Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 05

4.3 Uniform Circular Motion 874.3 Uniform Circular Motion A particle that moves around in a circle with a constant speed, like the car shown in Fig.4.13a, is said to experience a uniform

4.2 Projectile Motion 85

Example 4.4

A ball thrown from the top of a building has an initial speed of 20 m/s at an angle

of 30◦above the horizontal The building is 40 m high and the ball takes time t

before hitting the ground, see the Fig.4.12 Take g = 10 m/s2 (a) Find the time

t1for the ball to reach its highest point (b) How high will it rise? (c) How long

will it take to return to the level of the thrower? (d) Find the time of flight t (e)

What is the horizontal distance covered by the ball during this time? (f) What isthe velocity of the ball before striking the ground?

H = (10 m/s) × (1 s) −1

2(10 m/s2) × (1 s)2= 5 mThus, the maximum height of the ball from the ground is 45 m.

(c) When the ball returns to the level of the thrower, the y coordinate is zero again, i.e y = 0, and t = T To find the time T the ball takes to reach this location,

We can also use T = 2t1= 2 s for the symmetric part of the path

(d) To find t, we can use y = vy◦t−1

2gt2with y = −40 m and v y◦= 10 m/s

so that (after omitting the units):

−40 = 10 t− 5 t2 ⇒ 5 t2− 10 t− 40 = 0Solving this quadratic equation yields:

We reject the negative time and take only the positive root, i.e t= 4 s.

(e) The horizontal distance x covered by the ball at t= 4 s is:

x = (v◦cosθ◦) t= v x◦t

= (17.32 m/s)(4 s) = 69.28 m (f) The vertical component of the ball’s velocity at t= 4 s is given by:

The direction of →v at t= 4 s is indicated in the Fig 4.12by the angleθ Thus,

according to this figure we have:

4.3 Uniform Circular Motion 87

4.3 Uniform Circular Motion

A particle that moves around in a circle with a constant speed, like the car shown

in Fig.4.13a, is said to experience a uniform circular motion In this case, the

accel-eration arises only from the change in the direction of the velocity vector

We can use Fig.4.13b to find the magnitude and direction of this acceleration

In this figure, the particle is seen first at point P with velocity→vi at time ti and at

point Q with velocity v→ fat time tf, where→viandv→ f are different only in direction,i.e vi = vf = v In order to calculate the acceleration we start with the average

where v can be accomplished graphically as shown in Fig.→ 4.13c

The triangle OPQ in Fig.4.13b, which has sidess and r, is similar to the triangle

of Fig.4.13c, which has sides v and v This similarity enables us to write the

Fig 4.13 (a) Circular motion with constant speed v (b) Velocity vectors → vi and →vf at P and Q.

(c) Graphical method to obtain→ v

Whent is very small, the two points P and Q of Fig.4.8b becomes extremelyclose, and hences and θ are very small too In this limit,  v would point toward→

the center of the circular path, and because the acceleration is in the direction of→v ,

it will also be toward the center Consequently, in this limit the arc PQ (P Q = r θ)

will be equal tos and the ratio s/t approaches the speed v Thus, when t → 0,

the magnitude of the radial acceleration will be:

Fig 4.14 Directional change

of the velocity and

acceleration vectors in uniform

circular motion, where both

have constant magnitude

In addition, during this circular motion with constant speed, the particle travels

the circumference of the circle in a time T giving by:

T = 2πr

where T is called the period of revolution, or simply the period.

4.3 Uniform Circular Motion 89

Example 4.5

A satellite is circulating the Earth at an altitude h= 150 km above its surface,

where the free fall acceleration g is 9 4 m/s2 The Earth’s radius is 6.4 × 106m.

What is the orbital Speed and period of the satellite?

Solution: As shown in Fig.4.15, the radius of the satellite’s circular motion equals

the sum of the Earth’s radius R and the altitude h, i.e.

