Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Single and double projectiles. · Velocity diagrams of simple mechanisms. · Elastic and non-elastic collisions · Stepped rope and flywheel systems. · Angular momentum and impulse. · Clutches. · Balancing of rotating masses. · Governors
Typology: Lecture notes
1 / 115
This page cannot be seen from the preview
Don't miss anything!
Summary of Formulae
linear motion angular motion u + v = average velocity = ωωωω 1 + ωωωω 2 2 2 s = ( u + v) x t 2
θ = ( ωωωω 1 + ωωωω 2 ) x t 2 a = v – u = acceleration = αααα = ωωωω 2 - ωωωω 1 t t
s = ut + ½ at^2 θ (^) = ωωωω t + ½ αααα t 2
v^2 = u^2 + 2as (^) ωωωω 22 = ωωωω 12 + 2 αααα θ
S = θ x r
v = ωωωω x r
a = αααα x r
account the acceleration due to gravity. This, if we use our standard convention, means that the vertical acceleration will be –9.81m/s^2.
Since the acceleration is constant in both cases (constantly zero in the case of the horizontal motion), then the standard equations of linear motion can be applied. Remembering that we are dealing with vector quantities, and that the symbol ‘s’ is displacement, rather than distance travelled, we can make the following observations about projectiles.
There are a number of things we can deduce about the motion of the projectile. Considering the vertical motion:
Initial vertical acceleration = -9.81m/s^2
θθθθ
At maximum height , the projectile is no longer moving upwards, but has not started moving downwards. Its’ vertical velocity is zero. When the projectile returns to the same height it left from , the vertical displacement Sv = ZERO.
Using this information, it can easily be proved that: The time to reach maximum height is half the total time of flight ‘T ’. When the projectile returns to the same height it left from its vertical velocity is equal in magnitude but opposite in direction to its’ initial vertical velocity.
Considering the horizontal motion:
horizontal velocity. The horizontal displacement is thus simply calculated by multiplying the constant horizontal velocity by the total time of flight, T.
We can combine the constant horizontal velocity with the final vertical velocity to give the final total velocity. Since their magnitudes do not change if the projectile returns to the same height, then under this condition, the magnitude of the final velocity is the same as that of the initial velocity, and the projectile returns to the ground at the same angle that it left from.
v^2 = 12^2 - 2 x 9.81 x 0
v^2 = 144
v = - 12 m/s
Note that we are used to just saying that the square root of 144 is +12. However, it is equally true that the square root of 144 is –12. Here, we choose the solution that v = -12 m/s because the projectile is travelling…….. That’s right, downwards.
Example 2:- A ball is thrown vertically up in the air from the edge of a cliff 24.5 m high, so that when it falls down again it just misses the edge and falls on to the base of the cliff. If the initial vertical velocity was 19.6 m/s, how long will it take the ball to reach the foot of the cliff, and what is its velocity when it does so?
Solution
u = + 19.6 m/s, a = g = - 9-81 m/s^2 ,
s = - 24.5 m (ball ends up 24.5 m below starting point).
s = ut + ½at^2
-24.5 = 19.6 t - ½ x 9.81 x t^2
4.905t^2 - 19.6t - 24.5 = 0
t^2 - 4t - 5 = 0
(t - 5) (t + 1) = 0
t = 5 or -1 seconds
The negative answer is not applicable, (the ball cannot reach the ground before it has been thrown) so time, t = 5 seconds.
v = u + at
= 19.6 - 9.81 x 5 m/s
= 19.6 - 49.05 m/s
= - 29.45 m/s
The negative value indicates a downwards velocity
Note that many students given this question divide the travel up into three or two parts. There is no need to do this! We can consider the total journey in one movement, remembering that the value of ‘s’ in the formulae is that of displacement , which in this case is the height of the cliffs, 24,5m and negative because it is downwards.
Now lets try an example where we have both vertical and horizontal motion.
Example 3:- A stone is thrown upwards at 24.5 m/s in a direction inclined at 53° to the horizontal. find; (a) the time the stone is in the air (b) the greatest height reached (c) the horizontal distance covered (d) the angle at which it hits the ground.
The horizontal distance covered, will be given by the distance travelled at the constant value of the horizontal component of velocity (14.7 m/s), for the time the stone is in the air (4 seconds).
s = ut + ½at^2 ( or we could just use s = v.t. since velocity is constant, and acceleration is zero)
s = 14.7 x 4 + 0 m (horizontal acceleration = zero)
s = 58.8 m
The vertical component of velocity on landing is given from
v = u + at
= 19.6 - 9.81 x 4
= 19.6 - 39.
= - 19.6 m/s
i.e. equal and opposite to the initial vertical velocity,
The horizontal component will be constant throughout at 14.7 m/s, so the components of the velocity of the stone when it hits the ground are as shown.
