chapter 6 of cambridge textbook math 4 unit PDF

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6 Mechanics Chapter Overview: Some practical applications of calculus in mechanics are considered in this chapter. Simple mathematical models enable solutions to be found in terms of the familiar functions of this course. Problems later in the chapter require some knowledge of Newton’s laws of motion, and that acceleration be written as a function of x. Preparation for that is done in Section 6A. Next, simple harmonic motion is presented, firstly in terms of time in Section 6B, and then in terms of displacement in Section 6C. The work in Sections 6D and 6E takes the first step in making the theory of motion more realistic by introducing a resistive force. Horizontal motion with friction, and vertical motion in a resisting medium with constant gravity are considered. In the following section, projectile motion is first reviewed and then extended to include a resistive force proportional to the velocity, based on the work done in 6E. The chapter concludes with a collection of various problems which either extend the theory learnt to new situations or are harder examples of applications of that theory. One specific example investigated is the simple harmonic motion approximation for a pendulum.

6A Forces and Acceleration In the problems encountered in this chapter, the equations of motion are either specified explicitly or given indirectly, such as by a balance of forces. Whilst this is not a course in physics, a basic understanding of the laws of motion is required.

Some Assumptions: A significant simplification is the assumption that an object may be represented by a point mass, often called a particle. If the scale of the motion is large compared with the object, such as in the case of a ball bearing thrown 5 m, then this assumption is reasonable. A second significant assumption is that air is an ideal fluid and is not particulate in nature. At low to medium speeds this is a satisfactory assumption. A third assumption is that the forces due to the orbit of the earth around the sun and due to the rotation of the earth on its axis are negligible in the problems being considered. By way of example, in the problem of projectile motion without air resistance, the acceleration due to gravity at the surface of the earth is about 9·8 m/s2. The acceleration due to the orbit of the earth is about 6 × 10 −4 m/s2, and at the equator the acceleration due to the rotation of the earth is about 0·034 m/s2. Thus this third assumption seems reasonable.

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6A Forces and Acceleration

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Newton’s Laws of Motion: The equations of motion encountered in this course are all derived from Sir Isaac Newton’s laws of motion, contained in his book Principia, published in 1686. It is written in Latin, the scientific language of the day. An early translation of the laws is as follows: Law I Every body continues in its state of rest or of uniform motion in a straight line except in so far as it be compelled by impressed force to change that state. Law II The rate of change of momentum is proportional to the impressed force and takes place in the direction in which the force acts. Law III To every action there is an equal and opposite reaction. In the second law, momentum is defined to be mv, the product of mass with velocity. The standard units used are kilograms, metres and seconds. The unit of force is called the newton, with 1 N = 1 kg m s−2 . If these SI units are used then, in the second law, the constant of proportionality is 1. Thus d F = (mv ) , dt and if the mass is constant then F = ma where a = v˙ . If other units are used then these equations must be appropriately modified. It may seem strange to state that the mass is constant, but in many cases the mass is certainly not constant, such as a rocket as its fuel is burnt. In this course, however, the mass is always assumed to be constant.

1

NEWTON’S SECOND LAW OF MOTION: When a force F newtons is applied to a constant mass m kg, which is free to move, the resulting acceleration is a m s−2 , where F = ma

In many situations it is simply a matter of integrating the force equation in order to find the other details of the motion.

WORKED EXAMPLE 1: A body of mass 4 kg is acted upon by a variable force F = 48(5 − t) newtons for 5 seconds. If the body starts from rest at x = 0 then what is its final velocity and how far has it travelled? SOLUTION: From Newton’s second law, after dividing through by the mass, dv = 60 − 12t . dt Integrating, v = 60t − 6t2 + C . At t = 0 the velocity is v = 0, so C = 0 and v = 60t − 6t2 . Hence at t = 5, v = 150 m/s. Integrating again gives the displacement: x = 30t2 − 2t3 + D . At t = 0 the body is at x = 0, so D = 0 and x = 30t2 − 2t3 . Hence at t = 5, x = 500 m.

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6A

Resolution of Forces: When two or more forces act on a body, the problem can often be solved by resolving the forces. Typically the forces are resolved horizontally and vertically to determine the equations of motion. Usually a force diagram, called a free-body diagram , is helpful at this step. The sum of the components can then be used in Newton’s second law.

