student name grade centre components of the velocity vector of the faster particle at q, when t= 2 g...

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4 Unit Math Homework for Year 12 (Worked Answers)

Student Name: Grade:

Date: Score:

Table of contents

1 Topic 7 — Mechanics (Part 2) 1

1.1 Motion of a particle in two dimension . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Projection Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Circular Motion in a Horizontal Plane . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 Uniform circular motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

This edition was printed on June 17, 2013.Camera ready copy was prepared with the LATEX2e typesetting system.Copyright © 2000 - 2013 Yimin Math Centre (www.yiminmathcentre.com)

4 Unit Math Homework for Year 12

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Year 12 Topic 1 Worked Answers Page 1 of 14

1 Topic 7 — Mechanics (Part 2)

1.1 Motion of a particle in two dimension

1.1.1 Projection Motion

Definition:

• The particle to be moving on a number plane.

• F on the particle is the vector sum of all the physical forces acting on it.

•−→F = m−→a , ⇒ FH = maH , FV = maH .

• where FH and FV are the horizontal and vertical components of−→F respectively.

•−→F has magnitude mg and is directed vertically downward throughout the motion.

Example 1.1.1 A particle is projected upward over horizontal ground with velocity U inclined atan angle θ to the horizontal. A second particle is projected simultaneously from the same point inthe same direction with velocity V > U .

1. Show that through the motion, the line joining the positions of the particles makes a constant

angle with the horizontal.

Solution: Axes and origin: Initial condition: t = 0, x = y = 0, x = U cos θ, y = U sin θ.

Equation of motion: x = 0, y = −gHorizontal component: x = 0, x = U cos θ, ⇒ x = Ut cos θ

Vertical component: y = −g, y = −gt+ U sin θ, ⇒ y = −1gt2 + Ut sin θ

Hence after t seconds, the two particles are at positions:

A(Ut cos θ, −12gt2 + Ut sin θ) (slower particle)

B(V t cos θ, −12gt2 + V t sin θ) (faster particle).

Gradient AB = (V−U)t sin θ(VU )t cos θ

= tan θ.

Hence AB makes an angle θ with the horizontal throughout the motion.

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2. Given that the range and time of flight of the first particle are respectively 1gU2 sin 2θ and 2

gU sin θ,

find the position of the faster particle and the direction of its velocity vector when the slower

particle hits the ground. If this velocity vector is directed upward at an angle β = 12θ to the

horizontal. find UV

in terms of θ and deduce that 14< U

V< 1

2.

Solution: AB2 = (V − U)2t2 cos2 θ + (V − U)2t2 sin2 θ

AB = (V − U)t throughout the motion.

The slower particle hits the ground at P when the faster particle is at Q

and t = 2gU sin θ.

Hence QP makes an angle θ with the horizontal and QP = (V − U)2gU sin θ.

At Q, x = OP +QP cos θ = 1gU2 sin 2θ + 1

g(V − U)U sin 2θ

y = QP sin θ = 2gU(V − U) sin2 θ

∴ Q has coordinates (1gUV sin 2θ, 2

gU(V − U) sin2 θ).

The components of the velocity vector of the faster particle at Q,

When t = 2gU sin θ are x = V cos θ, y = (V − 2U) sin θ.

Note that if V = 2U, then −→v is horizontal, and the faster particle is at its

greatest height.

If β = 12θ and −→v is upward, then V > 2U and tan

(θ2

)=(1− 2U

V

)tan θ.

Let λ = tan θ2

then UV< 1

2and λ =

(1− 2U

V

)2λ

1−λ2

∴ 1− λ2 =(2− 4U

V

)λ2 = 4U

V− 1

UV= 1

4(λ2 + 1)

∴ UV= 1

4sec2

(θ2

)and 1

4< U

V< 1

2.


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Exercise 1.1.1

1. A particle of mass m is projected vertically upward under gravity in a medium in which the

resistance is mk times the square of the speed, where k is a positive constant. The speed of

projection is V . Find the maximum height.

Solution: Choose the point of projection as origin and ↑ as positive direction.

