Mechanics Flashcards
Particle assumptions
- negligible dimensions, so forces will all act on a particle at the same point
- this means air resistance and rotational forces don’t act on the object (they can be ignored)
Rod/beam (3)
Long rigid particle, only 1 dimension isn’t negligible
* Mass is concentrated along a line
* No thickness
* Rigid (does not bend)
Thin
Negligible thickness
Rigid
Doesn’t bend
Smooth pulley
String is wwrapped around a disc (pulley)
* “Smooth” = no frictional force between the surfaces
* Tension in the string is the same on either side of the pulley
Rough
Frictional force acts during movement
Light (1/2)
Has no mass
A light string has the same tension throughout
Inextensible string
Doesn’t stretch - acceleration is constant
String = thin
Lamina
Flat, thin, 2D object
Mass is distributed across the surface
Lamina example
Sheet of paper
Wire
Thin, inextensible, 1D length of metal
Bead (2)
Particle with a hole in it so it can move on a wire or string.
* Moves freely along a wire or string
* Tension is the same on either side of the bead
Peg
Fixed support for another object
Dimensionless
The velocity of a particle is given by v = 3i +5j m s−1. Find:
a) the speed of the particle
b) the angle the direction of motion of the particle makes with the unit vector i.
a) speed = √(32 + 32) = 5.83m/s
b) acute angle made between the vector and the x-axis
tan(a) = 5/3 so a = 59°
Draw a diagram - the axes must meet at the starting location. This will show you the angle that needs to be found
The acceleration of a motorbike is given by a = 3i − 4j m/s. Find the angle the direction of the acceleration
vector makes with the unit vector j.
tan(3/4) = 0.75
θ = 36.9° (to 3 s.f.)
Angle required = 180° – 36.9° = 143.1°
j acts in the +ve y direction so the angle will be obtuse
Draw a diagram - the axes must meet at the starting location. Using alternate angles will show you the angle that needs to be found
The acceleration of a boat is given by a = −0.05i + 0.15j m/s. Find the angle the direction of the acceleration vector makes with the unit vector i.
tan θ (0.15/0.05) = 3
θ =71.6° (to 3 s.f.)
Angle required = 180° – 71.6° = 108° (to 3 s.f.)
Draw a diagram - the axes must meet at the starting location. This will show you the angle that needs to be found
A plane flies from P to Q.
The displacement from P to Q is 100i + 80j m. Find the angle the vector PQ makes with the unit vector j.
tan(150/50) = 3
θ = 71.6° (to 3 s.f.)
Draw a diagram - the axes must meet at the starting location. Using alternate angles will show you the angle that needs to be found
A man walks from A to B and then from B to C. His displacement from A to B is 6i + 4j m. His displacement from B to C is 5i − 12j m.
What is the magnitude of the displacement from A to C?
AC = AB + BC
AC = (6i + 5i
4j - 12j)
AC = (11i -8j)
AC = √(112 + (-8)2) = 13.6 km
magnitude not total distance
A man walks from A to B and then from B to C. His displacement from A to B is 6i + 4j m. His displacement from B to C is 5i − 12j m.
What is the total distance the man has walked in getting from A to C ?
Total distance = |AB| + |BC|
|AB| = √(62 + 42) = 7.21 km
|BC| = √(52 + -122) = 13 km
Total distance = 7.21 + 13 = 20.21 km
total distance not magnitude
velocity
The rate of change of displacement
velocity of projection
The velocity with which the projectile is projected
time of flight
The total time that an object is in
motion from the time it is projected to the time it hits the ground
You throw a ball up in the air. It travels upwards, stops, then accelerated downwards until it hits the ground. In which direction is u/v positive?
Either direction - you can pick, but you must keep it the same.
What acceleration would you use when modelling the path of a rock thrown upwards until it hits the ground?
a = -9.8m/s2
How do you derive v = u + at?
- Sketch a velocity-time graph, with v and u as their corresponding velocities
- The change in velocity = v - u
- The acceleration = (v - u) / t
- Rearrange to get v = u + at
How do you derive s = (u + v) / 2 * t?
- Sketch a velocity-time graph, with v and u as their corresponding velocities (u isn’t 0)
- Displacement = area under graph = area of trapezium = (u + v) * t / 2
- Rearrange to get s = (u + v) / 2 * t
How do you derive v2 - u2 = 2as?
- Rearrange v = u + at for t and substitute into s = (u + v) / 2 * t
- t = (v - u) / a
- s = (u + v) / 2 * ((v - u) / a)
- s = (v2 - u2) / 2a
- Rearrange to get v2 - u2 = 2as
How do you derive s = ut + 1/2 at2?
- Substitute v = u + at into s = (u + v) / 2 * t
- s = (u + [u + at]) / 2 * t
- s = (2u + at) / 2 * t
- s = (u + 1/2at) * t
- s = ut + 1/2at2
How do you derive s = vt - 1/2 at2?
- Rearrange v = u + at for u and substitute into s = (u + v) / 2 * t
- u = v - at
- s = ([v - at] + v) / 2 * t
- s = (2v - at) / 2 * t
- s = (v - 1/2at) * t
- s = vt - 1/2ut2
How do you solve mechanics problems where an object is moving in one direction?
F = ma
Expression for the resultant force
Equate these two expressions
Explain Newton’s Third Law.
For every action there is an equal and opposite reaction.
When two bodies A and B are in contact, if body A exerts a force on body B, then body B exerts a force on body A that is equal in magnitude and acts in the opposite direction.
A scale-pan is hanging by a string and has two boxes stacked on top of each other - A on top of B. How would you find the force exerted by mass A on mass B?
Find the normal reaction force of A.
“by A on B” = resolve for A
A scale-pan is hanging by a string and has two boxes stacked on top of each other - A on top of B. How would you find the force exerted on mass B by the scale-pan?
Find the force on the scale-pan by B.
(The force exerted on B by the scale-pan
has the same magnitude but acts in the opposite direction as this force - Newton’s Third Law.)
“on mass B by the scale-pan” = resolve for the scale-pan
A scale-pan is hanging by a string and has two boxes stacked on top of each other - A on top of B. Can the boxes A and B considered by themselves?
Yes for A but no for B, as the scale-pan is acting on B but not A
A toy train travels along a straight track, leaving the start of the track at time t = 0. It then returns to the start of the track. The distance, s metres, from the start of the track at time t seconds is
modelled by
s = 4t2 − t3, where 0 ≤ t ≤ 4. Explain this restriction on t.
Motion begins at t = 0 therefore t can’t be less than 0.
s is distance from the start of the track (or s is a distance so cannot be negative) so s is ≥ 0
4t2 - t3 ≥ 0
t2(4 - t) ≥ 0
t2 ≥ 0 for all values of t and (4 - t) is non-negative for t ≤ 4
therefore t2(4 - t) is non-negative when t ≤ 4
Hence 0 ≤ t ≤ 4
Time can’t be negative, although the question mentions t starts at 0s.
How do you find the maximum/minimum velocity, given a formula for the velocity?
Differentiate the formula for the velocity (you must differentiate the thing you’re finding the max/min of)
How do you find the velocity, given a formula for the acceleration?
Integrate the formula for the acceleration
When do you do the modulus of a definite integration?
When the curve goes under the x-axis (otherwise it would result in a negative area)
