Chapter 8Rotational of Rigid Body

[ 8 Hours ]

8.1 Equilibrium of a uniform rigid body

8.2 Rotational Kinematics8.3 Rotational dynamics8.4 Work and energy of rotational

motion8.5 Conservation of Angular

Momentum.

Edited by Chongyl & NorainiHashim

2

8.1 Equilibrium of Uniform Rigid Body

Torque ( ) τ

• is the turning effect produced by a force .

• is measure of how much a force acting on an object causes that object to rotate.

• Also known as moment in engineering.

• It is a vector quantity.

• The unit of torque is N m(newton metre), a vector productunlike the joule (unit of work), also equal to a newton metre, which is scalar product.

Consider a force F acts at point P on a body that is free to rotate about an axis through O:

How to calculating torque ?

r : distance between the rotation axis and the point of application of force.

θ : angle between r and F

Point of action of a force

Rotation axis θ

The magnitude of the torque due to the force acting is given by:

dF=τwhere d – perpendicular distance from

point O to the line of action of the force.

Distance, d is called moment arm of force F

Refer to the figure, we have : d = r sin θ ; thus:

Fr

×=τIn vector notation form:

sinF rτ θ=rF

and between angle : θwhere

• Sign convention of torque:(1) Positive - turning tendency of

the force is anticlockwise.

(2) Negative - turning tendency of the force is clockwise.

• The value of torque depends on the rotation axis and the magnitude of applied force.

Caution :If the line of action of a force is through the rotation axis then

θFrτ sin=0=τ

and 0=θ

Difficult to open the door.

Easier to open the door.

From : dF=τ d∝τ

9

sinF rτ θ=rF

and between angle : θwhere

Point of action of a force

Rotation axis θ

ROTATION AXISTorques require point of referencePoint can be anywhereUse same point for all torquesPick the point to make problem least difficult

11

Example

(a) A mechanic applies a force of 200 N as shown in figure above. What torque is applied to the nut?

(b) Would it be easier to undo the nut with a longer or shorter spanner? Explain your answer.

(a) Torque,

(b) From: r ↑ ↑ . This makes it easier to undo the nut with a longer spanner.

τ

This is the point where we taking torque (axis of rotation)

r

sinF rτ θ=200(0.28)(sin 50)=

Nm43=τ

sinF rτ θ= rτ ∝

This is the point of application of force

θ = 130°

Example

Calculate the total torque at point A.

A

cm100cm80

+

Solution

Total torque at point A is due to force F1 and F2

21 FFA τττ +=

4(1)sin 90 6(0.8)sin 60= −

157.44−=

Nm157.0−=

1 1 1 2 2 2sin sinA F r F rτ θ θ= +

sinF rτ θ=Apply equation :

1

FOLLOW UP EXERCISE

(1) For the each figure below, calculate the torque at point O. Given the force applied F is 30 N and the length of the rod is 4 m.

O

F

O

F

30°

O

F

60°O

F

60°

2 m

O

F

OF

120 Nm 60 Nm

0 Nm –51.96Nm

0 Nm 0 Nm

O

F

1 mO

F

3.0 m

45°

F

2.5 m

135°O

F

2.0 mO

140°

F

2.5 mO

F

O60°

90 Nm 63.64 Nm

38.57 Nm–53.03 Nm

–75 Nm 103.92 Nm

Answer : –40 Nm ( clockwise)

m 5m 5

m 10

m 6O

Solution :

ForceTorque (N m), τo=Fd=Frsinθ

1F

( )( ) 90330 −=−

2F

( )( ) 50510 +=+

N 10=2FN 30=1F

N 20=3F

m 3=1d

m 5=2d

3F

0The resultant torque:

m N 405090 −=+−=∑ Oτ(clockwise)

A uniform metre rule of weight 2.0 N is pivoted at the 60 cm mark. A 4.0 N weight is suspended from one end, causing the rule to rotate about the pivot.At the instant when the rule is horizontal, what is the value of the resultant turning moment about the pivot?

