Acceleration of particle along the curve

[Tanya Sharma]

Homework Statement

Homework Equations

The Attempt at a Solution

Since the tangent moves at uniform angular speed ,the particle also moves with constant angular speed . Is it correct ?

If it is right then the particle has only radial acceleration given by $ω^2r$ ,where 'r' is the radius of curvature .

r = sec2x

r ( x= πa/4) = 1/√2 .

Hence resultant acceleration at x= πa/4 = 2√2 .

I don't have the answer key . Have I interpreted the question correctly and done it right ?

I would be grateful if somebody could verify my work .

 

[SammyS]

Homework Statement

Homework Equations

The Attempt at a Solution

Since the tangent moves at uniform angular speed ,the particle also moves with constant angular speed . Is it correct ?

If it is right then the particle has only radial acceleration given by $ω^2r$ ,where 'r' is the radius of curvature .

r = sec2x

r ( x= πa/4) = 1/√2 .

Hence resultant acceleration at x= πa/4 = 2√2 .

I don't have the answer key . Have I interpreted the question correctly and done it right ?

I would be grateful if somebody could verify my work .

It may be true that the particle moves with constant angular speed (not sure how you would define that), but not for the reason you give.

I think that the problem statement is saying that the angle the tangent line makes with the x-axis increases at a constant rate.

 

[Tanya Sharma]

Ok . So what should be the correct approach ?

 

[SammyS]

Ok . So what should be the correct approach ?

Get dy/dx.

Slope of tangent line = tan(θ), where θ is the angle the tangent line makes w.r.t the x-axis. Right ?

Start there.

 

[D H]

Since the tangent moves at uniform angular speed ,the particle also moves with constant angular speed . Is it correct ?

No, that isn't correct.

Angular velocity is not the way to solve this problem. As Sammy mentioned, determining dy/dx is the first step.

 

[D H]

Ok . So what should be the correct approach ?

The problem statement says the tangent to the curve rotates with a constant angular velocity. I would take that as meaning the unit tangent.

HINT: The magnitude of the time derivative of any unit vector is the rate at which that unit vector is rotating.

 

[Tanya Sharma]

No, that isn't correct.

Angular velocity is not the way to solve this problem. As Sammy mentioned, determining dy/dx is the first step.

Ok .

dy/dx = tan2x i.e tanθ = tan2x .

θ = 2x .

dx/dt = 1

Is it correct ? If yes , what next ?

 

[SammyS]

Ok .

dy/dx = tan2x i.e tanθ = tan2x .

θ = 2x .

dx/dt = 1

Is it correct ? If yes , what next ?

Where did the $\ a\$ go ?

 

[Tanya Sharma]

Sorry .

dy/dx = tan(x/a) i.e tanθ = tan(x/a)

θ = x/a

dθ/dt = (1/a)dx/dt

dx/dt=2a

Is it correct now ?

 

[SammyS]

Sorry .

dy/dx = tan(x/a) i.e tanθ = tan(x/a)

θ = x/a

dθ/dt = (1/a)dx/dt

dx/dt=2a

Is it correct now ?

Looks good!

 

[D H]

dx/dt=2a

What's that two doing there? (That's a devil's advocate kind of question.)

I find that it's best to work symbolically until the very, very end. If needed (sometimes it's not needed), I'll plug in the numerical values, but only at the very, very end.

 

[Tanya Sharma]

What's that two doing there? (That's a devil's advocate kind of question.)

I find that it's best to work symbolically until the very, very end. If needed (sometimes it's not needed), I'll plug in the numerical values, but only at the very, very end.

I have worked symbolically only . In fact what I did in post#7 is also correct , since 'a' is a constant .

 

[Tanya Sharma]

Is the tangential acceleration equal to $4m/s^2$ ?

Is net acceleration equal to $2\sqrt{6}m/s^2$ ?

 

[SammyS]

What's that two doing there? (That's a devil's advocate kind of question.)

I find that it's best to work symbolically until the very, very end. If needed (sometimes it's not needed), I'll plug in the numerical values, but only at the very, very end.

I don't mind that 2 so much, since early in the problem we're told the tangent line rotates with constant angular speed of 2 rad/s . Otherwise, we need to carry around some ω₀, which I suppose isn't so bad. Plugging in a = 1/2 so early in the game could be a bad idea.

 

[SammyS]

Is the tangential acceleration equal to $4m/s^2$ ?

Is net acceleration equal to $2\sqrt{6}m/s^2$ ?

I did not get that.

Can you show some intermediate steps?

 

[ehild]

The curve is given in Cartesian coordinates, so the acceleration can be expressed by $\ddot x(t), \ddot y(t)$. We also know that dx/dt =2a, constant.

 

[Tanya Sharma]

The curve is given in Cartesian coordinates, so the acceleration can be expressed by $\ddot x(t), \ddot y(t)$. We also know that dx/dt =2a, constant.

Is $4m/s^2$ the net acceleration ?

 

[ehild]

I think yes. Why aren't you sleep yet?

 

[Tanya Sharma]

Hooray! Now I can sleep peacefully .

Thanks ehild , Sammy , D H .

 

[D H]

That's the answer.

Good night!