a ,^r-h".1 a . f 3.? rad

C clocktrrise] sa

a)Fina

{r=O.r2-= 22.O n fsl; Q.os

91 ^O.21n

= O = 3+ t.?S rod x ItevZlr rad

the d ista sr ce fie Car frave [s:

d = L0,*'r)t=i(zz+o)(r)

d=91 ^and tt> de t<^r6n16q aq6le,

d=.e r=.58'n,- O.29mL

5Ll-3 revoluffarrl '

b) r/ = raj22 n/3o.L1m

(^) = 7 5.86 ,",.d xs

lrevF"od

T-o ftnd

= lo.o N= 9.oo t.l

= lz.o u

Pil:J'M = l0.o rjR.= o.25 nn

Iatru=IuR>= tLro)(o.2s)2

f =o,3125 {1,,r2

TL

Zt= TxTr(o,2D) -T, (o.zs) ,I, - T, = 1.25 "t

Yn.+ = I t a2 + T.J(o.25)(ro.o)srn?o -

=-2.5'2.25t1'2,3.55Nm

T = r FstnO

b.zs)(q.oo)sin9o + tu'o( o'r) singo4

/Dot^t b< {ooteA -F, is o'f 9o" tor\

a-o\

the

lF,

tr,

fs

b.g te. I Nz odd r.rp the indtvidra.Q torluraf

at o.25vn , bc* h clocEulise

Ac+fS a.t o'l ^, Counterr-clocLt^rise

net

qctrnJ

IT,

:. T1^e- not lor1t^e is 3.95 Nm'n fite c,locLnisa direcJta''n

Solrre 3 eXnationS

Sub @ a, d@ i^@(rr.o+-bo,)-(2"

lZa= Q,o14

tt,0?

a.nd

f, T, - /*,3 = M14.

T, - a'oSb = 2a' C

mr3 sing -Tz - /tnr3cos9 = nt&G)(a.x)(sin3o) -r2 - Lo'u)b\t?:t"

?k& -Tz= bo- @3 rln bnorrong.

0.3tz5xg. = ror-

e= O.3o1m ls'

+ a.bab) = 5obacl subshtute :

T, = ?.b?NTr'1"LZN