Q (1 Viewer)

[CM_Tutor]

a projectile is fired vertically upwards from the earths surface with velocity U m/s. The retardation due to gravity is given by the law k/x^2 where x is the distance of the projectile from the centre of teh earth, and k is a constant. The acceleration due to gravity on the earths surface is g. The earths radius is R. Neglecting the air resistance show that if U^2 = gR, then the projectile reaches the height R above the earths surface. What is the time for this journey?

Ok i do all the fundamental stuff in deriving this but i get R(gR-v^2) / (v^2 + gr). Wont cancel :/

As with the method from post #23, above, we have the origin of motion (x = 0) at the centre of the Earth. We have acceleration governed by:

x'' = -k / x², for x ≥ R

We require that x'' = -g at x = R, and so k = gR². Thus, we have x'' = -gR² / x², for x ≥ R.

Using this equation with the boundary condition that v = +U at x = +R and x'' = d/dx(v² / 2), we find that
v² = U² - 2gR + 2gR² / x

So, put v = 0 and U² = gR, and you get 0 = -gR + 2gR² / x, which gives x = 2R. That is, the max height (where the particle stops) is 2R from the origin, or R above the surface.

Are you also having a problem with the time part of the question?

 

[CrashOveride]

Oh dear, all t'was needed was to put v=0 in my expression above. I'll have a go at the time part

 

[ToO LaZy ^*]

y = f(x)...sketch y = Ln[f(x)]

how do you do these..? i hate these ones!

sorry about the diagram..it's meant to be a cubic

 

[CrashOveride]

As with the method from post #23, above, we have the origin of motion (x = 0) at the centre of the Earth. We have acceleration governed by:

x'' = -k / x², for x ≥ R

We require that x'' = -g at x = R, and so k = gR². Thus, we have x'' = -gR² / x², for x ≥ R.

Using this equation with the boundary condition that v = +U at x = +R and x'' = d/dx(v² / 2), we find that
v² = U² - 2gR + 2gR² / x

So, put v = 0 and U² = gR, and you get 0 = -gR + 2gR² / x, which gives x = 2R. That is, the max height (where the particle stops) is 2R from the origin, or R above the surface.

Are you also having a problem with the time part of the question?

Ok i took dx/dt from v² = U² - 2gR + 2gR² / x and then attempted to integrate the RHS from R to 2R but i got something which i dont think is easily integratable. I tried on integrator online and it was some large expression. :/

 

[Rorix]

how do you do these..? i hate these ones!

sorry about the diagram..it's meant to be a cubic

ln1=0
f(x)>0
as f(x) -> 0, ln f(x) -> -infinity
so you have asymptotes x=0, whatever the 2nd intercept is
ln f(x) = 0 in the middle at the peak there
then another asymtote at the 3rd intercept, crosses x axis when f(x)=1, then f(x) -> infinity rather slooooooowly

 

[CM_Tutor]

CrashOveride, have you tried:

Putting in expression for U² and making x the subject, then putting this into the expression for x'' (so you have a function of v), and using x'' = dv/dt, integrating with respect to v from v = √(gr) to v = 0?

Just a thought - I haven't done it, so it might not work, but it looks to be worth a try...

 

[CrashOveride]

Ill have a go now

 

[ToO LaZy ^*]

ln1=0
f(x)>0
as f(x) -> 0, ln f(x) -> -infinity
so you have asymptotes x=0, whatever the 2nd intercept is
ln f(x) = 0 in the middle at the peak there
then another asymtote at the 3rd intercept, crosses x axis when f(x)=1, then f(x) -> infinity rather slooooooowly

cool..that helped a lot. thanks1!

 

[CM_Tutor]

CrashOveride - I tried the method I suggested - it's messy, but it seems to work. I get the required time as

t = (π + 2)√R / 2√g = 2077.5 ... s

Does this match your answer / the answer given in the book?

 

[CrashOveride]

That is the correct answer.

I got x as a function of v and subbed back into x'' = -gR^2.m / x^2
And it appears im going to get a -ve time now....and also how did you eliminate the m ?

 

[CM_Tutor]

I did not have an 'm' anywhere, so I'm not sure where yours comes from. We started with x'' = -gR² / x², so there shouldn't be an 'm'.

When I wrote this out, I actually took a slightly different approach. I

* put U² = gR into v² = U² - 2gR + 2gR² / x
* rearranged to get 1 / x = (v² + U²) / 2gR²
* then showed that x'' = -(v² + U²)² / 4gR²
* By noting x'' = dv/dt, I then showed dt/dv = -4gR² / (v² + U²)²
* It follows that the required time is t = ∫ (U-->0) dt/dv dv = ∫ (0-->U) 4gR² / (v² + U²)² dv

So, we now just need to do the integration, and noting that U, g and R are all constant, that isn't too bad.

I used a the substitution v = Utan θ to show that

∫ 1 / (v² + U²)² dv = (1 / U³) * ∫ cos²θ dθ
= v / 2U²(v² + U²) + (1 / 2U³) * tan^-1(v / U) + C, for some constant C

From there, it isn't difficult to evaluate the required definite integral to get

t = gR²(π + 2) / 2U³

which simplifies to (π + 2)√R / 2√g by using U = √(gr)

 

[CrashOveride]

Ok i get up to a point where i have

t = 2gR²/U² [ 1 / (U² + v²) + (1/U).arctan(v/U) ] with limits 0 --> U

Evaluating that i end up getting a U⁴ on the denominator and i have a minus inconveniencing the numerator

 

[CM_Tutor]

The U⁴ has occured because there is a v missing from the numerator of the first term. As for it being negative, it isn't when I do it. The expression you have is correct, apart from the missing v. Here it is:

t = (2gR² / U²) [ v / (U² + v²) + (1 / U) * arctan(v / U) ] with limits 0 --> U
= (2gR² / U²) [ U / (U² + U²) + (1 / U) * arctan(U / U) ] - (2gR² / U²) [ 0 / (U² + 0²) + (1 / U) * arctan(0) ]
= (2gR² / U²) [ (1 / 2U) + (1 / U) * π / 4], as arctan(0) = 0
= (2gR² / gR) * (1 / U) * [(1 / 2) + (π / 4)], using U² = gR
= (2R / U) * [(1 / 2) + (π / 4)]
= (R / U) + (Rπ / 2U)
= R(2 + π) / 2U
= [R(2 + π) / 2] * 1 / √(gR), using U = √(gR)
= (π + 2)√R / 2√g, as required / expected

 

[ToO LaZy ^*]

circle question

can't do this one.

 

[mojako]

is angle BAC 90 degrees?

 

[ToO LaZy ^*]

it doesn't say whether or not it's a right angle...but i presume it isn't, or else it would have said so.

 

[mojako]

nope.. it cant be a right angle...
the diagram is delberately misleading
(or else there's a typo in the letters)

im still trying to solve it.. ill give it 10-20 minutes or so before i give up

 

[ToO LaZy ^*]

haha..well at least you lasted longer than me.

 

[mojako]

how many marks does it have??

 

[ToO LaZy ^*]

4.................