Earth

R

By using the centripetal acceleration given by Eq.4.29, we find that the magnitude

of the satellite’s acceleration can be written as:

ar=v2

r = v2

R + h

For the uniform circular motion of the satellite around the Earth, the satellite’s

centripetal acceleration is then equal to the free fall acceleration g at this altitude.

That is:

ar= g = 9.4 m/s2From the preceding two equations we have:

g= v2

R + h

Solving forv and taking the positive root gives:

v =g (R + h)

=



(9.4 m/s2)(6.4 × 106m+ 150 × 103m) = 7,847 m/s ≈ 28,000 km/h

With this high speed, the satellite would take T = 2πr/v = 1.46 h to make one

complete revolution around the Earth

4.4 Tangential and Radial Acceleration

When the velocity of a particle changes in both direction and magnitude, the particlecan move in a curved path as shown in Fig.4.16 In this situation, the velocityv→

is always tangent to the path and usually the acceleration→a makes an angle with

the velocity The vector→a can be resolved into two component vectors: a tangential

Fig 4.16 When the velocity →v of a particle changes in both direction and magnitude, the acceleration

→a can be decomposed to a radial component vector →arand a tangential component vector →at

The tangential acceleration at a particular point arises from the time rate of the speed

of the particle and has a magnitude given by:

at= d v

4.4 Tangential and Radial Acceleration 91

The radial acceleration at a particular point arises from the time rate of change in

the direction of the velocity vector and has a magnitude:

In the case of uniform circular motion, wherev is constant, we have at= dv/dt = 0

and acceleration is always radial, i.e.→a =→ar Furthermore, if the direction of the

velocity→v does not change, then ar= 0 and the motion will be in one dimension,i.e.→a =a→t.

4.5 Non-uniform Circular Motion

A particle that moves around a circle with a variable speed, is said to experience

a non-uniform circular motion In this case, the total acceleration arises from the

change in magnitude ofv (represented by→ →at) and the change in direction of v→

θ is a unit vector tangent to the circular path in the direction of increasing

θ (measured in a counterclockwise sense from the positive x axis).

Using this notation, we can write the particle’s total acceleration→a as:

Fig 4.17 (a) The acceleration

of a particle moving in a circle

with a tangential component

→atand a radial component →ar

directed toward the center of

the circle (b) Definitions of

the unit vectors ˆ →r and→θˆ

θ = 30◦with the vertical, as shown in Fig.4.18 At this instant, find its acceleration

in terms of tangential and radial components

Fig 4.18

ar

ata

Solution: When the cord makes an angleθ to the vertical line, the component of

the gravitational accelerationg→that is tangent to the circular path has a magnitude

g sin θ Thus the magnitude of the tangential acceleration is:

at= g sin θ = (9.8 m/s2)(sin 30◦) = 4.9 m/s2Since the speed of the sphere at this instant isv = 2 m/s and the radius of the circle that the sphere swings about equals the length of the cord, i.e r = L = 1 m, then

the magnitude of the radial acceleration is:

4.5 Non-uniform Circular Motion 93

ar= v2

r = (2 m/s)2

1 m = 4 m/s2

Therefore, T − mg cos θ = mar, where T is the cord’s tension From the relation

Eq.4.34we can find the magnitude of a at θ = 30◦as follows:

a=



(4.9 m/s2)2+ (4 m/s2)2= 6.32 m/s2The angleφ between the vector→a and the cord will be:

Section 4.1 Position, Displacement,Velocity, and Acceleration Vectors

(1) The initial position vector of a butterfly can be described in unit vectors by

→

ri= 3→i − 7→j + 4→k, and five seconds later by→rf= −2→i + 3→j −→k (allunits in meters) (a) What is the butterfly’s displacement vector? (b) What isthe butterfly’s average velocity?

(2) The position vector of a particle moving in two dimensions is given by→r =

x(t)→i + y(t)→j , where x(t) = 2 t + 1, y(t) = 2 t2, t is the time in seconds,

and all numerical coefficients have the proper units so that→r is in meters.