WORKED EXAMPLE 2: In a conical pendulum, particle P of mass m hangs from a ceiling on a wire of length ℓ. The wire makes a constant angle α with the vertical and P moves around a circle with constant angular velocity ω . Let r be the radius of the circular motion and let h be the height of the cone traced out by the wire.

a l h P

r

(a) Draw a free-body diagram of the situation.

(b) The horizontal force required to keep the particle in circular motion is mrω 2, directed towards the axis of the cone. By resolving the forces horizontally and vertically, obtain an expression for ω in terms of ℓ, α and g . g (c) The period of the motion is 2πω . Show that when ℓ = π2 (about 0·994 m) the period is always less than 2 seconds.

SOLUTION: Let T be the tension in the wire. (a) The diagram is shown on the right.

T

(b) The vertical component of tension is T cos α. The only other vertical force is due to gravity and is −mg, the minus sign indicating a downwards force. As there is no vertical acceleration, the sum of the vertical forces is zero. Thus

a P mg

T cos α − mg = 0 or T cos α = mg (1) The horizontal component of tension keeps the particle in circular motion. So T sin α = mrω 2 Now in the cone sin α = rℓ so from equation (2)

(2)

T rℓ = mrω 2 or T = mℓω 2 Substitute this result into (1) to get mℓω 2 cos α = mg g so ω2 = ℓscos α or

ω=

g . ℓ cos α

(c) From the given formula for the period, s ℓ cos α period = 2π g √ = 2 cos α (when ℓ = πg2 .) Hence the the period is always less than 2 seconds, with period → 2− seconds as α → 0+ .

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6A Forces and Acceleration

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Other Forms of Acceleration: There are some problems where the equation of motion is more naturally expressed using displacement. For example, it is easy to measure the displacement x of one end of a spring when a mass m is hung from it. It is found that x is proportional to the force so the resulting equation is v˙ = −kx

x m

(for some constant k.)

In order to do anything meaningful with this equation it is necessary to rewrite the acceleration v˙ as a derivative in x instead of t. This is done as follows. dv dv dx = × (by the chain rule) dt dx dt dv ×v (by the definition of velocity) = dx dv dv =v . thus dt dx Putting this into the equation of motion for the spring gives dv v = −kx dx which is a variable separable differential equation.

WORKED EXAMPLE 3: A spring is hung from a ceiling and extended a distance a metres then released from rest. It is found that the equation of motion for the spring is x ¨ = −x, where x metres is the displacement of the end of the spring at time t seconds. Show that −a ≤ x ≤ a. dv SOLUTION: Replace the acceleration x ¨ with v and double to get dx dv 2v = −2x dx which is variable separable. Next integrate both sides to get v 2 = C − x2 for some constant C. But at t = 0, x = a and v = 0 so 0 = C − a2 thus C = a2 , and rearranging the equation gives x2 + v 2 = a2 . Graphing v against x, this is the equation of a circle, and hence −a ≤ x ≤ a. Note that this does not prove that the end of the spring ever reaches x = −a. That will be done later in this chapter. Notice that the equation was doubled in the first line of the solution above. This made the integration easier. There is another approach which is often used. d 2x dv =v 2 dx dt   dv dv 1 + v = 2 v dx dx d (by the product rule) = 12 (v × v) dx d 2x d  1 2 that is v . = 2 dx 2 dt

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WORKED EXAMPLE 4: A particle starts at the origin with velocity v = 1 m/s. It is found that its displacement x metres at time t seconds satisfies x ¨ = e−2x . (a) Find an expression for v 2 in terms of x. (b) Explain why the velocity must always be positive. √ (c) Hence show that v → 2 as t → ∞.

SOLUTION: (a) From above d  1 2 v = e−2x dx 2 d  2 v = 2e−2x so dx thus v 2 = C − e−2x for some constant C. Now apply the conditions v(0) = 1 and x(0) = 0. 1=C −1 so C =2 and hence v 2 = 2 − e−2x .