Equation of motion: v = −g − kv2

Initial conditions: t = 0, x = 0, v = V.

Expression relating x and v: 12dv2

dx= −(g + kv2) ⇒ −2 dx = dv2

g+kv2

⇒ −2k dx = −k dv2g+kv2

⇒ −2kx+ c = ln |g + kv2|;x = 0, v = V, c = ln |g + kV 2| ⇒ x = 1

2kln |g+kV 2

g+kv2| . . . (1)

At maximum height, v = 0. Let the maximum height be h.

From (1) h = 12k

ln |1 + kgV 2|.

2. A particle of mass 0.5 kg is released from rest and moves vertically downward under gravity

in a medium which exerts a resistance to the motion of 110v2 . At time t after release it has

fallen a distance x and has velocity v. Taking g = 10ms−1, show that v2 = 50(1 − e0.4x) and

x = 10e−0.4x.

Solution: Origin is point of release. ↓ is positive direction F = mg − v2

10.

Equation of motion: x = g − v2

5.

Initial conditions: t = 0, x = 0, v = 0.

Expression relating v and x: v dvdx

= g − v2

5⇒ dx = 5v

5g−v2 dv

⇒ dx = 52

(1√

5g−v −1√

5g+v

)dv, ⇒ x = 5

2

(ln 1√

5g+v+ ln 1√

5g−v

)+ c,

∴ c = −52ln(5g − v2) + c c is constant.

x = 0, v = 0, ⇒ c = 52lg 5g ⇒ x = −5

2ln(1− v2

5g

)⇒ v2 = 5g(1−g−0.4x);

g = 10 ⇒ v2 = 50(1− e−0.4x).and x = 1

2.d

2vdx⇒ x = 10e−0.4x.


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Exercise 1.1.2

1. A particle of mass 1 kg is projected vertically upward under gravity with speed 2c in a medium in

which the resistance to motion is gc2

times the square of the speed, where c is a positive constant.

Find the time of ascent of the particle and the maximum height.

Solution: Choose a point of projection as origin and ↑ as positive.

Initial condition: t = 0, x = 0, v = 2c.

Equation of motion: x = −g − gc2v2.

Expression relating x and v: v dvdx

= −g − gc2v2, ⇒ −g dx = v dv

1+ v2

c2

⇒ −gx+ A = c2

2ln(1 + v2

c2

), A is a constant.

x = 0, v = 2c ⇒ A = c2

2ln 5 ⇒ x = c2

2gln 5c2

c2+v2. . . . (1)

Expression relating v and t: dvdt

= −g − gc2v2, ⇒ −g dt = dv

1+ v2

c2

,

−gt+ A = c. tan−1 vc, A constant;

t = 0, v = 2c ⇒ A = c. tan−1 2 ⇒ t = cg(tan−1 2− tan−1 v). . . . (2)

When the particle reaches its highest point, its velocity is zero. So v = 0

Form (2) t = c. tan−1 2g

is the time of ascent.

Let h e the distance between the point of projection and the highest point.

Then v = 0 ⇒ from (1) h = c2

2gln 5.

2. A particle is projected with speed V and angle of elevation α from a point O on the edge of a cliff

of height h. When the particle hits the ground its path makes an angle tan−1(2 tanα) with the

horizontal. Find the distance from the foot of the cliff to the point where it lands.

Solution: Axes, origin and path: After t seconds particle is at the position:

x = V cosα.t . . . (1)

y = h+ V sinα.t− gt2

2. . . . (2) O is the foot of the cliff. Oh = h.

OP is the distance from the foot of the cliff to the hit place.

As the particle hits the ground, y = 0, ⇒ from (2) gt2

2− V sinα.t− h = 0.

Solving this quadratic, we obtain the time of flight:

t =V sinα+

√V 2 sin2 α+2gh

g. . . (3)

OP = V cosα.

(V sinα+

√V 2 sin2 α+2gh

g

). . . (4)


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Exercise 1.1.3

1. A particle is projected from a point O at time t = 0 with speed V and angle of elevation α . It

moves under gravity and reaches its horizontal range R at time t = T . If the direction of motion

of the particle makes an angle β with the horizontal when t = 14T , show that tan β = 1

2tanα.