(a) Find the average velocity vector during the time interval from t = 0 to

t = 2 s (b) Find the particle’s velocity vector→v as a function of time, and find its magnitude and direction at t = 2 s (c) Find the particle’s acceleration

vector→a .

(3) The position vector of a particle moving in two dimensions is described by

→

r = (4 t3− 12 t + 9)→i + (6 t + 4)→j , where t is the time in seconds and

all numerical coefficients have the proper units so that →r is in meters.

(a) Find the average velocity vector between t = 1 s and t = 2 s (b) Find the

particle’s instantaneous velocity vector→v as a function of time, and then find its magnitude and direction at t = 1 s (c) Find the velocity and the speed of the particle at t = 3 s (d) Find the average acceleration vector between t = 1 s and t = 2 s (e) Find the instantaneous acceleration vector a→as a function of

time, and find its magnitude and direction at t = 2 s (f) Find the time at which

the x component of the particle’s displacement vector is at a relative

maxi-mum/minimum, showing how you can determine whether it is a maximum or

a minimum

(4) The position vector of a particle moving in three dimensions is given by

→

r = 3 t →i − 2t2 →j + 2→k, where t is the time in seconds and all numerical

coefficients have the proper units so that→r is in meters (a) Find the magnitude

of the particle’s position vector→r as a function of time, and then find its value

when t = 2 s (b) Find the particle’s velocity vector→v as a function of time, and find its magnitude and direction at t = 2 s (c) Find the particle’s acceleration

vector→a as a function of time, and find its magnitude and direction at t = 2 s.

Section 4.2 Projectile Motion

(Neglect air resistance and take g= 10 m/s2in all projectile Exercises)

(5) A small ball is projected horizontally from a tall building with a speedv◦of

10 m/s, see Fig.4.19 Find its position and its velocity components after12s.

Fig 4.19 See Exercise (5)

x

° 0

(6) A student running with a constant speedv◦goes straight over the cliff shown

in Fig.4.20 The student is at a height h= 45 m above water level once he

leaves the cliff The student lands in the water at a point where x = 39 m.

(a) How fast was the student running when he jumped over the cliff? (b) What

is his speed and what is the angle of his impact with the water?

(7) A long jumper leaves a cliff atθ◦= 45◦above the horizontal with an initial

speedv◦ and lands 6 m away, see Fig.4.21 The cliff is at a height h= 2 mabove sea level (a) What is the speed of the jumper? (b) How long will it takehim to reach the water?

4.6 Exercises 95

Fig 4.20 See Exercise (6) y

x

x h

Fig 4.21 See Exercise (7)

(a) What must be the minimum speed of the jumper to reach the cliff? (b) Howlong will it take him to reach the cliff?

Fig 4.22 See Exercise (8)

x and y components of→v◦ (b) When t = 2 s, find the position of the ball and

the magnitude and direction of its velocity→v (c) What is the value of the H (the highest point of the ball’s trajectory) and how much time t1has elapsed forthe ball to reach that point? (d) Calculate the values of the total time of the

ball’s flight T and the horizontal range R.

Fig 4.23 See Exercise (9)

°

x t=T

H

x

° 0

j ) m/s, see Fig.4.24 (a) What are the vertical and horizontal velocity

com-ponents after time t= 1 s? (b) What are the time and the coordinates of theprojectile when it reaches the highest point? (c) What are the values of the total

time of the projectile’s flight T and the horizontal range R?

Fig 4.24 See Exercise (10)

(11) A ball is launched from the ground with an initial speedv◦of 40 m/s at an angle

θ◦= 60◦towards a cliff of height h, see Fig.4.25 The ball strikes the cliff after

5 s Find: (a) the height h of the cliff, (b) the maximum height H, (c) the speed

of impact, and (d) the horizontal distance between the cliff and the firing point

is 15 m and the vertical height of the tip of the ramp is 1.5 m above the cars.

