Non-Uniform Circular Motion Problem

[creechur]

Homework Statement

A car starts from rest on a curve with radius 150m, with an acceleration of 1.8 m/s². How many revolutions will the car make before the magnitude of its acceleration is 2.40 m/s²?

Homework Equations

a_t² + a_r² = total acceleration²

theta = omega (t) + .5 (acceleration) (t)²

The Attempt at a Solution

I solved for velocity by plugging in v²/ R into the acceleration triangle equation. Then I solved for "t" by dividing my velocity by the acceleration and got 15 m/s for velocity and 8.3 seconds for time. I plugged that into the position equation and got 62.5 radians, divided by 2pi and got 9.95 revolutions. That was wrong - tips please?

 

[Hootenanny]

This equation is incorrect,

theta = omega (t) + .5 (acceleration) (t)²

If we look at the dimensions we find that,

[θ] = 1 (i.e. dimensionless)
[ωt] = T^-1T = 1 (i.e. dimensionless)
[at²] = LT^-2T² = L (i.e. a length)

Do you know any rotational kinematic equation? (They are analogous to the linear kinematic equations).

 

[Fightfish]

This equation is incorrect,

If we look at the dimensions we find that,

[θ] = 1 (i.e. dimensionless)
[ωt] = T^-1T = 1 (i.e. dimensionless)
[at²] = LT^-2T² = L (i.e. a length)

Do you know any rotational kinematic equation? (They are analogous to the linear kinematic equations).

I presume he meant angular acceleration, $$\alpha$$ by "acceleration" in the equation though?
The problem with using that equation is that the angular acceleration of the car is not constant, depending on the velocity of the car at any given time.

 

[Hootenanny]

I presume he meant angular acceleration, $$\alpha$$ by "acceleration" in the equation though?
The problem with using that equation is that the angular acceleration of the car is not constant, depending on the velocity of the car at any given time.

Perhaps the OP did mean angular acceleration. However, a constant tangential acceleration implies that the angular acceleration is also constant, since the two are directly proportional.

 

[Fightfish]

Perhaps the OP did mean angular acceleration. However, a constant tangential acceleration implies that the angular acceleration is also constant, since the two are directly proportional.

Oh right, must have confused angular acceleration and centripetal acceleration for a moment there :(
That's probably what creechur must've done too.

I don't think I would bother with rotational kinematics for this question though, the relation arc length traveled s = vt, where v is the tangential velocity along with translational kinematics should suffice.

 

[creechur]

I did mean angular acceleration by the "a" in my equation, I just don't know how to make the symbols. If the approach I have taken isn't correct, what else would you do?

 

[Fightfish]

Consider the linear distance traveled by the car during the period. In one revolution, the car would have transversed a distance of $$2 \pi r$$. Thus, the number of revolutions can be obtained.

 

[creechur]

How do you take into account the change in acceleration?

 

[Fightfish]

How do you take into account the change in acceleration?

From the question, it is implied that tangential acceleration does not change; it is the centripetal or radial acceleration that increases as the car speeds up. Didn't you assume that when you solved for the final velocity as well?