WileyPLUS Assignment 3 is available · The golf ball must travel at least a distance equal to its...

WileyPLUS Assignment 3 is availableChapters 6 & 7

Due Wednesday, November 11 at 11 pm

Next WeekTutorial and Test 3

Chapters 6 & 7See web page for tutorial questions

1Friday, October 30, 2009

Equations for rotational motion

Analogous to the equations for linear motion.

!= !0+"t

!=12("0+")t

!= "0t+12#t2

!2 = !20+2"#

v= v0+at

x=12(v0+ v)t

x= v0t+12at2

v2 = v20 + 2ax

ω ⇔ v

α ⇔ a

θ ⇔ x

arc length, l = rθspeed, v = rωπ radians = 180o


8.33/27: A child is running around a stationary merry-go-round at 0.25 rad/s. At the moment he sees his favourite horse, one quarter turn away, the merry-go-round starts to move in the same direction as the child is running, accelerating at 0.01 rad/s2.

What is the shortest time it takes the child to catch up with the horse? Child

Horse

!c = 0.25 rad/s

" = 0.01 rad/s2What are the angular positions of child and horse?


8.26/21: A spinning top is made to spin by pulling on a string that is wrapped around it. The length of the string is L = 64 cm and the radius of the top is r = 2 cm. The string is pulled so that the angular acceleration of the top is 12 rad/s2.

What is the final angular velocity of the top?

ω2 = ω20 + 2αθ

What is θ?

The number of turns of string around the top is N = L/(2πr),so θ= N x 2π = (L/2πr) x 2π = L/r = 64/2 = 32 rad

ω = 28 rad/s

(think v2 = vo2 + 2ax)


8.75/14: A golf ball passes through a windmill,which has 8 blades and rotates at ω = 1.25 rad/s.

The opening between successive blades is equal to the width of a blade.

A golf ball is of diameter d = 0.045 m. What must be the minimum speed of the golf ball so that itpasses through an opening between blades?

There are 8 blades and 8 gaps between blades. The angular width of each gap is 2π/16 rad.

The golf ball must travel at least a distance equal to its diameter while the blade moves the width of one gap.

v vThat is, t =

d

v=

2π/16ω

→ v =8dω

π

v = 0.143 m/st = θ/ω

d


Chapter 9: Rotational Dynamics

Sections 1, 2, 3, 6 only

• Action of torques

• The two conditions of equilibrium

• Centre of gravity

• Conservation of angular momentum


A

B

r

r

!

lv

Circular Motion

l = r!v = r"

" = #$/#t! = #!/#t

! in radians: 1800 = % rad.

" in rad/s! in rad/s2


Torques

Lever arm, l = wsin90◦

w = width of door

Torque, != Fl = Fw

Torque = (magnitude of force) ! (lever arm)


Torques

w = width of door

!

Torque, != Fl = Fwsin"

l = wsin!

Lever

arm

Lever arm:

Pivot, axis of rotation


Torques

w = width of door

Torque, != Fl = Fwsin"= 0

!= 0


= 45 N

9.3/1: The torque applied to the bolt is:

!= Fl

l

!= (45 N)× (0.28sin50◦ m)

= 9.65 N.m

l = 0.28sin50◦ m

Line of action of the force

lever arm


Torque generated by Achilles tendon about ankle joint:

l = (0.036 m)× cos55◦ = 0.0207 m

!= Fl = (720 N)(0.0207 m) = 14.9 N.m

F = 720 N

Sign convention: torque is positive when tending to rotate counterclockwise about reference point. So, torque = –14.9 N.m.

Line o

f

actio

n of t

he

force


Two ways to calculate torques

Lever arm, l

Line of action of the force

d

F$

P

F#

900

$

1) Torque = (force) & (lever arm):

2) Torque = F# & d : τ = F⊥d = F sin θ × d

Torque about point P:

F# = component of F perp. to line to P

Point of application of F

900

τ = Fl = F × d sin θ

Choose the method that is the more convenient...


