Proving Constant Angular Velocity and Acceleration in Circular Motion

[jiboom]

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 the centre of its path.

so i have

R=[2cos(at),2sin(at)]

V=[-2asin(at),2acos(at)]

A=[-2a^2cos(at),-2a^2sin(at)]



not sure how to show angular is constant.

i can take |V| and that s constant but I am not using angular velocity.

for the second part:

if if i take direction of A= (d2y/dt2)/(d2x/dt2)=tan(at) where d2y/dt2 is second derivative of y wrt t

does this answer the question as motion is circle centre (o,0) so any radius will have tan (at) as its gradient?

 

[tiny-tim]

hi jiboom!

(try using the X² button just above the Reply box )

not sure how to show angular is constant.

i can take |V| and that s constant but I am not using angular velocity.

that's fine … now use v = rω (and r is constant)

for the second part:

if if i take direction of A= (d2y/dt2)/(d2x/dt2)=tan(at) where d2y/dt2 is second derivative of y wrt t

does this answer the question as motion is circle centre (o,0) so any radius will have tan (at) as its gradient?

erm …

isn't it obvious that A is parallel to (x,y)? ​

 

[jiboom]

now you point it out, yes it is:)

i was going to use v = rω but as i am taking |v| am i not just getting angular speed not velocity?

 

[jiboom]

think I am being thick.

is it that as angular velocity only depends on the angle we have |ω|=ω

just as v=|v| for 1-d motion

 

[jiboom]

sorry but I am 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?

 

[HallsofIvy]

sorry but I am now confused again.

the question actually says
x=2cosωt
y=2sin ωt

so

V=V=[-2ωsin(ωt),2ωcos(ωt)]

so |V|=2ωso using v=rω only gives me r=2?

No, you knew from the original position vector that the mass was moving in a circle with radius 2: if $x=2 cos(\omega t)$and $y= 2 sin(\omega t)$, then $x^2+ y^2= 4 cos^2(\omega t)+ 4 sin^2(\omega t)= 4$.

The calculation you have done is the derivation of $v= r\omega$!

 

[jiboom]

No, you knew from the original position vector that the mass was moving in a circle with radius 2: if $x=2 cos(\omega t)$and $y= 2 sin(\omega t)$, then $x^2+ y^2= 4 cos^2(\omega t)+ 4 sin^2(\omega t)= 4$.

The calculation you have done is the derivation of $v= r\omega$!

so what do i need to say to justify angualr velocity is constant?

 

[tiny-tim]

hi jiboom!

(been out for the evening )

yup, x=2cosωt y=2sin ωt means r = 2 and angular speed = ω !

so what do i need to say to justify angualr velocity is constant?

again, it's obvious!

go to sleep on it :zzz:, and it'll all be obvious in the morning!
​

 

[jiboom]

hi jiboom!

(been out for the evening )

yup, x=2cosωt y=2sin ωt means r = 2 and angular speed = ω !


again, it's obvious!

go to sleep on it :zzz:, and it'll all be obvious in the morning!
​

lol. won't be sleeping til i get this.

i don't 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
angular velocity = constant

think its going to be long night unless i get a big nudge here,its not obvious to me due to a lack of understanding about velocity v speed |v| angular velocity ω and angular speed | ω|

 

[tiny-tim]

v=rω

now is ω angular velocity or angular speed?

what's the difference??

we're in 2D, so they're the same

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
angular velocity = constant

but you know that v is constant (from the formula), so it does

… its not obvious to me due to a lack of understanding about velocity v speed |v| angular velocity ω and angular speed | ω|

i think you're over-thinking this

v is velocity, v is speed, v = |v|

ω is angular velocity, ω is angular speed, ω = |ω|

v = rω

v = ω x r ​

 

[jiboom]

i see what you are saying but I am 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 I am trying to show

and i don't understand

v = ω x r

lhs is a vector,but rhs is vector product of 2 scalars ?

 

[jiboom]

No, you knew from the original position vector that the mass was moving in a circle with radius 2: if $x=2 cos(\omega t)$and $y= 2 sin(\omega t)$, then $x^2+ y^2= 4 cos^2(\omega t)+ 4 sin^2(\omega t)= 4$.

