Mechanics 2 Revision Notes, Exercises of Mechanics

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Mechanics 2

Revision Notes

November 2012

• 1 Kinematics
  • Constant acceleration in a vertical plane
  • Variable acceleration
  • Using vectors
• 2 Centres of mass
  • Centre of mass of n particles
  • Centres of mass of simple laminas
  • Centres of mass of composite laminas................................................................................
  • Laminas suspended freely under gravity
  • Toppling on a slope
  • Centres of mass of wire frameworks.
• 3 Work, energy and power
  • Definitions
  • Work done by a (constant) force.
    • Forces parallel to the displacement
    • Forces at an angle to the displacement
    • Work done by gravity
  • Work-energy equation
  • Potential energy
  • Power
• 4 Collisions
  • Impulse and momentum – using vectors
    • Impulse = change in momentum
    • Conservation of linear momentum...............................................................................................................
  • Newton’s law of restitution
    • Coefficient of restitution
    • Collisions with a plane surface
    • Multiple collisions
  • Kinetic energy and impulses/collisions
• 5 Statics of rigid bodies
  • Moment of a force
  • Equilibrium
  • Three non-parallel forces in equilibrium
  • Triangle of forces
• Appendix
  • Centre of mass of n particles
  • Medians of a triangle
  • Centre of mass of a triangle

1 Kinematics

Constant acceleration in a vertical plane

We can think of the horizontal and vertical motion separately and use the formulae:

v = u + at , s = ut +

ଵ ଶ at

(^2) , v (^2) = u (^2) + 2 as , s = ଵ ଶ ( u^ +^ v ) t

Example 1: A stone is thrown at a speed of 20 ms -1^ at an angle of 35 o^ to the horizontal. Find ( a ) the greatest height reached ( b ) the direction in which it is moving after 1 second (c) the height of the stone after it has travelled a horizontal distance of 25 m.

Solution: ( a ) Vertical motion u = 20 sin 35o^ , a = −9.8, v = 0, s = h v^2 = u^2 + 2 as ⇒ 0 = (20 sin 35o^ ) 2 − 2 × 9 ⋅ 8 h

⇒ h = 6⋅ 71408017 ⇒ greatest height reached is 6⋅71 m to 2 S .F.

( b ) Horizontal motion u = 20 cos 35o Vertical motion u = 20 sin 35 o^ , a = −9.8, t = 1, v =?

v = u + at = 20 sin 35o^ − 9 ⋅ 8 × 1 = 1⋅ 671528727

⇒ tan ߠ

ଵ·଺଻ଵ… ଶ଴ ୡ୭ୱ ଷହ

⇒ stone is moving at 5⋅ 8 o^ above the horizontal, 2 S .F.

( c ) Horizontal motion u = 20 cos 35o^ , s = 25, a = 0

⇒ t = (^) ଶ଴ ୡ୭ୱ ଷହଶହ ൌ 1 · 525968236

Vertical motion u = 20 sin 35 o^ , a = −9.8, t = 1⋅5259…, s =?

s = ut +

ଵ ଶ at

(^2) = 20 sin 35o (^) × 1 ⋅5259… − ଵ ଶ ×^9 ⋅^8 ×^ (1⋅525…)^

2

⇒ s = 6⋅ 095151075 ⇒ stone is at a height of 6⋅1 m after it has travelled 25 m horizontally 2 S.F.

35 o

20

1 ⋅67…

v

θ 20 cos 35

Variable acceleration

When a is given as a function of t v = (^) ݐ ݀ ܽ׬ do not forget the + c

s = ݐ ݀ ݒ ׬ do not forget the + c

When s (or v ) is given as a function of t

v = ௗ௦ௗ௧

and a =

ௗ௩ ௗ௧ or^ a^ =^

ௗ మ^ ௦ ௗ௧ మ

Note that s is the displacement (the distance from the origin), which is not necessarily the same as the distance travelled (the particle may have moved forwards and backwards).