9.-/8: One end of a metre stick is pinned to a table, so that the stick can rotate freely around the surface of the tabletop. Two forces, both parallel to the tabletop, are applied in such a way that the net torque is zero. Where along the stick must the 6 N force be applied?

x


9.2/3: The engine applies a torque of τeng = 295 N.m to the wheel of

a car, which does not slip against the road surface because the static friction force applies a countertorque. The car is travelling at constant velocity.

What is the static friction force?

v

'eng

Fs

r = 0.35 m

As the wheel is turning at constant rate, thenet torque applied to it must be zero.

Therefore, 'eng = Fs & r

Fs = 'eng/r = 295/0.35 = 843 N

O


The Two Conditions of Equilibrium

Balance is important!


The Two Conditions of Equilibrium

For an object to be in equilibrium:

1) The sum of forces acting on the object must be zero:

!�Fi = �F1+�F2+ . . . = 0

Then the acceleration is zero by Newton’s first law.

2) Balance: the sum of torques about any point must be zero:

!"i = "1+ "2+ . . . = 0

Then the angular acceleration is zero.


Are the three forces shown sufficient to keep the rod in equilibrium?

A

For the rod to be in equilibrium, the sum of the forces and of the torques about any point must be zero.

A) It is not possible to adjust the magnitudes of the forces so that the net force is zero.

B) It is possible to adjust the magnitudes of the forces so that the net force is zero, but the net torque cannot be zero.

C) It is possible to adjust the magnitudes of the forces so that the net force and the net torque are both zero.

Clickers!


A

The sum of forces could be made to be zero by adjusting the magnitudes of the forces.

That is, F1 + F2 + F3 = 0

For the rod to be in equilibrium, the sum of the forces and of the torques about any point must be zero.

Second condition of equilibrium? Calculate torques about A.

F1 and F2 exert no torque about A, but F3 does.

$ the net torque cannot be zero unless F3 = 0, but then the first condition of equilibrium fails (unless the other forces are zero too!).

$ the rod cannot be in equilibrium if only these forces act.

First condition of equilibrium is OK


Focus on Concepts, Question 6

Five hockey pucks are sliding across frictionless ice. The drawing shows a top view of the pucks and the three forces that act on each one. The forces can have different magnitudes (F, 2F, or 3F), and can be applied at different points on the puck. Only one of the five pucks could be in equilibrium. Which one?

A) 1B) 2C) 3D) 4E) 5

D) 4 The net force is zero and the net torque is zero.


Conditions of equilibrium:

1) Forces acting on the board:

−F1+F2−W = 0

2) Torques about the axis:

That is, F2 =3.9W1.4

= 2.786W

0×F1+1.4F2−3.9W = 0

Forces acting on the diving

board


If the weight of the diver is W = 530 N, then

F1 = 947 N,

F2 = 2.786W = 1480 N

F1 = F2 - W = (2.786 - 1)W = 1.786W

-F1 + F2 - W = 0 (first condition)

F2 = 2.786W (second condition)


What weight can the muscle lift?

Maximum tension in muscle = 1840 N


Calculate torques about the shoulder joint to eliminate Sx, Sy from the torque equations.

Forces on arm from shoulder

joint

(Wa acting at centre of gravity of arm)


= 1840 N

= 31 N

= 0.28 m

= 0.62 m

0.15 m

Torques about shoulder joint (axis):

lM = 0.15sin13◦ = 0.0337 m

1840×0.0337−31×0.28−Wd×0.62= 0

Wd = 86 N


= 1840 N

= 31 N

= 86 N

Forces:

x : Sx−M cos13◦ = 0 Sx = 1840cos13◦ = 1793 N

y : M sin13◦+Sy−Wa−Wd = 0

Sy =−1840sin13◦+31+86=−297 N

(shoulder joint exerts a downward force on the arm)26Friday, October 30, 2009

A

Ladder is 8 m long. No friction at wall.

WF = 875 NWL = 355 N

A

Forces acting on ladder

50◦


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