Solution: After seconds the particle is at the position:

x(t) = V cosα.t, . . . (1)

y(t) = V sinα.t− gt2

2, . . . (2)

tan β = y′x =yx, hence from (1) and (2) we get tan β = V sinα−gt

V cosα,

where t = T4⇒ tan β = tanα− g

V cosα× T

4;

y = 0, t = T ⇒ from (2) T4= V sinα

2g.

And hence tan β = tanα− gV cosα

× V sinα2g

= tanα− 12tanα = 1

2tanα.

2. A and B are two points on level ground, 40m apart. Simultaneously a particle is projected from

A towards B and another particle is projected from B towards A, each with speed 20ms−1 at an

angle of elevation of 45◦ . Given that the two particles collide, find the time and the height above

AB at which this occurs.

Solution: O is the centre of AB, i.e. AO = OB = 20m

After t seconds the particle projected from A is at he position:

x = −AO + V cosα.t . . . (1)

y = V sinα− gt2

2. . . (2)

Because of the symmetry of conditions of the problem the particle collide over

the point O.

x = 0, ⇒ from (1) t = AOV cosα

⇒ t = 2020 cos 45◦

=√2 seconds.

t =√2 ⇒ from (2) y = 20× sin 45◦ ×

√2 ⇒ y = (20− g) metres.


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Exercise 1.1.4

1. A particle is projected under gravity horizontally with the speed 30ms−1 from a point B 45m

vertically above a point O on horizontal ground. Taking g = 10ms−2, find the time taken for the

particle to reach the ground and the horizontal distance it has then traveled.

Solution: Initial conditions: t = 0, x = 0, y = h, x = U = 30ms−2, y = 0, h = 45m.

After t seconds the particle are at position:

x = Ut . . . (1)

y = h− gt2

2. . . (2)

As B reaches the ground, y = 0 ⇒ from (2) t =(

2hg

) 12

∴ t =(2×4510

) 12 = 3 seconds.

Let R be the horizontal distance travelled by B.

Hence when t = 3, ⇒ x = R,

From (1) ⇒ R = 3× U ⇒ R = 3× 30 = 90 metres.

2. A projectile is fired with speed V at an angle of elevation α from a point O and hits a stationary

target at a distance d from O on the same level. Find the value of V .

Solution: Equation of motion:

x = V cosα.t . . . (1)

y = V sinα.t− gt2

2. . . (2)

Let the time of collision be T. Hence:

t = T, x = d ⇒ from(1) d = V cosα× T ⇒ T = dV cosα

.

t = T, y = 0, ⇒ from (2) 0 = V sinα× T − gT 2

2⇒ V = gT

2 sinα.

∴ V = g2 sinα× ×

dV cosα

=(

dgsin 2α

) 12


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Exercise 1.1.51. A projectile is fired with speed V at an angle of elevation α from a point O. At the instant of

projection the target is fired from a point at a distance d from O on the same level with speed u

and angle of elevation β in the plane of the path of the projectile and away from O. Given that

the projectile hits the target, find the time at which this occurs.

Solution: Equation of motion:

Projectile

{x1 = V cosα.t (1)

y1 = V sinα.t− gt2

2(2)

target

{x2 = U cos β.t+ d (3)

y2 = U sin β.t− gt2

2(4)

At the time of collision the coordinates of the projectile and teh target are equal.

Hence x1 = x2 from (1) and (3) V cosα.t = U cos βt+ d

∴ t = dV cosα−U cosβ

; (5)

y1 = y2 from (2) and (4) V sinα = U sin β ⇒ V = U sinβsinα

.

Substituting this into (5) we obtain: t = d sinαU(cosα sinβ−cosβ sinα)

∴ t = d sinαU× 1

sin(β−α)

2. A and B are two points on level ground 110m apart. A particle is projected from A towards

B with speed 60ms−1 at an angle of elevation of 30◦. At the same instant another particle is

projected from B towards A with speed 50m−1 . Given that the two particles collide, find the

angle of projection of the second particle and the time of collision.