Example 1 : A particle is moving along the x -axis with an acceleration 5 − 2 t ms-2^. At time t = 0, the particle moves through the origin with speed 6 ms-1^ in the direction of the positive x -axis. ( a ) Find the displacement of the particle after 9 seconds. ( b ) Find the distance travelled in the first 9 seconds.

Solution:

( a ) a =

ௗ௩ ௗ௧ = 5^ −^2 t^ ⇒^ v^ =^ ݐ ݀ ݐ2 െ 5 ׬^ =^5 t^ −^ t^

(^2) + c

v = 6 when t = 0, ⇒ c = 6 ⇒ v = 5 t − t^2 + 6

s = (^) ݐ ݀ ݒ ׬ ൌ (^) 6 ׬൅ 5 ݐെ ݐ ݀ ଶ^ ݐ ൌ ൅ ݐ

ହ ଶ ݐ^

ଷ ݐ^

s = 0 when t = 0, ⇒ c ′ = 0

⇒ s = ൅ ݐ6 ହଶ ݐ ଶ^ െ ଵଷ ݐ ଷ

When t = 9, s = 6 × 9 + ହଶ ൈ 9ଶ^ െ ଵଷ ൈ 9ଷ^ = 13⋅ 5

The displacement after 9 seconds is 13⋅5 m.

( b ) Note: the particle could have gone forwards then backwards, in which case the distance travelled would not be the same as the final displacement. We must first find t when the velocity is zero. v = 5 t − t^2 + 6 = (6 − t )(1 + t )

⇒ v = 0 when t = (−1) or 6. The particle is moving away from the origin for 0 ≤ t < 6, and towards the origin for 6 < t 9, so we want the sum of the two distances d 1 + d 2. When t = 6, s = d 1

⇒ d 1 = 6 × 6 +

ହ ଶ ×^6

ଷ ×^6

3

⇒ d 1 = 54

⇒ d 2 = 54 − 13 ⋅5 = 40⋅ 5 ⇒ total distance travelled = d 1 + d 2 = 94⋅5 m

Using vectors

This is just combining horizontal and vertical motion into one expression.

Example 1: A particle moves with velocity v = ቀ ݐ^

ଶ ݐെ

ቁ ms -1^. It is initially at the point (6, 3).

Find its acceleration after 2 seconds, and its displacement at time t.

Solution: a =

࢜ௗ ௗ௧

ௗ൬ ݐ3^2 ݐെ

൰ ௗ௧

⇒ at t = 2, a = ቀ^12 െ

ቁ ms-

s = ݀ ࢜׬ ݐ = ቀ ׬ݐ^

ଶ ݐെ

ݐ ଷ^ ܿ ൅ ଵ

ݐെ2 ଶ^ ܿ ൅ ଶ

Particle is initially at (6, 3) ⇒ ܿቀ ܿଵ ଶ

ቁ = ቀ^6

⇒ s = ൬ ݐ^

ݐെ2 ଶ^ ൅ 3

൰ m

d 1

O t = 9

d 2

x t = 6

13 ⋅ 5

2 Centres of mass

Centre of mass of n particles

The centre of mass, (ҧݔ, ݕത), of n particles, which have masses m 1, m (^) 2, … , m (^) n , at points ( x 1 , y 1 ,), ( x 2 , y 2 ), …, ( xn , yn ) is given by

M ൬ݔ

⇔ M ҧݔ = ݉∑ ௜ ݔ௜ and M ݕത = ݉∑ ௜ ݕ௜

where M is the total mass, M = ݉∑ ௜

or M g = ݉∑^ ࢘ ௜ ௜

Centres of mass of simple laminas

1. The centre of mass of a uniform rod is at its mid point.

2. The centre of mass of a uniform rectangular lamina (sheet) is at its point of symmetry.

3. The centre of mass of a uniform triangular lamina ( a ) G is at the point where the three medians meet. G divides each median in the ratio 2: ஺ீீ ஽