A. zero C. –1.4 NmB. 1.4 Nm D. 1.6 Nm

Answer : B * г = 4(0.4)-2(0.1) = 1.4 Nm

Rigid body is defined as a body with definite shape that doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is exerted on it.

There are two conditions for the equilibrium of forces acting on a rigid body.

(1) The vector sum of all forces acting on a rigid body must be zero.

Condition for equilibrium of uniform rigid body

0netF F= =∑

OR

0 , 0 x yF F= =∑ ∑

(2) Summation of all torques about any point on the body is zero.

any point 0τ =∑

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((1) Draw a free body diagram showing & labeling all external forces acting on the body.

(2) Resolving the forces into x & y components. And apply:

0;0 =Σ=Σ yx FF

(3) Take torque about any convenient point in the system & equate the sum of the torque to zero.

0=Στ(4) Solve for the unknown

quantities.

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Some useful tips for you !

If there are 2 or more unknown forces, chooses the point where unknown forces pass through as rotation axis. Such choice simplifies the torque equation because the torques produced by these forces will equal to zero.

Weight of the object is acting at its center of gravity. For uniform object, weight always acting at the center of the object.

The words ‘rough’ & ‘smooth’ have a particular meaning in physics.‘rough’ need to consider friction force.‘smooth’ No friction force involved

Example

A 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 a negligible weight and is bolted down at the left end while being supported 1.40 m away by a fulcrum. Find the forces exert on the board by the bolt and the fulcrum.

Free body diagram of the diving board

Taking torques at the left end of the board:

–

+90

90

1W

530N−

1F

Force y comp. Torque, τO= Fr sinθ

2F

1F−

2F

2(1.4)sin 90F+0

530(3.9)sin 90−

0=Σ yF

1 2 530 0F F− + − =

1 2

1

530

1476.43 530 946.43N

F FF

= −

= − =

Since board in equilibrium, sum of vertical forces must be zero.

0=Στ2(1.4)sin 90 530(3.9)sin 90 0F+ − =

2530(3.9) 1476.43

1.4F N= =

1W2W

A BO35 cm 75 cm

Figure 5.16

Example

A hanging flower basket having weight, W2 =23 N is hung out over the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is counterbalanced by a body of weight, W1as shown in Figure 5.16. If the mass of the beam is 3.0 kg, calculatea. the weight, W1 needed,b. the force exerted on the beam at point O.(Given g =9.81 m s –2)

The free body diagram of the beam :

Let point O as the rotation axis.

Solution

A B

O

1W

2W

N

gm

0.35 m 0.75 m0.20 m

0.55 m 0.55 m

1W

1W− ( )1 10.75 sin 90 0.75W W− = −

2W

23− ( )( )23 0.35 sin 90 8.05+ =

gm ( )( )29.4 0.20 sin 90 5.88− = −29.4−=

( )( )9.813−

N

N 0

Force ycomp. Torque, τO= Fd = Fr sinθ

––+

Since the beam remains at rest thus the system in equilibrium. a. Hence

b. and

N 2.89=1W

0Oτ =∑10.75 8.05 5.88 0W− + − =

0=∑ yF

N 55.3=N

029.423 =+−−− NW1

( )2.89 23 29.4+ 0N− − − =

FOLLOW UP EXERSICE

Given: W=50 N, L=0.35 m, x=0.03 mFind the tension in the muscle.

Solution

Given: W=50 N, L=0.35 m, x=0.03 mTaking point O as rotation axis

0(0.03)sin 90 50(0.35)sin 90 0

583 N

O

FF

τΣ =− =

=

+–

32

FOLLOW UP EXERCISE

The see saw in the diagram is balanced. Determine the value of weight, W

+ – –

N

Since system in equilibrium, thus we have:

0=Σ pivotτ

NW 750=

11255.1 =W

(1.5)sin 90 300(1.0)sin 90550(1.5)sin 90 0 0

W −− + =

Taking torque about the pivot.

Solution

ForceTorque (N m), τA= Fd = Fr sinθ

W +W(1.5)sin90300N –300(1.0)sin90550N – 550 (1.5)sin90

N 0

Example

A uniform rod of mass 30 kg and length 3 m is smoothly hinged at A. The rod is held in a horizontal position by a rope. The rope is attached to the rod at a point B, that is 2 m from A. The angle between the rope and rod is 60°. A mass of 100 kg is suspended from the end of the rod at C. Find the tension in the rope.

