Mechanics 2

Question 0

Sine rule?

Answer 0

a/Sin(A) = b/Sin(B) = c/Sin(C)

Question 1

SUVAT equations:
s = ?

                           v = ?

                       v^2 = ?

s (no acceleration) = ?

Answer 1

SUVAT equations:
s = ut + 1/2at^2

                           v = u + at

                       v^2 = u^2 + 2as

s (no acceleration) = [(u+v)/2]t

Question 2

Cosine rule?

Answer 2

a^2 = b^2 + c^2 - 2bcCos(A)

^^letters (corresponding to sides) may be swapped around as long as the term on the left is consistent with the angle in the Cos function.

Question 3

Projectiles rules & assumptions & things to look out for (horizontal/vertical travel, acceleration, distance)

Answer 3

• ALWAYS resolve Horizontal and Vertical motion SEPARATELY, but remember that the time taken for both before the projectile hits the ground is the same.
• NO ACCELERATION HORIZONTALLY.
• Acceleration vertically is normally just g=9.8, unless extra forces are stated as well.
• Total Vertical distance CAN BE NEGATIVE if the projectile ends up lower than it started off. Otherwise distance vertically is ZERO if the projectile lands at the height it started. (Think of this as DISPLACEMENT).
• Path travelled is SYMMETRIC, AS LONG AS the projectile lands at the same height it started at.

Question 4

”s” to “v” to “a”?

Answer 4

s = f(t)

v = (ds/dt)

a = (dv/dt) = (d^2s/dt^2)

Distance (s) is a function of time - f(t)

Distance differentiates to get speed
v =(ds/dt)

Speed differentiates to get acceleration
a = (dv/dt) = (d^2s/dt^2)

Question 5

Kinematics of a particle in a plane?

Answer 5

• Treat i components and j components separate when resolving.
• Position vectors at time t are given as functions of time.
• Accelerations may be given as i and j components with a time coefficient in there as well - bear in mind that the particle may not have started it’s motion at the origin.
• Velocities may be given in i and j components with a t^2 coefficient if they are accelerating.

Question 6

Force, mass, acceleration?

Answer 6

F = ma

Question 7

How do you find the centre of mass for objects on a straight line?

Answer 7

Σ[m(i)y(i)] = ŷΣ[m(i)]

Where (i) denotes individual masses/displacements. (0,ŷ) is the position of the centre of mass of the system
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MASSES ON A STRAIGHT LINE:
To find the centre of mass of some masses on a straight line:
1) Find the distance of the masses from the origin.
2) Take moments about the origin - the added moments of each individual mass is equal to the total moment that appears to act at the centre of mass (with this mass being the sum of all the smaller masses).
3) You now have
“moments” = “total mass” x “d”
4) [Σmoments]/total mass = distance from origin that centre of mass appears to act. (May be negative, i.e. To the left)

Question 8

Define “centre of mass”

Answer 8

Centre of mass is the point at which the whole mass of a body can be considered to be concentrated (where the whole mass appears to act).

REMEMBER IT IS POSSIBLE FOR CENTRE OF MASS TO ACT OUTSIDE OF THE OBJECT

Question 9

How do you find the centre of mass of some masses in a plane?

Answer 9

METHOD 1:

Take moments about the x-axis, as with the method for masses on a single line, equating them to Mx to get the x value (after dividing moments by total mass, M). THEN take moments about the y axis and equate to My to get the y coordinate.
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METHOD 2:

Do in a column format:

  X1            X2                             X m1 ---   + m2 ---  +... = (m1+m2...)---
  Y1            Y2                             Ŷ

Multiply the masses by the actual x and y coordinate values, then sum the coordinates on the left. Divide the large resultant coordinate’s x and y values by the sum of the individual masses (as appears on the right hand side) to get the x and y coordinates for the centre of mass.

Question 10

Centre of mass in a plane - what if the question doesn’t specify axes or coordinates?

Answer 10

Choose your own! Usually the lowest point, furthest to the left. Label all vertices of the object relative to this point.

REMEMBER IT IS POSSIBLE FOR CENTRE OF MASS TO ACT OUTSIDE OF THE OBJECT

Question 11

Must the centre of mass act inside the object?

Answer 11

NO!! Centre of mass can act outside of the object.

Question 12

The shortcut for centre of mass of a triangle?

Answer 12

TAKE THE MEAN OF THE 3 COORDINATES!! - if a WHOLE SHAPE, NOT A FRAMEWORK.

G = {([x1+x2+x3]/3) , ([y1+y2+y3]/3)}

For any triangle, the centre of mass is 2/3 each median from each vertex. So draw lines from each corner to the middle (median) of the opposite side of the triangle. The 3 lines you end up with will cross at this 2/3 point, and this is where the centre of mass acts.

Question 13

Centre of mass of a uniform sector of a circle?

Answer 13

IF IT IS A BLOCK SHAPE: A uniform sector of radius r and angle 2α, measured in RADIANS, has centre of mass along “axis of symmetry” (line down middle) at a distance of 2rSin(α)/3α from the centre.

IF IT IS A FRAMEWORK centre of mass is a distance of rSin(α)/α from the centre, along the axis of symmetry (so it’s only different by a factor of 2/3).

DON’T FORGET THAT A SEMICIRCLE IS A UNIFORM SECTOR!!

Question 14

Centre of mass of a uniform rectangular lamina?

Answer 14

Directly in the middle, draw lines of symmetry and they cross at G, the centre of mass.