஻ீீ ா

஼ீீ ி

ଶ ଵ

( b ) If A , B and C are the points ( a 1 , a 2 ), ( b 1 , b 2 ,) and ( c 1 , c 2 ), then the centre of mass, G , is at the point

G (

ଵ ଷ ( a^1 +^ b^1 +^ c^1 ),^

ଵ

ଷ ( a^2 +^ b^2 +^ c^2 )^ )

or g =

ଵ ଷ ( a^ +^ b^ +^ c )

4) The centre of mass of a uniform sector of a circle with angle 2 α

Circle centre O with radius r

Angle of sector is 2 α

G lies on the axis of symmetry and

OG =

ଶ௥ ୱ୧୬ ఈ ଷఈ

A

B C D

F E

G

G

G O

α α

r

r

Example 1: A uniform triangular lamina has mass 3 kg, and its vertices are O (0, 0), A (5, 0) and B (4, 3). Masses of 2, 4 and 5 kg are attached at O , A and B respectively. Find the centre of mass of the system.

Solution: First find the centre of mass of the triangle, G (^) T.

gT = ଵଷ ( a + b + c ) = ଵଷ ቆቀ^00 ቁ ൅ ቀ^50 ቁ ൅ ቀ^43 ቁ ቇ

⇒ G (^) T is (3, 1).

We can now think of 4 masses, 2 kg at O , 4 kg at A , 5 kg at B and 3 kg (the triangle) at G (^) T.

M g = ݉∑ ࢘ ௜ ௜ ⇒ (2 + 4 + 5 + 3) g = 2ቀ^00 ቁ + 4ቀ^50 ቁ + 5ቀ^43 ቁ + 3ቀ^31 ቁ

⇒ 14 g = ቀ^4918 ቁ

⇒ G is at ቀ

ସଽ

ଵସ ,^

ଵ଼

Centres of mass of composite laminas

In the following examples ρ is the surface density, or mass per unit area.

Example 3: Find the centre of mass of a rectangle attached to a semi circle, as shown in the diagram. O is the centre of the semi-circle.

Solution: The centre of mass will lie on the line of symmetry, OC , so we only need to find the horizontal distance of the centre of mass from O. By symmetry the centre of mass of the rectangle is at G 1, as shown.

The semi-circle is a sector with angle 2 ×

గ ଶ

⇒ OG 2 =

ଶൈଶ ୱ୧୬ ߨ 2 ଷൈߨ 2 =

଼ (^) ଷగ

Note OG 1 = −3 and OG 2 =

଼ ଷగ

negative and positive

x

2 O A (5, 0)

B (4, 3)

GT (3, 1) 4

5

3

y

A B

E D

C O

2 4

6

(^3) G 1 3 G 2

Laminas suspended freely under gravity

Example 5: The lamina in example 4 is suspended from A , and hangs in equilibrium. Find the angle made by AB with the downward vertical.

Solution: G must be vertically below A. This must be stated in any solution (method). From example 4 we know that

FG = 8⋅588… and AF = 5 The angle made by AB with the downward vertical is ∠ GAB = ∠ FGA = arctanቀ

ହ଼ .ହ଼଼… ቁ^ = 30⋅^20761248 ⇒ angle made by AB with the downward vertical is 30⋅ 2 o to the nearest 0⋅ 1 o^.

Toppling on a slope

Example 6: What angle of slope would cause a 4 × 6 uniform rectangular lamina to topple (assuming that the friction is large enough to prevent sliding).

Solution: G must be vertically above A when the lamina is on the point of toppling. This must be stated in any solution (method).

By angle theory ∠ AGM = θ , the angle of the slope.

tan θ =

ଶ

ଷ ⇒^ θ^ = 33⋅^69006753

The lamina will topple when the angle of slope exceeds 33⋅ 7 o^ , to nearest 0⋅ 1 o^.