(b) Now at t = 0, x = 0 and v = 1, so the particle begins moving to the right. That is, x becomes positive and v is positive. In order for v to change sign, it must first be zero, which cannot happen for x ≥ 0. Hence v remains positive. (c) Since v 2 = 2 − e−2x , v ≥ 1 for all time and hence as t → ∞ so too x → ∞. Thus lim v 2 = lim v 2 t→∞

x→∞

= lim 2 − e−2x x→∞

= 2, √ and since v > 0, take the positive square root to get lim v = 2. t→∞

OTHER FORMS OF ACCELERATION: 2

The four common forms of acceleration are

d2 x d  1 2 dv dv v . = =v = 2 dx dx 2 dt dt

In each problem, choose the form most suitable for integration. d  1 2 v , integration will often dx 2 yield v 2 as a function of x, such as v 2 = a2 − x2 in Worked Example 3. Rewriting this equation:  2 dx = a2 − x2 , dt

Integrating Twice: When acceleration is written as

which is a non-linear first order differential equation. Such equations are usually too difficult to solve in this course, and no further progress is possible. If, however, the sign of v can be established then the appropriate square root can be taken and a second integration performed. The sign of v may be determined either mathematically (as in Worked Example 4) or, more typically, from the physics of the situation.

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WORKED EXAMPLE 5: Experiments suggest that acceleration due to gravity is inversely proportional to the square of the distance to the centre of the planet. Thus, if x is altitude and R is the radius of the earth then d 2x −k = 2 (x + R )2 dt

where k is a positive constant.

The negative sign indicates the acceleration is downwards. An object is dropped from x = R. Let g be the acceleration due to gravity at the surface of the earth. p (a) Show that the object hits the ground with speed Rg. (b) Find an expression for the time taken in terms of x, R and g .

SOLUTION: Given that x ¨ = −g at x = 0, it follows that −g = −kR−2 . 2 Hence k = R g . (a) Rewrite the given differential equation as d  2 v = −2k(x + R)−2 dx so v 2 = 2k(x + R )−1 + C Using the initial condition v = 0 at x = R , C = −kR −1 . k 2k − Hence v2 = x+R R k 2R − (x + R) = × x+ R  R R−x = Rg (from the value of k above.) R +x p Clearly at x = 0, v 2 = Rg and hence the speed is Rg .

(b) From the physical situation, the velocity is always negative. Thus s p R−x v = − Rg × R +x so taking reciprocals and rearranging yields s p R+x dt =− . Rg × dx R−x Z p R +x √ dx Integrating, Rg × t = − 2 2  Z R − x R x =− dx √ +√ 2 2 R − x2 R 2 − xp p so Rg t = −R sin−1 ( xR ) + R 2 − x2 + D . But x = R at t = 0 so 0 = −R sin−1 1 + D . and D = πR  s2    s 2 R π x x − sin−1 Hence t= + 1 − 2 . g 2 R r

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Exercise 6A 1. In each part, the velocity v is Express: (i) t in terms of x, (a) v = 6 (b) v = −6x−2

given as a function of x. It is known that x = 1 when t = 0. (ii) x in terms of t. (e) v = 1 + x2 (f ) v = cos2 x

(c) v = −6x3 (d) v = e−2x

2. In each motion of the previous question, find x ¨ using the result x ¨=

d 1 2 ( v ). dx 2

3. In each part, the acceleration x ¨ is given as a function of x. By replacing x ¨ with express v 2 in terms of x given that v = 0 when x = 0. 1 1 (a) x ¨ = 6x2 (b) x ¨= x (c) x ¨= 2x + 1 e

(d) x ¨=

d 1 2 ( v ), dx 2

1 4 + x2 dv , dt

4. In each part, the acceleration x ¨ is given as a function of v. By replacing x ¨ with express t in terms of v . 2 (a) x ¨ = 2 and when t = 0, v = 0 v (b) x ¨ = v 2 and when t = 0, v = 12

(c) x ¨ = 2 + v and when t = 0, v = 1

5. In each part, the acceleration x ¨ is given as a function of v. By replacing x ¨ with v

dv , dx

express x in terms of v . v2 (c) x ¨ = 2 + v and when x = 0, v = 0 and when x = 0, v = 1 (a) x ¨= 4 3 (b) x ¨ = and when x = 0, v = 6 v 6. A particle of mass m moves in a straight line subject to a force F . At time t, the displacement of the particle is x and the velocity is v. The particle was initially at rest at the origin. (a) If F = 6t − 4 and m = 2, find x when t = 4. (b) If F = 2x + 1 and m = 1·5, find the positive value of v when x = 3. 1 and m = 0·25, find t when v = 4. (c) If F = v +2 1 and m = 0·5, find x when v = 3. (d) If F = v +2 DEVEL OPMENT