The calculation you have done is the derivation of $v= r\omega$!

tiny-tim has gone off-line. can you help explain what they were getting at?

 

[tiny-tim]

hi jiboom!

(just got up :zzz:)

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 I am trying to show

v = |v| is constant because the coordinate formula for v shows it is

and i don't understand

v = ω x r

lhs is a vector,but rhs is vector product of 2 scalars ?

no, ω and r are both vectors, the rhs is a vector product of two vectors

(technically, ω is a pseudovector )

v = ω x r is the definition of the angular velocity vector for a rigid body

 

[jiboom]

hi jiboom!

(just got up :zzz:)


v = |v| is constant because the coordinate formula for v shows it is


no, ω and r are both vectors, the rhs is a vector product of two vectors

(technically, ω is a pseudovector )

v = ω x r is the definition of the angular velocity vector for a rigid body

im afraid I am 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 I am misunderstanding, is equal to ω


so i don't see how the constancy of v follows without assuming what I am trying to show, ie ω is constant.


v = ω x r: still don't see what the vector is for ω. i thought the r was radius,but i now think its the postion vector of the particle.

 

[tiny-tim]

hi jiboom!

… so i don't see how the constancy of v follows without assuming what I am trying to show, ie ω is constant.

it follows from the following coordinate equations …

V=[-2asin(at),2acos(at)]

… what is |v| ? ​

v = ω x r: still don't see what the vector is for ω. i thought the r was radius,but i now think its the postion vector of the particle.

ω is the vector of magnitude ω pointing along the axis of rotation

and yes, r is the position vector of the particle

important

v = ω x r is an equation you need to be completely familiar and happy with

you need to study it in your books or wikipedia or other internet site (google "angular velocity" and "rigid body") until you are ​

 

[jiboom]

hi jiboom!


it follows from the following coordinate equations …

… what is |v| ? ​

ω is the vector of magnitude ω pointing along the axis of rotation

and yes, r is the position vector of the particle

important

v = ω x r is an equation you need to be completely familiar and happy with

you need to study it in your books or wikipedia or other internet site (google "angular velocity" and "rigid body") until you are ​

as i say |v|=2ω, but if ω is the angular velocity then i can't 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.

 

[jiboom]

anyone ?

 

[jiboom]

∂

hi jiboom!

(been out for the evening )

yup, x=2cosωt y=2sin ωt means r = 2 and angular speed = ω !


again, it's obvious!

go to sleep on it :zzz:, and it'll all be obvious in the morning!
​

hi. don't 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 => angular velocity is.

 

[tiny-tim]

in your original post, you wrote …

V=[-2asin(at),2acos(at)]

so |V| = … ?

 

[jiboom]

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 can't say |v| is constant UNLESS i know omega in the question is constant?

maybe i should be clearer, when in question i have

R=(2cos wt,2sinwt)

as this w is not said to be constant in the question i could have w=w(t) and should really have

V=(-2sin(wt)(d(wt)/dt),2cos(wt)(d(wt)/dt)

 

[jiboom]

in your original post, you wrote …


so |V| = … ?

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)?

 

[tiny-tim]

if this omega (a in my original post) is the angular speed i can't say |v| is constant UNLESS i know omega in the question is constant?

maybe i should be clearer, when in question i have

R=(2cos wt,2sinwt)

as this w is not said to be constant in the question i could have w=w(t) and should really have

V=(-2sin(wt)(d(wt)/dt),2cos(wt)(d(wt)/dt)

oh i see! :tongue2:

no, i think it's safe to assume that the ω in the question is constant (in exactly the same way as you would if it was a instead of ω ) …

it never occurred to me to think otherwise!​

stop imagining that the questions are more difficult than they really are!

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)?

yes!

 

[jiboom]

oh i see! :tongue2:

no, i think it's safe to assume that the ω in the question is constant (in exactly the same way as you would if it was a instead of ω ) …

it never occurred to me to think otherwise!​

stop imagining that the questions are more difficult than they really are!


yes!

great,thanks for your time with this.