Solution: Initial conditions: t = 0, AB ≡ d = 100, u = 60, α = π6, V = 50, β =?

Particle from A:

{x2 = d, x2 = −V cosβ

y2 = 0, y2 = −V sinβ.Particle from B:

{x1 = 0, x1 = u cosα

y1 = 0, y1 = u sinα

After t seconds, the particles are at positions:

{x1 = u cosα.t x2 = d− V cosβ.t

y1 = u sinα.t− gt2

2 V sinβ.t− gt2

2When the particle collide their coordinates are equal. Hence x1 = x2 and y1 = y2.

∴ u sinα = V sinβ ⇒ sinβ = uV sinα ⇒ sinβ = 6

5 sinπ6 ⇒ sinβ = 3

5 , cosβ = 45

∴ tanβ =3545

⇒ tanβ = 34 ⇒ β = tan−1 3

4 .

x1 = x2 ⇒ (u cosα+ V cosβ)t = d ⇒ t = 110

60×√3

2+50× 4

5

⇒ t = 113√3+4

∴ t = 113√3+4× 3√3−4

3√3−4 = 3

√3− 4 seconds.


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1.2 Circular Motion in a Horizontal Plane

Definition:

• The linear velocity of P is a vector −→v in the direction of motion and its directionalong the tangent at P.

• If ` is the arc length from the x-axis to P, then v = d`dt

= ddt(rθ) = r dθ

dt= rω

• The linear velocity −→v , measured in ms−1,

• angular velocity ω, measured in radians s−1.

Tension forces: When particles are connected by a taut string, the string exerts a tensionforce on each particle.

Reaction forces:

• A body P in contact with a surface exerts a force on the surface and the surface exertsan equal and opposite force on P.

• These forces are termed an action-reaction pair and the force on P by the surface iscalled a reaction force.

• The reaction force acts at right angles to the surface.

Example 1.2.1 A mass of 2 kg is revolving at the end of a string 2m long on a smooth horizontaltable with uniform angular speed of 1 revolution per second. Find the tension in the string.

Solution: Observed acceleration is a = `ω2,

whereω = 2π rad s−1, and ` = 2m is the length of a string.

The resultant has horizontal component ma = m`ω2 ⇒ T = m`ω2.

m = 2, ` = 2, ω = 2π, ⇒ T = 16π2 N.


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1.2.1 Uniform circular motion

If P is moving in a circle of radius r with constant speed v and constant angularvelocity ω, the motion is described as uniform circular motion and the followingresults apply:

• Linear velocity is v = rω, directed along the tangent.

• Linear acceleration is a = rω2, directed towards the centre.

• the vector sum of all the physical forces on P (the resultant force) is:F = mrω2, ⇒ F = mv2

rdirected towards the centre.

• If f is the number of revolutions per second (frequency) and T is the time taken forone revolution (period), ω = 2πf = 2π

T.

Exercise 1.2.1

1. A particle of mass mkg is travelling at constant speed v ms−1 round a circle of radius rm .

If v = 8 and r = 2 , find the magnitude of the linear acceleration.

Solution: a = v2

ris obtained acceleration.

v = 8, r = 2 ⇒ a = 82

2a = 32ms−2.

2. A particle of mass 0.5 kg is attached to one end of a light inextensible string of length 2m. The

other end is fixed to a point A on a smooth horizontal table. The particle is set in motion in a

circular path. If the speed of the particle is 12ms−1 , find the tension in the string.

Solution: The resultant force is mv2

`to A where ` = 2m

Let T be the tension in the sting.

∴ T = mv2

`T = 0.5×122

2= 36N.


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Exercise 1.2.2

1. A particle of mass 0.1 kg moving on a smooth horizontal table with constant speed v ms−1 de-

scribes a circle with center O and radius rm . The particle is attracted towards O by a force of

magnitude4v N and repelled from O by a force of magnitude krN where k is a constant. Given

that v = 40 and the time of one revolution is π10

seconds, find the values of r and k.