The resultant velocity is seen to be 24.5 m/s at an angle of 53° to the horizontal, directed towards the ground (a mirror image of the initial velocity).
Example 4
A shell is fired horizontally from the top of a cliff towards the sea, the surface is 19.6 m below the cliff top. If the muzzle velocity of the shell is 1200 m/s, how far from the base of the cliff will the shell strike the surface of the sea.
Solution. First, draw a sketch.
Consider the vertical motion:-
u = 0, a = g = - 9.81 m/s^2 , s = - 19.6 m
-19.6m
1 A body is projected vertically upwards at a velocity of 320 m/s. Find the height attained and the total time taken to return to its initial position.
Ans: 5219 m 65.24 s
2 A body is projected vertically upwards at 40 m/s whilst at the same time another body is allowed to fall freely from a height of 70 m. Determine the height above ground at which they meet and the time taken.
Ans: 54.98 m 1.75 s
3 A shell is fired vertically upwards with an initial velocity of 60 m/s. Later, another shell is fired with an initial velocity of 100 m/s and after 2 s in flight passes the first shell. Calculate the time of flight of the first shell.
Ans: 6.8 s or 5.3 s
4 A projectile is fired with an initial velocity of 100 m/s at an angle of 30o^ to the horizontal. Determine its velocity, in magnitude and direction, after 2 s and 4 s respectively.
Ans: 92 m/s at 19.3o^ to the horizontal 87.3 m/s at 7.1o^ to the horizontal
To find the range of the projectile
Example. 5: A shell is fired at 30° to the horizontal with velocity 20 m/s from the top of a cliff 50 m high. Calculate the distance from the foot of the cliff to the point where the shell strikes the ground.
Initial velocity V 1 ;
Horizontal component = V 1 cosθ = 20 cos30 = 17.32 m/s
Vertical component = V 1 sinθ =20sin30 = 10 m/s
Time taken for shell to travel from E to B then fall to F
At B v 2 = v 1 + at = 0 = 10 - 9.81 t t = l.02s
It should be noted that the trajectory, i.e. the path of a particle projected from a point A on a horizontal plane and striking the plane at some point B, is symmetrical. The time of flight from A to B is twice the time taken by the particle to reach the highest point T. Hence the time of flight may be calculated by doubling the time taken to reach the highest point of the path. At the highest point T, the vertical velocity is zero and the particle is instantaneously travelling horizontally.
The direction of motion at any other time is found by considering the horizontal and vertical components of the velocity. During the first half of the motion, the particle is travelling at an angle above the horizontal and in the second half, below the horizontal.
General results Certain standard results can be established regarding the motion of a particle which is projected from a point 0 on a horizontal plane, with an initial velocity of U at an angle θ above the horizontal.
The following examples illustrate how this may be done.
Example
Find the time of flight T and range R on the horizontal plane.
Consider the motion of the particle from A to B.
horizontal motion vertical motion u 1 = Ucosθ u 1 = Usinθ s = R s = 0 t = T t = T no acceleration a = g (+ve down) =-g(-ve up)
using s = u 1 t + ½(at^2 ) for vertical motion: 0 = (Usinθ)T –1/2 (gT^2 ) ∴∴∴∴ T = 0 at A or T = 2Usinθ at B g using distance = velocity x time, for horizontal motion: R = (Ucosθ)T = 2U^2 sinθ cosθ g and using 2 sinθ cosθ =sin2θ (^) ∴∴∴∴ R = U^2 sin2θ g
The time of flight is 2Usinθ and the range is U^2 sin2θ g g
R t = T
θ
Projectile test
http://www.upscale.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/TwoBallsGravity/TwoBal lsGravity.html
g positive
Angle θ positive clockwise
The following section is just to remind you about angular motion. You will need to know the relationship between linear and angular motion, and how to convert between revolutions, radians and degrees. If you are already familiar with this work, you may like to move on to the next section on Resultant and Relative velocity.
May be measured in units of "DEGREES", RADIANS or REVOLUTIONS:
Note! Arc length = Angle (in radians) x Radius
For a circle, the arc length is 2πR meters and the angle is 360^0. 2 πR (m) 3600 Angle (in Radians) = = 2π radians R (m)
Hence 1 radian is approximately 57.3^0. Note also that the metres on the top and bottom of this equation cancel out, so radians are dimensionless
Angular Displacement (θ) [theta] Basic unit is the radian.
Angular Velocity (ω) [omega] The basic units of angular velocity are radians/second (rad/s). Often revolutions are used in either revs/min or revs/sec. Each revolution is worth 2π radians or 3600
Note! For N revs/min
2 π N Angular velocity ω = rads/s 60
Angular Acceleration (α) [alpha] This is the rate of change of angular velocity, where the units are radians per second^2 or rads/s^2.
