Circular Motion Q's (1 Viewer)

[vds700]

with q2, i couldnt do the part in a red box, i was fine with the first part of (ii), just using the sine rule.

Thanks for any help

 

[lolokay]

for question 2 just do implicit differentiation with respect to @ and substitute in udt/a for d@. I'll do the working if you want

for question 1, I'm assuming the @ in (1/2 @²) is the d@/dt one?
d(1/2 @'²)/d@' = @'
d(1/2 @²) = @' d@'

so L(1/2 @²)/d@
= L[d@/dt * d(d@/dt)]/d@

now ds = Ld@
so L[d@/dt * d(d@/dt)]/d@
= L[ds/Ldt * d(ds/dt)/L]/ds/L
= [ds/dt * d(ds/dt)]/ds
= d²s/dt²

 

[Trebla]

for q1:
s = Lθ
ds/dt = L.dθ/dt (since θ is a variable (dependent on t) and L is a constant)
= Lω (NB: ω is theta dot)
So: d²s/dt² = L.dω/dt (ω is a variable which is dependent on t)
= L.(dω/dθ)(dθ/dt) by the chain rule (just "multiply and divide by dθ")
= L.ω.(dω/dθ)
Use the fact that ω = d(ω²/2)/dω (which is just differentiating ω²/2 with respect to ω, rather than another variable)
= L.[d(ω²/2)/dω][(dω/dθ)]
= L.d(ω²/2)/dθ (as the "dω cancels out")

 

[vds700]

Thankyou so much Trebla and Lolokay

To Lolokay, I tried using implicit diff last night but it got really messy, maybe i made a mistake- I will try again today.

 

[lolokay]

if you haven't solved it yet

d/d@ a sin(@ - y) - (x + a)sin y = 0
= a(1 - dy/d@)cos(@ - y) - dy/d@(x+a)cos y = 0
a cos(@ - y) = dy/d@ [a cos(@ - y) + (x+a)cos y]
dy/d@ = a cos(@ - y)/[a cos(@ - y) + (x+a)cos y]

d@/dt = u/a
d@ = u dt/a

a dy/u dt = a cos(@ - y)/[a cos(@ - y) + (x+a)cos y]
dy/dt = u cos(@ - y)/[a cos(@ - y) + (x+a)cos y]

 

[vds700]

if you haven't solved it yet

d/d@ a sin(@ - y) - (x + a)sin y = 0
= a(1 - dy/d@)cos(@ - y) - dy/d@(x+a)cos y = 0
a cos(@ - y) = dy/d@ [a cos(@ - y) + (x+a)cos y]
dy/d@ = a cos(@ - y)/[a cos(@ - y) + (x+a)cos y]

d@/dt = u/a
d@ = u dt/a

a dy/u dt = a cos(@ - y)/[a cos(@ - y) + (x+a)cos y]
dy/dt = u cos(@ - y)/[a cos(@ - y) + (x+a)cos y]

Thanks mate- I appreciate it

 

[vds700]

Need help with another part of this q i asked ages ago (only came back to it today). I'm not sure how to do (iv), tried subbing in pi/2 and 0 in d(phi)/dt, but no idea what to next.

 

[vds700]

anyone???

Wheres Lolokay lol

 

[lyounamu]

anyone???

Wheres Lolokay lol

that guy is busy studying for 5 unit. that guy is accelerated like light years early by himself. I have never seen anyone smarter than (or of similar intelligence to) donza.

 

[lolokay]

you know that if x=a, then when @=0, y=0, and when @ = pi/2, sin[y] = 1/rt5 and cos[y] = 2/rt5 (just from looking at the right angle triangle NTO)

so subbing into the equation for dy/dt;
when @ = 0
dy/dt = ucos(0-0)/(2acos(0) + acos(0-0))
= u/3

when @ = pi/2
dy/dt = ucos(pi/2 - y)/(2acos[y] + acos[pi/2 - y])
= usin[y]/(2acos[y] + asin[y]
= u(1/rt5)/2a(2/rt5) + a(1/rt5)
= u/5
= 3/5 * u/3
= 3/5 dy/dt when @=0

 

[rlmck]

id rather have common sense than be able to solve maths equations . The time you spent posting these comments could go into study ie for the HSC

 

[lolokay]

solving questions on here is pretty efficient study imo. I'm not doing the HSC yet so can't exactly study for it anyway

 

[lyounamu]

id rather have common sense than be able to solve maths equations . The time you spent posting these comments could go into study ie for the HSC

wtf. seriously piss off. making a comment like that really suggests that you don't have a common sense at all.

the time he spend posting these comments help the other bosers and by doing those questions he improves his mathematical knowledge. you obviously haven't got a clue as to why people post questions up here.

 

[rlmck]

fair call. but did you get 94 in your trial for 2unit mathematics

 

[lyounamu]

fair call. but did you get 94 in your trial for 2unit mathematics

me? I got 119/120 in my 2008 cssa trial. I am not sure what you mean by 94...

 

[rlmck]

Wow thats excellent . i meant to say 94% in 2unit mathematic in my NSW independent Trial Examination .
I hope you do well in the HSC

 

[lyounamu]

Wow thats excellent . i meant to say 94% in 2unit mathematic in my NSW independent Trial Examination .
I hope you do well in the HSC

Thanks. I think the NSW Independent is harder than CSSA paper.

 

[vds700]

you know that if x=a, then when @=0, y=0, and when @ = pi/2, sin[y] = 1/rt5 and cos[y] = 2/rt5 (just from looking at the right angle triangle NTO)

so subbing into the equation for dy/dt;
when @ = 0
dy/dt = ucos(0-0)/(2acos(0) + acos(0-0))
= u/3

when @ = pi/2
dy/dt = ucos(pi/2 - y)/(2acos[y] + acos[pi/2 - y])
= usin[y]/(2acos[y] + asin[y]
= u(1/rt5)/2a(2/rt5) + a(1/rt5)
= u/5
= 3/5 * u/3
= 3/5 dy/dt when @=0

once again thanks so much for your help, you are insanely smart.