A mass on the end of a spring which is hanging vertically is raised up and let go. It then oscillates between 2m and 1.5m above the floor and completes 32 cycles in one minute. The height, h metres, of the mass above the floor after t seconds can be modelled by the function h=acos(pi t / 180)...
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Pile foundations are the part of a structure used to carry and transfer the load of the structure to the bearing ground located at some depth below ground surface.
A concrete pile for a domestic dwelling has a radius of 1000 mm at its base, and 2000 mm at the top. It is 2000 mm deep and...
Find the second moment of area about the vertical axis for the cross-section of the follwing shape.
i cant get the pic to paste so will describe it:
it is a rectangle of horizontal length 8 and vertical height 6.
if we let (0,0) be at the lhs bottom corner, then a circle has been cut...
i am such an idiot!! I put the point on BC and not AP!! your post made me scratch my head when you mention a right angle triangle! this led me to discover i had the wrong point!! DOH
i now have answer as i get the quadratric i was after :)
thanks
i know ABC is 90. the triangle left is not a nice one which is why i am saying com of APC can be worked using com of APC and ABP which as they are right angled i know com is 1/3 of the distance of base and height from right angle.
then as i dont care where on BC just that it is on BC i let x...
ok,ill try and outline my idea a bit better/.
consider the shape with APC cut out.
i know BG (G=com) is vertical if shape is suspended at B.
if BC is vertical then G lies on BC
then the shape has com (0,y bar) if i measure from B
so i need to find the com off the new shape and set the...
a triangle ABC with AB=BC=2a and AC=2a rt(2) is drawn on a uniform lamina. A semi-circle is drawn on BC as diameter,on the opposite side of BC from A and the area enclosed by the triangle and the semi-circle is cut out.
The resulting lamina is suspended freely from B,show that AB makes an angle...
cheeky bump/
having a lot of trouble with rel motion agaisnt a current. got all other questions in the exercise but these allude me. my methods must be ok if im getting the non current questions right so only need a nudge in right direction.
a motorboat moving at 8km/h relative to the water travels from A to a point B 10km whose bearing from A is 150. it then travels to C,10km from B and due west of B. if there is a current of constant speed 4km/h from north to south,find the 2 courses to be set.
having a bit of trouble with...
maybe an easier question would be to clear up what the w is in the question
so from
R=(2coswt,2sinwt)
is the w the same as |w| the angular speed?
if so,is it just to be assumed its a number and not a function of some variable such as time as it's not writtem w(t)?
hi,thanks for coming back.
forget the a,the letter used in the question was omega.
i know what |V| is,it is just 2omega
but....
if this omega (a in my original post) is the angular speed i cant say |v| is constant UNLESS i know omega in the question is constant?
maybe i should...
∂
hi. dont give up on me just yet :)
i do appreciate your help and just need one last shove to get there.
all i need convincing off now is that the omega=angular speed =|v| is constant.
does this follow from the question? if this is so,then for sure i am happy |v|=constant =>...
as i say |v|=2ω, but if ω is the angular velocity then i cant say this is constant without assuming what i want to prove.
i should say,i have changed that a in my op to ω as that is the letter the question uses.
im afraid im totally lost here.
v = |v| is equal to 2ω so how are you saying this is constant?
im trying to show ω is constant which you seem to be saying,forgive me if im misunderstanding, is equal to ω
so i dont see how the constancy of v follows without assuming what im trying...
i see what you are saying but im thinking along these lines:
ω =ω
then how can i then say |v| is constant? |v| is 2ω so is only constant if ω is but that is what im trying to show
and i dont understand
v = ω x r
lhs is a vector,but rhs is vector product of 2 scalars ?
lol. wont be sleeping til i get this.
i dont have much here to play with.
v=rω
now is ω angular velocity or angular speed?
the v i have is not angular velocity and not constant
r is 2
now in your first post you said to use v=rω but then i just get r=2 so that does not show...
sorry but im now confused again.
the question actually says
x=2cosωt
y=2sin ωt
so
V=V=[-2ωsin(ωt),2ωcos(ωt)]
so |V|=2ω
so using v=rω only gives me r=2????
a particle moves in the xy plane such that at a time t it is at a point (x,y) where
x=2cos(at)
y=2sin(at)
prove particle moves in a circular path with constant angular velocity. prove that the acceleration of the particle at time t is in the direction of the radius from the particle to...