Centres of mass of wire frameworks.

In the following examples ρ is mass per unit length.

1. The centre of mass of a uniform straight wire is at its centre.

2. The centre of mass of a uniform circular arc, of radius r and angle at

the centre 2 α, lies on the axis of symmetry and OG =

௥ ୱ୧୬ ఈ

G

A

B

5

8 ⋅588…

F

O G α

α

r

A

M

C

θ

θ

6

4

3

2

G

Example 7: A uniform wire framework is shown in the diagram. OABC is a rectangle and the arc is a semi-circle. OA = 10 and OC = 4. Find the position of the centre of mass.

Solution: With frameworks it is often easier to use vectors. Take the origin at O.

The semi-circle is an arc of angle 2 × గଶ

⇒ centre of mass is at G 4 , where OG 4 =

ହ ୱ୧୬ ߨ 2 ߨ 2

ଵ଴ గ

⇒ semi-circle has mass 5 πρ at ቀ4 ൅ 10 ߨ , 5ቁ.

We now consider each straight wire as a point mass at the midpoint of the wire.

OA OC AB semi-circle

mass 10 ρ 4 ρ 4 ρ 5 πρ

centre of mass ቀ^0 5

ቁ ቀ^2

ቁ ቀ^2

M g = ݉∑^ ࢘ ௜ ௜ ⇒ (18 + 5 π) ρ g = 10 ρ ቀ^0

ቁ + 4 ρ ቀ^2

ቁ + 4 ρ ቀ^2

ቁ + 5 πρ ቆ4 ൅^

ଵ଴ గ 5

⇒ g =

ଵ ଵ଼ାହగ ቀ

ቁ to 3 S .F.

O

G 1

G 2

G 3

G 4

A B

C

Work done by gravity

If a particle of mass m falls a vertical distance h , then the work done by gravity is mgh , force and displacement are in the same direction If a particle of mass m rises a vertical distance h , then the work done by gravity is − mgh , force and displacement are in opposite directions

When a particle is moving on a slope, it is usual to consider the vertical distance moved and multiply by mg to calculate the work done by gravity.

From A to B the particle moves a distance s , but its

vertical movement is h = s sin θ

⇒ work done by gravity = mgh = mgs sin θ

Work-energy equation

The equation Fs =

ଵ ଶ mv

ଶ mu

(^2) can be re-arranged as

ଵ ଶ mv

ଶ mu

(^2) + Fs

which can be thought of as Final K.E. = Initial K.E. ± Work done

Notice the ±

• If a force increases the K.E. then the work done is positive
• If a force decreases the K.E. then the work done is negative

mg

A

B

s h

θ

Example 1: A particle of mass 5 kg is being pulled up a rough slope by a force of 50 N parallel to the slope. The coefficient of friction is 0⋅2, and the slope makes an angle of

α = tan-1ቀ

ଷ ସ ቁ^ with the horizontal. The particle is observed to be moving up the slope with a speed of 3 ms-1^. Find its speed when it has moved 12 m up the slope.

Solution:

AB = 12 ⇒ h = 12 sin α = 6

Resolve R = 5 g cos α = 4 g

Moving ⇒ F = μ R = 0.2 × 4 g = 0⋅ 8 g

Work done by R = 0 ( to motion)

Work done by F = − 0 ⋅ 8 g × 12 = − 9.6 g J reduces K.E. so negative Work done by 50 N = 50 × 12 = 600 J increases K.E. so positive

Work done by gravity = mgh = − 5 g × 6 = − 30 g uphill, reduces K.E. so negative

Work-energy equation Final K.E. = Initial K.E. ± Work done

ଵ ଶ ×^5 v

ଶ ×^5 ×^3

(^2) − 9 ⋅ 6 g + 600 − 30 g

⇒ v^2 = 93⋅ 678 ⇒ v = 9⋅683387837 = 9⋅ 7 to 2 S .F. Particle is moving at 9⋅7 ms-1^ after it has travelled 12 m.