Solution

Free body diagram of the suspended rod:

T

R

RW LW

°60

m2

m5.1

A B C

30(9.81)100(9.81)

R

L

WW

==

Rotation axis

––

Concept : In equilibrium 0τ⇒ Σ =Taking torque at point A:* R pass through point A torque

produce by R are zero.

ForceTorque (N m), τA= Fd = Fr sinθ

R 0T + T (2)(sin 60 ) = 1.732T

WR – 30(9.81) (1.5)sin90 = – 441.45WL – 100 (9.81) (3)sin90 = – 2943

0=Σ Aτ1.732 441.45 2943 0T − − =

NT 1.1954=

45.3384732.1 =T

Example

Figure shows a load, of mass M = 430 kg, hanging by a rope from a boom. The boom consists of a hinged beam and a horizontal cable that connects the beam to the wall. The uniform beam has a mass m = 85 kg; the mass of the cable & rope are negligible.(a) What is the tension in the

cable ?

(b) Find the magnitude & direction of the force acting on the hinge.

Solution

Free body diagram for the beam

T

F

1W 2W

1.9m

A

B

C

+

––

θ = 37.2

52.8

52.8

θ = 37.2

3.14AC m=

System in equilibrium :

0 ; 0AF τΣ = Σ =

Let point A as the rotation axis

Force x-comp y-comp Torque

τA=Fd=FrsinθF Fx Fy 0T –T 0 T(3.14)sin 37.2

W1 0 85(9.81)= 833.85

–833.85 (1.57)sin 52.8

= –1042.31W2 0 430(9.81)

= 4218.3–4218.3 (3.14)sin

52.8= –10545.75

The system in equilibrium, thus

0Aτ =∑(1.9) 1042.31 10545.75 0T − − =

6098.98T N=

(a)

0=Σ xF0=−TFx

0=Σ yFNFx 98.6098=

833.85 4218.3 0yF − − =5052.15yF N=

(b)

The magnitude of the force is 2 2

x yF F F= + ( ) ( )2 26098.98 505215= +

7919.71F N=

and its direction is given by

1 1 5052.15tan ( ) tan ( )6098.98

y

x

FF

θ − −= =

39.64θ = °

x

y

F

39.64°

Example

A

Bsmooth wall

rough floor

A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure above. The height of the end A of the ladder is 8.0 m from the rough floor.Determine the horizontal and vertical forces the floor exerts on the end B of the ladder when a firefighter of mass 60 kg is 3.0 m from B.

Solution

A

B

2N

sf

1N

m 8.0m 10

m 6.0

α

Free body diagram of the ladder :

kg 60 ;kg 5.0 == fl mm

lm g

5.0 m

β 3.0 m

fm g β

β

0.8108sin ==α

0.6106sin ==β

α

+

+

–

Let point B as the rotation axis.

Forcex-

comp. (N)

y-comp.

(N)

Torque (N m), τB=Fd=Frsinθ

gml

1N1N

sf−

gm f

49.1−

2N

sf

0

589−0

2N

0

0

( )( )49.1 5.0 sin β147=

0

( )( ) βsin3.05891060=

( ) αN1 sin10−1N8−=0

0

0=∑ xF

0=− s1 fNN 151=sfHorizontal force:

0=∑ yF

058949.1 =+−− 2NN 638=2NVertical force:

Since the ladder in equilibrium thus

0=∑ Bτ

081060147 =−+ 1N

N 151=1N

FOLLOW UP EXERCISE

(1) A see-saw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 5.20. The fulcrum is under the centre of gravity of the board. Determine

a. the magnitude of the force exerted by the fulcrum on the board,b. where the father should sit from the fulcrum to balance the system.

ANS. : 1128 N; 1.31 m

SOLUTION

1.

