# Mechanics 3Watch

This discussion is closed.
Thread starter 13 years ago
#1
ok, so these questions i get alittle confused on and as there is no official mark scheme just wondered if i could get some help with it. Parts i and ii and some of iii are fine in the first question, and in the 2nd question i can do part i.

1) A simple pendulum consists of a light inextensible string AB of length l with the end A fixed and a particle mass m attached to B. The pendulum oscillates with period T
i) It is suggested that T is proportional to a product of powers of m, l and g. Use dimensional analysis to find this relationship (4marks)

The angle of the string makes with the downward verticle at time t is θ. The particle is released from rest with the string taut and θ=θ0
ii) Use the equation for motion of the particle to find the angular acceleration (θ double dot), in terms of θ, l and g (3marks)

The angle θ0 is chosen so that θ remains small throughout the motion,
iii) Use the small angle approximation for sinθ to show that the particle performs approximate angular simple harmonic motion. State the period of the motion and verify that it is consistent with your answer to part (i) (4marks)

iv) Calculate the proportion of time for which the particle travels faster than half of its maximum speed (4marks)

Now next question

2)
Michael is attempting to make a small car do a 'loop-the-loop' on a smooth toy racing track. He propels a car of mass m kg towards a section of the track in the form of a vertical cirle of radius 0.2m and the car enters the circle at it's lowest point with a speed of 2.8ms-1 . During the motion around the circle the angle the car has turned through is denoted by Ø, as shown in diagram below (Attached)

i) show that the speed v ms-1 of the car is given by
v2=3.92(1+cosØ) Hence show that hte reaction of the track on the car, R Nm is given by R= 9.8m(2+3cosØ)

The car leaves the track at point P, where Ø= λ
ii) calculate λ

iii) Calculate the speed of the car at P and hence calculate the greatest height of the car above the level of P

The car hits the track at the point Q which is 22/135 m below the level of the centre of the circle.

iv) Calculate the speed with which the car hits teh track at Q

Any help will be appreciated.
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13 years ago
#2
How many marks are there for Q2 (i) ?

Edit: it seems a bit complex.
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13 years ago
#3
also can u tell me which examining board is this q. cos i am also attempting qs of this nature tonite. i hope its OCR old specification
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Thread starter 13 years ago
#4
Part i) is worth 7marks ii) is worth 2 iii) is worth 3 and iv) is worth 3

It is the OCR (MEI) old spec, but they should be similar right?
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13 years ago
#5
for Q2 ii), use the eqn for the Reaction at the track.

where is the inital velocity = 2.8 m/s

at R = 0, solve for

Is that the eqn you have?
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13 years ago
#6
oh yea MEI that is rite. i mite actually check my exam apers for this one.
also which paper is that? aint there from jan2002 to june 2004.
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Thread starter 13 years ago
#7
Yeah it is, it's January 2005,
And Fermat i hadnt done that part of the question.
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13 years ago
#8
wow. jan 2005!!!! i forgot. ia m gonan graba copy of it in the mei website. btw have u dome all of jan 2005 except for those parts? cos then we can compare ans.
i got loads stuff to do so will try the jan 2005 paper this weekend.
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Thread starter 13 years ago
#9
I did it a while ago, but it's only those parts that confused me, question 4 seems abit too easy, question one confused me abit but it's good that our tutor gave us notes on masses on two springs.
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13 years ago
#10
i shall take your word on it for those qs. anyway i will bleep u this weekend sometime to check the answers. ta
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Thread starter 13 years ago
#11
Well i mean a whole question on springs, one question on SHM (with one part of dimensional analysis), one question on circular motion and a question on lamina's

all in all a question which is quite nice.
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13 years ago
#12
(Original post by geekypoo)
...
And Fermat i hadnt done that part of the question.
???

If you did Q2 i), then didn't you derive the eqns of motion for the particle?
That should have meant that an eqn for the reaction of the track would have come up as a part of that derivation.

Did you get the ans to part i) wihtout using the track reaction?

Or, are we talking at cross purposes
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Thread starter 13 years ago
#13
Um i'm just plain confused now, i think i might have another go at those questions now.
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Thread starter 13 years ago
#14
Oh yeah, i remember now, you just equate the equation to zero when R=0
so then 2+cosx=0 => cosx=-2/3 => x = 2.3rads or 131.8degrees?
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13 years ago
#15
(Original post by geekypoo)
Oh yeah, i remember now, you just equate the equation to zero when R=0
so then 2+cosx=0 => cosx=-2/3 => x = 2.3rads or 131.8degrees?
yep, that's right.
here are my other answers.

iii) = 1.31 m/s, h = 4.98 cm
iv) = 2.74 m/s
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Thread starter 13 years ago
#16
umm i got iii) as 0.06 m now and part iv) as 2.67m/s
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13 years ago
#17
(Original post by geekypoo)
umm i got iii) as 0.06 m now and part iv) as 2.67m/s
mmm, I get slight variations in my answers. Now I get a slightly smaller value for h.
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Thread starter 13 years ago
#18
At P cosθ=-2/3
v²= 3.92(1-2/3) = 3.92 x 1/3
v = 1.14m/s

mgh=½mv²
h = (½x 3.92x 1/3)/9.81
h= 0.067m
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13 years ago
#19
v²=3.92(1+cosØ)

but I used

v=3.92(1+cosØ) !!

oh ****. I guess I'll give things a break.
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Thread starter 13 years ago
#20
Yay i've done the last part, correct me if im wrong Fermat but it uses the conservation of energy doesnt it, so:

½mu² = ½mv² + mgh
½u²= ½v² gh

since h height of Q above start then
h= 0.2-22/135

(2.8)²= v² + 2g(0.2-22/135)
v²= 2.8² - 2g(0.2- 22/135)
v = 2.67m/s
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