(a) What is the minimum speed that he must drive off the horizontal ramp

of Fig.4.26a? (b) When the ramp is tilted upwards with angleθ◦= 8◦, as in

Fig.4.26b, what is the new minimum speed?

Fig 4.26 See Exercise (12)

(13) Figure4.27 shows a blasted chunk of a solid rock ejected from a volcano

It is safe to live at the foot of the volcano, but away from its center by 9 km (or

more), i.e x = 9 km when θ◦= 45◦and the speedv◦is maximum (a) At whatmaximum initial speed would a rock need to be ejected from the volcano’s

mouth and reach x= 9 km? (b) What would be its time of flight? (c) Does themaximum initial speed of part (a) increases or decreases when air resistance istaken into account?

(14) Figure4.28shows a fighter plane that has a speed v◦of 300 km/h, flying at

an angleθ◦= 15◦below the horizontal when a decoy rocket is released The

horizontal distance between the release point and the point where the decoy

strikes the ground is x = 600 m (a) How long was the decoy in the air?

(b) How high was the plane when the decoy was released?

Fig 4.27 See Exercise (13)

Fig 4.28 See Exercise (14)

0 y

(16) A girl sitting on a hill aims her cork slingshot at a boy hanging from a tree at

a horizontal distance L away and a vertical distance h up from the slingshot,

see Fig.4.29 At the instant the cork-projectile is fired, the boy releases himselffrom the tree, hoping to avoid being hit Ignoring air resistance, show that the

ycork = yboyand thus the cork collides with the boy

(17) An inclined plane makes an angleφ with the horizontal At the lowest point of

the incline, a projectile is fired with a speedv◦that makes an angleθ◦above thehorizontal, see Fig.4.30 At what angleθ◦does the range R along the incline

reach its maximum?

(18) A projectile is fired with a speedv◦that makes an angleθ◦above the horizontal.

Find the horizontal range R when the projectile is at a vertical position y = +h,

Section 4.3 Uniform Circular Motion

(20) Calculate the magnitude of the acceleration of a particle moving in a circle of

radius r = 0.5 m with a constant speed v of 10 m/s.

(21) A boy attaches a stone to the end of a rope of length r = 0.25 m, and rotates the

stone at a constant speed in a circular fashion Find the stone’s radial

acceler-ation when the period T is 2 s.

(22) As an approximation, assume the moon revolves about the Earth in a perfectly

circular orbit with a radius r = 3.85 × 108m and takes 27.3 days (which is

2.36 × 106s) to make a complete revolution, see Fig.4.32 (a) What is thespeed of the moon? (b) What is the magnitude of the radial acceleration of themoon toward the Earth’s center?

Fig 4.32 See Exercise (22)

(23) A car moves in a circle of radius r = 15 m with a constant speed v = 30 m/s,

see Fig.4.33 (a) What is the change in velocity (magnitude and direction) whenthe car goes around an arc ofθ = 60◦, as shown in the right part of Fig.4.33?

(b) What is the magnitude of the radial acceleration?

f

r r

s

π

Fig 4.33 See Exercise (23)

(24) In the model of the hydrogen atom proposed by Niels Bohr, an electron

circu-lates a stationary proton in a circle of radius r = 5.28 × 10−11m with a speed

-+

(25) A point P is located on the latitude that passes through Egypt’s soil at exactly

30◦N, and at a distance r = 6.4×106m away from Earth’s center, see Fig.4.35

As the Earth revolves about its axis, calculate the magnitude of the acceleration

of the point P.

Fig 4.35 See Exercise (25)

Top view

Rotation of P about the Earth's axis

gravity, g.

(27) A jet pilot performs a vertical loop when the speed of his aircraft is 1,200 km/h.

Find the smallest radius of the circle when the centripetal acceleration at the

lowest point does not exceed 6 g

Sections 4.4 and 4.5 Tangential and Radial Acceleration,

Non-uniform Circular Motion

(28) The speed of a particle moving in a circle of radius r = 4.5 m increases with a

constant rate of 2 m/s If at some instant, the magnitude of the total acceleration