7. Three forces act on an object of mass 2 kg. These forces are represented by the vectors 12i +23j , 9i − 7j and −5i +14j . Calculate the magnitude and direction of the acceleration e e e e e e of the object. 8. The diagram on the right shows two forces of magnitude −→ −−→ 20 N and 15 N represented by the vectors OA and OB. −→ −−→ (a) Express OA and OB as component vectors. (b) Calculate the magnitude of the resultant of the two forces, correct to the nearest newton. (c) Determine the direction of the resultant, correct to the nearest degree.

B

15N 54°

20N

A

32° O

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9. Tom can paddle his canoe with a force of 20 N. He starts paddling from a point on the south bank of a river and steers the canoe at 90◦ to the bank. He experiences a force of 6 N acting due east due to the current, and he also has to contend with a force of 4 N due to a breeze blowing from the north-east. (a) Express the resultant force on Tom and his canoe as a component vector. (b) Hence find the magnitude (in newtons to one decimal place) and direction (in degrees to one decimal place) of the resultant force. 10. [A formula from physics] A particle moves with constant acceleration a, so that its equation of motion is x ¨ = a. Its initial velocity is u. After t seconds, its velocity is v and its d 1 2 ( v ) for acceleration to show that v 2 = u2 + 2as. displacement is s. Use dx 2 11. A ball is thrown vertically upwards at 20 m/s. Taking g = 10m/s2, upwards as positive, and the ground as the origin of displacement, the equation of motion is then x ¨ = −10. (a) Show that v 2 = 400 − 20x, and find the greatest height. √ (b) Explain why v = 400 − 20x while the ball is rising. (c) Integrate to find the displacement–time function, and find how long it takes the ball to reach its greatest height. 12. Assume that a bullet, fired at 1 km/s, moves through the air with deceleration proportional to the square of the velocity, so that x ¨ = −kv 2 for some positive constant k . dv (a) If the velocity after 100 metres is 10 m/s, use x ¨=v to find x as a function of v, dx then find how far the bullet has travelled when its velocity is 1 m/s. dv to find at what time the bullet has (b) If the velocity after 1 second is 10 m/s, use x ¨= dt velocity 1 m/s. 13. A particle has acceleration x ¨ = e−x , and initially v = 2 and x = 0. Find v 2 as a function of x, and explain why v is always positive and at least 2. Then briefly explain what happens as time goes on. 14. A particle has velocity v = 6 − 2x, and initially the particle was at the origin. (a) Find the acceleration at the origin.   (b) Show that t = − 12 ln 1 − 31 x, and hence find x as a function of t. (c) Describe the behaviour of the particle as t → ∞.

15. An object is initially at rest at the origin. It moves in a straight line away from the origin with acceleration 2(1 + v) m/s2. Find, correct to 3 significant figures: (a) how long it takes for the velocity to reach 20 m/s, (b) the distance travelled when the velocity reaches 20 m/s. 16. A particle P of mass m starts from the origin O with velocity u and moves in a straight x line. When OP = x, where x ≥ 0, the velocity v of P is given by v = u + , where k is a k positive constant. (a) Prove that the force acting on P is at all times proportional to v, and state the constant of proportionality. (b) Given that the velocity is 3u at the point A, find: (i) the distance OA in terms of k and u. (ii) the time, in terms of k, taken by P to move from O to A.

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  17. A particle of mass 0·5 kg is acted upon by a force F = x − 12 newtons. Initially the particle is at rest 5 metres on the positive side of the origin. (a) Find v 2 in terms of x, and hence explain why the particle can never be at the origin. √ (b) Find where the speed of the particle is 2 5 m/s, justifying your answer, and describe the subsequent motion. 2 18. A particle of mass 2 kg √ is subject to a force of 6x newtons. Initially the particle is at x = 1 with velocity − 2 m/s. (a) Find v 2 as a function of x. (b) Then find the displacement–time function, and briefly describe the motion.

19. The accelera...