M3 exercise paper at back of textbook Watch

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lgs98jonee
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Hi
I am hoping some of you have the Heinamann textbook as I have a few queries as to some answers.

Q2. Do we use conservation of energy for this question?

Q5. a. I get radius as 24.8cm..(g/(4π²))

b. I dont know how to do this one

6b. How do I do this? It looks like an integrating question to me.


Thanks everyone who can help.
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Christophicus
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(Original post by lgs98jonee)
Hi
I am hoping some of you have the Heinamann textbook as I have a few queries as to some answers.

Q2. Do we use conservation of energy for this question?

Q5. a. I get radius as 24.8cm..(g/(4π²))

b. I dont know how to do this one

6b. How do I do this? It looks like an integrating question to me.


Thanks everyone who can help.
I've got to go out, so i'll be back in about half an hour. If noone's answered your query by then, i'll have a go.
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Christophicus
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Q2. Do we use conservation of energy for this question?
Impulse = mv
6 = 0.5v
v = 6/0.5 = 12ms-1

At lowest point of circle
KE = ½mu² = ½ x ½ x 12² = 36J
GPE = 0

At highest point of circle
KE = ½mv² = ½ x ½ x v² = ¼v²
GPE = mgh = ½ x 9.8 x 4 = 19.6J

By the work-energy principle
Initial(KE+GPE) = Final(KE+GPE)
36 + 0 = ¼v² + 19.6
¼v² = 36 - 19.6 = 16.4
v² = 65.6
v = √65.6 = 8.1ms-1

v > 0 at top of circle, therefore the particle completes vertical circles.

Q5. a. I get radius as 24.8cm..(g/(4π²))
speed of particle = 2rev/s = 2 x 2π rad/s = 4π rad/s
T = 4mg

Resolve vertically: Tcosθ = mg
cosθ = mg/T = mg/4mg = ¼

sin²θ + cos²θ = 1
sin²θ = 1 - cos²θ = 1 - (¼)² = 15/16
sinθ = √15/16 = ¼√15

Resolve horizontally:
Tsinθ = ma = mrω²

r = Tsinθ/mω²
= [4mg x ¼√15]/[m x [4π]²]
= [4g x ¼√15]/[16π²]
= [4g√15]/[64π²]
= g√15/16π²
= 0.240m
= 24cm (2sf)

b. I dont know how to do this one
radius of circle
= APsinθ = 24.0354cm
=> AP = 24.0354cm/sinθ
= 24.0354cm/¼√15
= 96.142cm/√15
= 24.8cm
= 25cm (2sf)

6b. How do I do this? It looks like an integrating question to me.
From 7)a C.O.M of S = (2a/3, 0)

Cylinder
Mass
= πr²h (ignore rho as it cancels)
= π(2a)² x 4a
= 16πa³

y-coordinate of centre of mass (from plane face)
= 2a

Mass x y
= 32πa4

Shape S
Mass
= π∫y² dx (between x-coordinates a and 0)
= π∫4ax dx
= 4πa∫x dx
= [4πa x ½x²] (between x-coordinates a and 0)
= [2πax²] (between x-coordinates a and 0)
= 2πa³ - 0
= 2πa³

y-coordinate of centre of mass (from plane face)
= 4a + (a - a/3)
= 13a/3

Mass x y
= 26πa4/3

∑my = y-bar.∑m
32πa4 + 26πa4/3 = y-bar[2πa³ + 16πa³]
122πa4/3 = y-bar[18πa³]
y-bar = 122πa4/54πa³
y-bar = 122a/54 = 61a/54
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lgs98jonee
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thanks alot...I still cant do 6b though ...
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Christophicus
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(Original post by lgs98jonee)
thanks alot...I still cant do 6b though ...
darn it I did 7)b.

I'll just be a sec.
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lgs98jonee
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(Original post by Widowmaker)
darn it I did 7)b.

I'll just be a sec.
lol thanks....you are really good at this and it is quite worrying for me
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ziya_the_king
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(Original post by lgs98jonee)
lol thanks....you are really good at this and it is quite worrying for me
lgs... m3 on tue like me?.. i wish i was good as widowmaker too lol
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Christophicus
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6)b

Find equilibrium position
F = ma
T - 3g = 0 (in equilibrium)
T = 3g
T = λe/L
=> λe/L = 3g
=> e = 3gL/λ = 3g(2)/12g = 6g/12g = ½m

At instantaneous rest,
3g - T = ma
3g - λ(e+x)/L = 3a
3g - 12g(½ + x)/2 = 3a
3g - 12g/4 - 12gx/2 = 3a
3g - 3g - 6gx = 3a
-6gx = 3a
a = -2gx

a in the form a = -ω²x therefore SHM.

ω² = 2g
ω = √2g

Vmax occurs when the particle passes through the equilibrium position, that is when P is a distance 2½m from O.

v² = ω²(a² - x²)
vmax = ωa
vmax = (2-½) x √2g = 6.64ms-1
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lgs98jonee
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(Original post by ziya_the_king)
lgs... m3 on tue like me?.. i wish i was good as widowmaker too lol
yup i do :/...i am wishing i had started doing papers before today ...luckily we are doing 13 modules at my school so I can mess up one. ;-)
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Christophicus
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(Original post by lgs98jonee)
yup i do :/...i am wishing i had started doing papers before today ...luckily we are doing 13 modules at my school so I can mess up one. ;-)
hey come on guys. I'm sure you'll do well.
An A in January was 53/75 so don't panic if you get a question wrong.
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lgs98jonee
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(Original post by Widowmaker)
hey come on guys. I'm sure you'll do well.
An A in January was 53/75 so don't panic if you get a question wrong.

good good...i can generally do it all except for centre of mass questions and circular motion questions...so I am going to try and do as much as revision exercise 2 as possible
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(Original post by Widowmaker)
hey come on guys. I'm sure you'll do well.
An A in January was 53/75 so don't panic if you get a question wrong.
confident about 100 % this tuesday widow?
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Christophicus
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(Original post by Phil23)
confident about 100 % this tuesday widow?
100% raw? no way. 90%+ UMS yes.
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im37
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Usually how many marks in M3 will u get an A??
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Feria
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Im hoping to do well too.. 90+ UMS... whcih would also give more insurance for my other modules... :P gl everyone!
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