# M3 exercise paper at back of textbookWatch

This discussion is closed.
#1
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/(4πÂ²))

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.
0
13 years ago
#2
(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/(4πÂ²))

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.
0
13 years ago
#3
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 = Â½muÂ² = Â½ x Â½ x 12Â² = 36J
GPE = 0

At highest point of circle
KE = Â½mvÂ² = Â½ x Â½ x vÂ² = Â¼vÂ²
GPE = mgh = Â½ x 9.8 x 4 = 19.6J

By the work-energy principle
Initial(KE+GPE) = Final(KE+GPE)
36 + 0 = Â¼vÂ² + 19.6
Â¼vÂ² = 36 - 19.6 = 16.4
vÂ² = 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/(4πÂ²))
speed of particle = 2rev/s = 2 x 2π rad/s = 4π rad/s
T = 4mg

Resolve vertically: Tcosθ = mg
cosθ = mg/T = mg/4mg = Â¼

sinÂ²θ + cosÂ²θ = 1
sinÂ²θ = 1 - cosÂ²θ = 1 - (Â¼)Â² = 15/16
sinθ = √15/16 = Â¼√15

Resolve horizontally:
Tsinθ = ma = mrωÂ²

r = Tsinθ/mωÂ²
= [4mg x Â¼√15]/[m x [4π]Â²]
= [4g x Â¼√15]/[16πÂ²]
= [4g√15]/[64πÂ²]
= g√15/16πÂ²
= 0.240m
= 24cm (2sf)

b. I dont know how to do this one
= APsinθ = 24.0354cm
=> AP = 24.0354cm/sinθ
= 24.0354cm/Â¼√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
= πrÂ²h (ignore rho as it cancels)
= π(2a)Â² x 4a
= 16πaÂ³

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

Mass x y
= 32πa4

Shape S
Mass
= π∫yÂ² dx (between x-coordinates a and 0)
= π∫4ax dx
= 4πa∫x dx
= [4πa x Â½xÂ²] (between x-coordinates a and 0)
= [2πaxÂ²] (between x-coordinates a and 0)
= 2πaÂ³ - 0
= 2πaÂ³

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[2πaÂ³ + 16πaÂ³]
122πa4/3 = y-bar[18πaÂ³]
y-bar = 122πa4/54πaÂ³
y-bar = 122a/54 = 61a/54
0
#4
thanks alot...I still cant do 6b though ...
0
13 years ago
#5
(Original post by lgs98jonee)
thanks alot...I still cant do 6b though ...
darn it I did 7)b.

I'll just be a sec.
0
#6
(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
0
13 years ago
#7
(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
0
13 years ago
#8
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 = Â½m

At instantaneous rest,
3g - T = ma
3g - λ(e+x)/L = 3a
3g - 12g(Â½ + x)/2 = 3a
3g - 12g/4 - 12gx/2 = 3a
3g - 3g - 6gx = 3a
-6gx = 3a
a = -2gx

a in the form a = -ωÂ²x therefore SHM.

ωÂ² = 2g
ω = √2g

Vmax occurs when the particle passes through the equilibrium position, that is when P is a distance 2Â½m from O.

vÂ² = ωÂ²(aÂ² - xÂ²)
vmax = ωa
vmax = (2-Â½) x √2g = 6.64ms-1
0
#9
(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. ;-)
0
13 years ago
#10
(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.
0
#11
(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
0
13 years ago
#12
(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?
0
13 years ago
#13
(Original post by Phil23)
confident about 100 % this tuesday widow?
100% raw? no way. 90%+ UMS yes.
0
13 years ago
#14
Usually how many marks in M3 will u get an A??
0
13 years ago
#15
Im hoping to do well too.. 90+ UMS... whcih would also give more insurance for my other modules... :P gl everyone!
0
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