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Eesa268
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#1
A mass suspended from one end of an ideal spring undergoes vertical simple harmonic motion with an amplitude of 2.0cm. Three complete oscillations are made in 4.0s
Find the acceleration of the mass at "equillibrium position"
Find the magnitude of the acceleration of the mass at the position of maximum displacement
Help with this PLZ
Find the acceleration of the mass at "equillibrium position"
Find the magnitude of the acceleration of the mass at the position of maximum displacement
Help with this PLZ
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#2
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#2
Acceleration at the equilibrium position is easy - SHM occurs when there is a restoring force pointing towards the equilibrium position AND the magnitude of the restoring force is proportional to the distance from the equilibrium position.
when the mass is at the equilibrium position the distance from the equilibrium position is zero... and F=ma tells us that the acceleration here must also be zero.
for the other part you will need to select an appropriate formula from the data sheet and convert the information given by the question into the appropriate SI units
when the mass is at the equilibrium position the distance from the equilibrium position is zero... and F=ma tells us that the acceleration here must also be zero.
for the other part you will need to select an appropriate formula from the data sheet and convert the information given by the question into the appropriate SI units
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Eesa268
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#3
(Original post by Joinedup)
Acceleration at the equilibrium position is easy - SHM occurs when there is a restoring force pointing towards the equilibrium position AND the magnitude of the restoring force is proportional to the distance from the equilibrium position.
when the mass is at the equilibrium position the distance from the equilibrium position is zero... and F=ma tells us that the acceleration here must also be zero.
for the other part you will need to select an appropriate formula from the data sheet and convert the information given by the question into the appropriate SI units
Acceleration at the equilibrium position is easy - SHM occurs when there is a restoring force pointing towards the equilibrium position AND the magnitude of the restoring force is proportional to the distance from the equilibrium position.
when the mass is at the equilibrium position the distance from the equilibrium position is zero... and F=ma tells us that the acceleration here must also be zero.
for the other part you will need to select an appropriate formula from the data sheet and convert the information given by the question into the appropriate SI units
A car of mass 𝑚=1000kg is driven round a smooth circular track of radius 𝑟=250m and takes a time 𝑇=30s to complete one lap. At what angle 𝜃 must the track be banked to counteract the tendency of the car to slip sideways?
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#4
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#4
is there a circular motion formula that can help you find the amount of centripetal force for the car?
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Eesa268
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#5
(Original post by Joinedup)
is there a circular motion formula that can help you find the amount of centripetal force for the car?
is there a circular motion formula that can help you find the amount of centripetal force for the car?
But it asks for the angle ?
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Eesa268
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#6
(Original post by Joinedup)
is there a circular motion formula that can help you find the amount of centripetal force for the car?
is there a circular motion formula that can help you find the amount of centripetal force for the car?
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#7
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#7
(Original post by Eesa268)
??helloo
??helloo
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Eesa268
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#8
(Original post by Joinedup)
So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?
So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?
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#9
(Original post by Joinedup)
So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?
So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?
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#10
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#10
Have a look at this - you should have done a sketch that looks something like diagram 3 to help you.
https://study.com/academy/lesson/cir...lar-track.html
https://study.com/academy/lesson/cir...lar-track.html
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Eesa268
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#11
(Original post by Joinedup)
Have a look at this - you should have done a sketch that looks something like diagram 3 to help you.
https://study.com/academy/lesson/cir...lar-track.html
Have a look at this - you should have done a sketch that looks something like diagram 3 to help you.
https://study.com/academy/lesson/cir...lar-track.html
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#12
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#12
(Original post by Eesa268)
So am I using the formula Nsintheta=mv^2/R
So am I using the formula Nsintheta=mv^2/R
you can keep going like it shows till you get
tan θ = v2/(gR)
or you can just look at the sketch diagram you've made which looks something like this
and realise that you can get tan θ already because you already know Ny... because it's the size of the weight force on the car
and you already know Nx because it's the centripetal force on the car.
so you can just read off the sketch that tan θ = centripetal force / weight force
and you already know numerical values for both those.
That's why drawing sketches is a really useful thing to remember to do.
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Eesa268
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#13
f
for v do i use angular speed or normal speed?
(Original post by Joinedup)
That's just a stepping stone... it's got N and sin θ in it - and you don't know N yet but you want θ
you can keep going like it shows till you get
tan θ = v2/(gR)
or you can just look at the sketch diagram you've made which looks something like this
![Name: bankedtrack.jpg
Views: 127
Size: 13.8 KB]()
and realise that you can get tan θ already because you already know Ny... because it's the size of the weight force on the car
and you already know Nx because it's the centripetal force on the car.
so you can just read off the sketch that tan θ = centripetal force / weight force
and you already know numerical values for both those.
That's why drawing sketches is a really useful thing to remember to do.
That's just a stepping stone... it's got N and sin θ in it - and you don't know N yet but you want θ
you can keep going like it shows till you get
tan θ = v2/(gR)
or you can just look at the sketch diagram you've made which looks something like this
and realise that you can get tan θ already because you already know Ny... because it's the size of the weight force on the car
and you already know Nx because it's the centripetal force on the car.
so you can just read off the sketch that tan θ = centripetal force / weight force
and you already know numerical values for both those.
That's why drawing sketches is a really useful thing to remember to do.
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#14
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#14
in the circular motion types of formulae, angular velocity is always ω (lower case omega)
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Eesa268
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#15
(Original post by Joinedup)
it's the linear speed in m/s, careful about using normal to mean usual cos normal also means perpendicular
in the circular motion types of formulae, angular velocity is always ω (lower case omega)
it's the linear speed in m/s, careful about using normal to mean usual cos normal also means perpendicular
in the circular motion types of formulae, angular velocity is always ω (lower case omega)
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#16
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#16
(Original post by Eesa268)
I have tried working it out I get very low answer between 0-1
I have tried working it out I get very low answer between 0-1
circumference of the rack is 2πr = 1570m
speed of the vehicle is d/t = 52.3 m/s
tan θ = v2/(gr)
=2740/(250*9.81)
=1.12
1.12 is the tangent of the angle we're looking for so...
tan-1 1.12 = 48.2°
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#17
(Original post by Joinedup)
OK
circumference of the rack is 2πr = 1570m
speed of the vehicle is d/t = 52.3 m/s
tan θ = v2/(gr)
=2740/(250*9.81)
=1.12
1.12 is the tangent of the angle we're looking for so...
tan-1 1.12 = 48.2°
OK
circumference of the rack is 2πr = 1570m
speed of the vehicle is d/t = 52.3 m/s
tan θ = v2/(gr)
=2740/(250*9.81)
=1.12
1.12 is the tangent of the angle we're looking for so...
tan-1 1.12 = 48.2°
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Eesa268
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#18
(Original post by Eesa268)
Thanks I was using different formulas to work it out and I didn’t take into account the circumference
Thanks I was using different formulas to work it out and I didn’t take into account the circumference
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#19
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#19
(Original post by Eesa268)
tan^-1 does not equal 48.2 it is 0.84
tan^-1 does not equal 48.2 it is 0.84
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#20
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