# Acceleration of particle along the curve

• Tanya Sharma
In summary, the student attempted to solve a homework equation that stated the tangent line to the curve rotates at a constant angular velocity of 2 rad/s. However, he reached a wrong conclusion after incorrectly calculating the resultant acceleration.
Tanya Sharma

## 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 .

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Tanya Sharma said:

## 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 ?

Tanya Sharma said:
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.

Tanya Sharma
Tanya Sharma said:
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.

Tanya Sharma
Tanya Sharma said:
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.

D H said:
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 ?

Tanya Sharma said:
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 ?

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 ?

Tanya Sharma said:
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!

Tanya Sharma said:
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.

D H said:
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 .

Last edited:
Is the tangential acceleration equal to ##4m/s^2## ?

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

D H said:
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.

Tanya Sharma said:
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?

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 and SammyS
ehild said:
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 ?

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

SammyS
Hooray! Now I can sleep peacefully .

Thanks ehild , Sammy , D H .

SammyS

Good night!

Tanya Sharma

## 1. What is the acceleration of a particle along a curve?

The acceleration of a particle along a curve is the rate of change of its velocity with respect to time. It is a vector quantity that describes how the speed and direction of the particle's motion is changing.

## 2. How is the acceleration of a particle along a curve calculated?

The acceleration of a particle along a curve can be calculated using the formula a = dv/dt, where a is the acceleration, v is the velocity, and t is the time. This formula can also be written in terms of the particle's position along the curve, as a = d²r/dt², where r is the position vector.

## 3. What factors affect the acceleration of a particle along a curve?

The acceleration of a particle along a curve can be affected by several factors, such as the shape of the curve, the speed of the particle, and the presence of external forces. Additionally, the direction of the acceleration can also be influenced by the direction of the curve and the particle's initial velocity.

## 4. How does the acceleration of a particle along a curve relate to its motion?

The acceleration of a particle along a curve is directly related to its motion. If there is no acceleration, the particle will continue to move at a constant speed along the curve. However, if there is acceleration present, the particle's speed and direction of motion will change as it moves along the curve.

## 5. Can the acceleration of a particle along a curve be negative?

Yes, the acceleration of a particle along a curve can be negative. This indicates that the particle is slowing down in its motion along the curve. It is important to note that the negative sign does not necessarily mean the particle is moving in the opposite direction, as the direction of acceleration is determined by the direction of the curve and the particle's initial velocity.

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