Example 2: Tarzan swings on a rope, lets go and falls to the ground. If Tarzan was initially 7 m above the ground and not moving, and if he lets go when he is 3 m above the ground, with what speed does he hit the ground?

Solution:

The only forces acting on Tarzan are the tension in the rope, T , and gravity, mg. The work done by T is 0, since T is always perpendicular to the motion

Work done by gravity = mgh = mg × 7 downwards, increases K.E. so positive

Work-energy equation Final K.E. = Initial K.E. ± Work done

ଵ ଶ ×^ mv

(^2) = 0 + 7 mg

⇒ v^2 = 14 g ⇒ v = 11.71324037 = 12 ms-1^ to 2 S.F.

B

A

h

α

3

v

5 g

R

F

50

mg

T

7 m 3 m

You are expected to understand the terms loss in P.E. and gain in P.E., but all you can always use work done by gravity if you wish.

Example 4: A block, of mass 20 kg, is pulled up a rough slope of angle α, tan-1ቀଷସ ቁ, by a

rope. The block starts from rest and is moving at 6 ms-1^ when it has moved a distance of

10 m. If the coefficient of friction is μ = 0⋅3, find

the tension in the rope.

Solution:

R R = 20 g cos α = 16 g

moving ⇒ F = μ R = 0⋅ 3 × 16 g = 4⋅ 8 g

Work done by R = 0 ( to motion) Work done by F = Fs = 4⋅ 8 g × (−10) = − 48 g decreases K.E. so negative

Gain in P.E. = mgh = 20 g × (−10 sin α) = − 120 g decreases K.E. so negative

Work done by tension = Ts = 10 T increases K.E. so positive

Work-energy equation Final K.E. = Initial K.E. ± Work done

ଵ ଶ ×^20 ×^6

(^2) = 0 − 48 g − 120 g + 10T

⇒ T = 200⋅ 64 = 200 N to 2 S .F.

Notice that we could have thought of work against by gravity, mgh , instead of gain in P.E.

α 20 g

F

R T

10 m

Power

Power is the rate of doing work. The units are Watts = Joules/second. For a constant force, F , moving a distance s the work done is W = Fs

⇒ the power, P =

ௗ ௗ௧ ( Fs ) =^ F^

ௗ௦ ௗ௧ =^ Fv^ since^ F^ is constant ⇒ the power developed by a constant force F moving its point of application at a speed v is P = Fv.

Example 5: A car of mass 900 kg is travelling up a slope of 5 o^ at a constant speed. Assume that there is no resistance to motion – other than gravity. ( a ) If the engine of the car is working at a rate of 20 kW, find the speed of the car. ( b ) The car later travels up a slope of 8 o^ at the same speed. Find the power developed by the engine.

Solution: ( a ) Power, P = 20 000 = Dv

⇒ v =

ଶ଴ ଴଴଴ ஽ Constant speed ⇒

R ⇒ D = 900 g sin 5o

⇒ v =

ଶ଴ ଴଴଴

ଽ଴଴௚ ୱ୧୬ ହ ೚^ = 26⋅01479035 = 26 ms

-1 to 2 S .F.

( b ) v =

ଶ଴ ଴଴଴ ଽ଴଴௚ ୱ୧୬ ହ೚^

from part ( a )

R ⇒ D ′ = 900 g sin 8o

⇒ power developed = D ′ v

⇒ P = 900 g sin 8 o^ ×

ଶ଴ ଴଴଴ ଽ଴଴௚ ୱ୧୬ ହ೚

⇒ P = 20 000 ×

sin଼ ೚

ୱ୧୬ ହ೚^ = 31936⋅^64504

⇒ power developed by the engine is 32 kW to 2 S .F.

5 o

R D

900 g

v

(^8 900) g o

R ′ D ′

v