Solution: Forces on the particle: Where T1 = 4v, T2 = kr, k > 0. If t = π

10is

the time of one revolution,

Then the angular velocity ω = 2πt⇒ ω = 20 rad s−1.

v = ω.r and v = 40 ⇒ r = vω⇒ r = 40

20= 2m.

The resultant force on the particle is T1 − T2 = 4v − kr

to O,

Hence 4v − kr= mv2

r⇒ k = 4vr −mv2;

v = 40, r = 2, m = 0.1, ⇒ k = 4× 40× 2− 0.1× 402 = 160N .

2. A particle P of mass 0.2 kg moving on a smooth horizontal table with constant speed v ms−1

describes a circle with center O such that OP = rm . The particle is subject to two forces, one

towardsO with magnitude 8v N and one away fromO with magnitude kr2N , where k is a positive

constant. Given that k = 75 and r = 1 , find the possible values of v.

Solution: Forces on the particle P : F1 = 8v, F2 = kr2, OP = r. Observed

acceleration is a = v2

r.

Hence the vector sum of force is ma = mv2

r, and directed of O.

The horizontal component of the resultant force is mv2

r⇒ F1 − F2 =

mv2

r.

⇒ 8v − kr2

= mv2

r⇒ v2 − 8r

mv + k

mr= 0 ⇒ V± = 4r

m±√

(4rm)2 − km

r;

k = 75, r = 1, m = 0.2 ⇒ v+ = 25 and v− = 15.


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Exercise 1.2.3

1. A particle of mass mkg is travelling at constant speed v ms−1 round a circle of radius rm. If

v = 3, r = 6, and the force acting towards the centre of the circle is of constant magnitude 6N ,

find the value of m.

Solution: The observed acceleration is: a = v2

r.

The resultant force F is ma = mv2

r. Hence F = mv2

r⇒ m = rF

v2

So we have v = 3, r = 6, F = 6, ⇒ m = 6×632⇒ m = 4 kg.

2. A particle of mass 0.25 kg is attached to one end of a light inextensible string of length ` = 0.5m.

The other end is fixed to a point A on a smooth horizontal table. The particle is set in motion in a

circular path. If the speed of the particle is 8ms−1 , find the tension in the string and the reaction

with the table.

Solution: The force on P is shown above, here T is the tension in the inextensible string

whose length is ` = 0.5m, and N is the reaction force.

Observed acceleration is: a = v2

`.

Hence the vector sum of the force o P is mv2

`towards A.

The resultant has vertical component zeroN = mg; m = 0.25 ⇒ N = 14g N .

The resultant has horizontal component ma = mv2

`⇒ T = mv2

`;

m = 0.25, ` = 0.5, v = 8, ⇒ ∴ T = 0.25×820.5

= 32N.

3. A particle of mass 0.5 kg is attached to one end of a light inextensible string of length 2m. The


circular path. If the string breaks when the tension in it exceeds 64N , find the greatest speed at

which the particle can travel.

Solution: T = mv2

`, ⇒ mv2

`= 64

∴ v =(64×`m

) 12 ⇒ v =

(64×20.5

) 12 = 16ms−1.


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Exercise 1.2.4

1. A mass of 2 kg is revolving at the end of a string 2m long on a smooth horizontal table with

uniform angular speed. If the string would break under a tension equal to the weight of 20 kg,

find the greatest positive speed of the mass.

Solution: Observed acceleration is: a = `ω2,

where ω = 2π rad s−1, and ` = 2m is the length of a string.

The resultant has horizontal component ma = m`ω2 = mv2

`⇒ T = mv2

`;

The string breaks if T > 20 g. As T = mv2

`we have mv2

`≤ 20 g

∴ v ≤(20×g×`m

) 12 , ` = 2, m = 2, ⇒ v ≤ (20g)

12 ms−1.

2. A particle P of mass 0.2 kg moving on a smooth horizontal table with constant speed describes

a circle with centre O such that . The particle is subject to two forces, one towards O with

magnitude 8v N an one away from O with magnitude N, where k is a positive constant. If , find

the set of possible values of k.