BW DWFW

N

m5.3

m75.1x+ −

NNN

WWWNF

DBF

y

11280)81.9(45)81.9(10)81.9(60

0

0

==−−−

=−−−

=∑

(b) Force Torque = Fr sin θ (Nm)

00BW

FW

DW

N

( sin 90 )60(9.81)( )

588.6

FW xx

x

+== +

(1.75sin 90 )45(9.81)(1.75)772.54

DW−= −= −

Taking torque at Fulcrum

mxxx

fulcrum

31.154.7726.588

054.7726.588

0

==

=−

=∑τ

(2) A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure below. A cable at an angle 30° with the beam helps to support the light.a. Sketch a free body diagram of the

beam.b. Determine

i. the tension in the cable,ii. the force exerted on the beam by

the pole.Answer:(i) 490 N(ii) 427 N at 6.61°

above +x

SOLUTION(a)

BW LW

R

T

30

+

− −L21 L

21

A B

Taking torque at Hinge (point A)Force x-comp

(N)y-comp (N) = Fr sin θ

(Nm)0

0

0

BW

LW

T

R xR yR

1( )(sin 90 )2

110(9.81)( )2

49.05

BW L

L

L

−

= −

= −

1.98)81.9(10

−=−=

− BW

2.196)81.9(20

−=−=

− LW

TT

866.030cos

−=−

TT

5.030sin

+=+ ( )(sin 30 )

0.5 ( )T L

T L+= +

( )(sin 90 )20(9.81)( )196.2

LW LL

L

−= −= −

τ

(b)(i)

(ii) NT

TLLTLLL

R

5.4905.025.245

05.02.19605.4900

==

=+−−

=∑τ

NRR

TRTR

F

x

x

x

x

x

773.424)5.490(866.0

866.00866.0

0

===

=−

=∑

NRR

TRF

y

y

y

y

05.49

)5.490(5.03.294

05.02.1961.98

0

=

−=

=+−−

=∑

NRR

RRR yx

6.427)05.49()773.424( 22

22

=

+=

+=

The force exerted on the beam by the pole is

and its direction is given by

1

tan

49.05tan ( )424.773

6.6 ( )

y

x

RR

above x

θ

θ

θ

−

=

=

= ° +

54

A cat walks along a uniform plank that is 4.0 m long and has a mass of 7 kg. The plank is supported by 2 sawhorses one 0.5 m from the left end of the board and other 1.50 m from its right end. When the cat reaches the right end, the plank just begin to tip. What is the mass of the cat ?

Answer : 2.33 kg

(3)

SOLUTION

mc(9.81)7(9.81)

NRNL

1.5 m

0.5 m 1.5 m

Just begin to tip NL = 0 N

0τ =∑

c

7(9.81)0.5sin 90 (9.81)(1.5)sin 90 0m =2.33kg

cm− =

While standing on a long board resting on a scaffold, a 70 kg painter paints the side of a house, as shown in figure above. If the mass of the board is 15 kg, how close to the end can the painter stand without tipping the board over ?

{ answer : 1.2 m from left end of board }

(4)

SOLUTION

15(9.81)70(9.81)

NRNL

2.5 m

1.5 m

Just begin to tip NR = 0 N

0τ =∑70(9.81) sin 90 15(9.81)(1.25)sin 90 0

0.26x

x m− =

=

1.25 mx

From the left end, x = 1.5 – 0.26 = 1.2 m

(5) A beam of length 10 m andweight of 60 N is supported atone end by a pivot and theother end by an egg. If a robotof weight 80 N run along thebeam. The egg can with standa maximum force of 50N.

eggpivot

(i) Draw ALL the forces acting on the beam.

(ii) How far from the pivot the kid can reach without breaking the egg.

Answer : 2.5 m

RE

10 m

WK WB

x

RP

5 m

Arrows with correct direction …..1 marks

Correct position of the forces and label …..1 marksNo arrow for forces drawn

…..0 marks

Solution:RE = 50 N; WK = 80 N; WB = 60 N

Taking torque at the pivot,

2.5m 0 (50)(10)(60)(5)(80)(x)0)(10)(R)(5)(W)(x)(W

0

Ebk

==+−−=+−−

=∑

x

τ …….K1

…….RG1

…….JU1