I don't have that paper. What's the question?
About normal VS tangential acceleration, Normal acceleration is also known as radial acceleration and it is towards the centre (centripetal force).
Tangential acceleration is along the tangent; perpendicular to the radius.
The site here should help.
In the 1st on that site, an is the tangential acceleration and the v for velocity occurs because of the tangential acceleration.
I don't think we need to know the value of the normal acceleration to be honest; i just know what it is but i haven't come across it in a paper yet but if it does come up it'll be probably be given.Heres the link http://4unitmaths.com/hsc1981-1989.pdf So the normal acceleration is w^2r, then whats the tangential acceleration?
I don't think we need to know the value of the normal acceleration to be honest; i just know what it is but i haven't come across it in a paper yet but if it does come up it'll be probably be given.
Tangential acceleration = -gsintheota
Thanks for the links to the paper ^^
Here are the questions:
8. (i)(a) Given that ω is a complex root of the equation x^3 = 1, show that ω^2 is also a root of this equation.
(b) Show that 1 + ω + ω2 = 0, and 1 + ω2 + ω4 = 0.
(c) Let α, β be real numbers.Find, in its simplest form, the cubic equation whose roots are α + β,αω + βω−1, αω2 + βω−2.
(ii) Using induction, show that for each positive integer n there are unique positive integers Pn and Qn such that (1 +√2)^n = Pn + Qn√2.
Show also that (Pn)^2 − 2(Qn)^2 = (−1)n.
For 8ia - since w is a root --> w^3 = 1
now w^2 = (w^3)^2/3 = 1^2/3 = 1
Therefore w^2 is also a root to the equation.
For 8ib since w^3 = 1
w^3 - 1 = 0
(w -1) ( w^2 + w + 1) = 0
Therefore w^2 + w + 1 = 0
Since w is a root, w^2 is also a root (proven before)
So sub w = w^2 we get w^4 + w^2 + 1 = 0
Haven't figured out the rest yet :/
ill do the other ones nowI don't think we need to know the value of the normal acceleration to be honest; i just know what it is but i haven't come across it in a paper yet but if it does come up it'll be probably be given.
Tangential acceleration = -r x d(1/2 x(d@/xt)/dt^2) but they'll ask you to derive it
Thanks for the links to the paper ^^
Here are the questions:
8. (i)(a) Given that ω is a complex root of the equation x^3 = 1, show that ω^2 is also a root of this equation.
(b) Show that 1 + ω + ω2 = 0, and 1 + ω2 + ω4 = 0.
(c) Let α, β be real numbers.Find, in its simplest form, the cubic equation whose roots are α + β,αω + βω−1, αω2 + βω−2.
(ii) Using induction, show that for each positive integer n there are unique positive integers Pn and Qn such that (1 +√2)^n = Pn + Qn√2.
Show also that (Pn)^2 − 2(Qn)^2 = (−1)n.
For 8ia - since w is a root --> w^3 = 1
now w^2 = (w^3)^2/3 = 1^2/3 = 1
Therefore w^2 is also a root to the equation.
For 8ib since w^3 = 1
w^3 - 1 = 0
(w -1) ( w^2 + w + 1) = 0
Therefore w^2 + w + 1 = 0
Since w is a root, w^2 is also a root (proven before)
So sub w = w^2 we get w^4 + w^2 + 1 = 0
Haven't figured out the rest yet :/
Ah, well let the circle have a centre 0 and a radius of R at a fixed point A on the circumference. For Tangential acceleration:6. A point P is moving in a circular path around a centre O. Define the angularvelocity of P with respect to O at time t. Derive expressions for the tangential and normal components of the acceleration of P at time t. Theres the question with tangential and normal.. :S
You are awesomeill do the other ones now
in uniform circular motion the angular velocity is constant or independant of time so dw/dt will just = 0 and the tangential acceleration is 0, which holds by the definition of tangential accelerationAh, well let the circle have a centre 0 and a radius of R at a fixed point A on the circumference. For Tangential acceleration:
Let P be a point on the circle which is moving.
Let x = AP = r@ where @ is <P0A
dx/dt = v = rw where w = d@/dt
For acceleration, dv/dt = r dw/dt = r d^2@/dt^2
We know that normal acceleration is equal to centripetal force towards the centre.
So F = m x a(normal) = mv^2/r
So a(normal) = v^2/r = rw^2 (as v = rw as shown above)
Ah i see. So for that question should i derive it as i did and then just say it equals 0 or leave it?in uniform circular motion the angular velocity is constant or independant of time so dw/dt will just = 0 and the tangential acceleration is 0, which holds by the definition of tangential acceleration
Awesome, thanksi'd probably derive it just to be safe as the question specifically asks it
(ii) Using induction, show that for each positive integer n there are unique positive integers Pn and Qn such that (1 +√2)^n = Pn + Qn√2.Cheers guys. Shouldn't the mathetmatical induction specific an initial condition of Pn = Qn = 1 though.. because you just assumed its 1?
ill do the other ones now
Because, we require that P_k+1 and Q_k+1 be integers. Now, since P_k and Q_k are integers, it follows that P_k + 2Q_k and P_k + Q_k are also integers. We're just saying that P_k+1 = P_k + 2Q_k and Q_k+1 = P_k + Q_k.
Think of it as simplifying the expression.
How do you do Q8c (i) and (ii)and what's the difference between tangential and normal acceleration?