Solution: The force on P shown above: F1 = 8v, F2 =kr2, OP = r.

Observed acceleration is a = v2

r

Hence the vector sum of forces is: ma = mv2

rand directed towards O.

The horizontal component if the resultant force is:mv2

r= F1 − F2 ⇒ 8v − k

r2= mv2

r

∴ k = 8vr2 −mv2r; r = 1, m = 0.2, ⇒ k = 8v − 0.2v2

The function k(v) = 8v − 0.2v2 has the derivative k′(v) = 8− 0.4v.

Hence k′(v) = 0 ⇒ v = 20 and k”(v) = −0.4 < 0

⇒ At the point v = 20 the function k(v) has its maximum value k(20) = 80.

Therefore 0 ≤ k ≤ 80.


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Exercise 1.2.5

1. A particle of mass 0.1 kg moving on a smooth horizontal table with constant speed v ms−1

describes a circle with centre O and radius rm. The particle is attracted towards O by a force of

magnitude 4v N and repelled from O by a force of magnitude krN , where k is a constant. Given

that k = 30 and r = 1, find the possible values of v.

Solution: Forces on the particle P are show above, where T1 = 4v, T2 =kr, k > 0

The resultant force on the particle is T1 − T2 = 4v − kr

towards O.

Hence 4v = −kr= mv2

r⇒ v2 − 4r

mv + k

m= 0;

k = 30, r = 1, m = 0.1 ⇒ v2 − 40v + 300 = 0, ⇒ v = 20±√400− 300;

∴ v = 30ms−1 or v = 10ms−1.

2. A particle moves with constant angular velocity ω in a horizontal circle of radius r on the inside

of a fixed smooth hemispherical bowl of internal radius 2r. Show that ω2 = g

r√3.

Solution: P performs uniform circular motion about C,

hence the resultant force is directed towards C.

The resultant has a vertical component zero⇒ N = cos θ = mg . . . (1)

The resultant has a horizontal component mrω2 ⇒ N sin θ = mrω2 . . . (2)

Dividing (2) by (1), rω2

g= tan θ.

But tan θ = r√r42−r2 = 1√

3⇒ ω2 = g

r√3.


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Exercise 1.2.6

1. A particle of mass 0.25 kg is attached to one end of light inextensible string of length 0.5m. The


circular path. If the string breaks when the tension in it exceeds 50N , find the greatest angular

velocity at which the particle can travel.

Solution: Forces on P shown above:

here T is the tension in the inextensible string whose length is ` = 0.5m,

and N is the reaction force.

Observed acceleration is a = v2

`.

Hence the vector sum of forces on P is mv2

`towards A.

The resultant has horizontal component ma = mv2

`⇒ T = mv2

`;

The string breaks if T > 50N

Hence from mv2

`≤ 50 ⇒ v ≤

(50−`m

) 12 .

But ω = v`⇒ ω ≤

(50m−`

) 12 ;

m = 0.25, ` = 0.5 ⇒ ω ≤ 20 rad s−1.

2. A mass of 1 kg is fastened by a string of length 1m to a point 0.5m above a smooth horizontal

table and is describing a circle on the table with uniform angular speed of 1 revolution in 2

seconds. Find the force exerted on the table and the tension in the string.

Solution: Observed acceleration is a = rω2, where r =√`2 − h2.

Hence the vector sum of forces on P is ma = mrω2.

The resultant has a horizontal component mrω2 ⇒ T sin θ = mrω2

⇒ T = mrω2

sin θ. But sin θ = r

`⇒ T = mω2` . . . (1)

m = 1, ` = 1, ω = π ⇒ T = π2N

The resultant has a vertical component zero⇒ T cos θ +N = mg,

from (1) N = mg −mω2` cos θ.

But cos θ = h`⇒ N = mg −mω2h; m = 1, h = 0.5, ω = π ⇒ N = g − π2

2N.


student name grade centre components of the velocity vector of the faster particle at q, when t= 2 g...

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