# Dynamics/Mechanics problem

1. Nov 3, 2009

### paul_harris77

I am attempting to solve the dynamics question below:

"A car of mass 600kg is driven in a circle of constant radius (100m). It is known that the car becomes unstable under a side force of 6800N. Determine the minimum time allowable for the car to accelerate uniformly from 80 to 100km/h."

m= 600kg, r =100m and F=6800N.

I am assuming that the overall aim is to obtain the tangential acceleration (aθ) i.e. the acceleration in the car's direction of travel, and from that work out the time using the change in velocity of 20km/h.

I have worked out the "resultant" acceleration on the side of the car, using a=F/m = 6800/600 = 11.33ms^-2.

The trouble I am having is splitting this overall "resultant" acceleration into tangential (aθ)

I know the equation for resultant acceleration is derived from simple pythagoras:

a = sqrt((aθ^2)+(ar^2))

I also have the following equations:

ar = wr^2 (where w is omega and ar is radial acceleration)
aθ = r*alpha (where alpha is angular acceleration)
vθ=wr

Since, w and vθ are changing, I have started by trying to work out the acceleration components in terms of changes in vθ and w over time.

The answer should be 4.9 seconds, but I am stuck with how to get this answer.

Any help would be greatly appreciated.

Many thanks

Paul

2. Nov 4, 2009

### Staff: Mentor

Those are the equations I would start with, and use them to calculate the forces in the radial and tangential directions. Add those up vectorially to find the magnitude of the total force. Presumably the tangential acceleration will have to be limited to some value to keep the total force from exceeding the value given.

3. Nov 4, 2009

### paul_harris77

Thanks for reply. I believe that what the question means by side force is the overall resultant force acting on the side of the car from the radial and tangential forces combined. Hence the the magnitude of the total "resultant" force on the car is the side force of 6800N. So this force and hence the acceleration of 11.33ms^-2 needs to be split up into tangential and radial components.

I can work out aθ in terms of the minimum time, t, but I cannot do the same for the radial acceleration. Radial acceleration is equal to wr^2, but since w is accelerating, I cannot use this and cannot get it into terms of t.

Any ideas?

Thanks

Paul

4. Nov 4, 2009

### Staff: Mentor

I'm not seeing the difficulty. You have an w(t) which starts out at some w(0) and increases according to the acceleration. At any instant, the radial force is just mw^2/R, as you have stated. The force to accelerate the car in the tangential direction is F=ma, which you can relate to the angular acceleration as you have indicated....

5. Nov 5, 2009

### paul_harris77

Sorry but I still do not understand. Since the resultant force ( side force of 6800N) is constant and the tangential force is constant (uniform acceleration), surely the radial force has to be constant? But it isn't since w is increasing.

6. Nov 5, 2009

### Staff: Mentor

The tangential force cannot be constant. In order to optimize the acceleration (minimize the time to the higher speed, as asked in the OP), you will have to reduce the tangential acceleration as the angular speed increrases. You can accelerate harder in the beginning, but less so near the end, or you will exceed the max total vector acceleration magnitude.

7. Nov 5, 2009

### paul_harris77

Thanks for the reply. So, presumably, I need to work out the angular acceleration (alpha) for w(0) and w(t). Do I do this by using the vector sum of the accelerations, substituting in w(0) and w(t) and rearranging for alpha?:

11.33^2=((w^2r)^2)+((alpha*r)^2)

If I do this for w(0) I get alpha= 0.1133 rads^-2. This means that the tangential acceleration is aθ = r*alpha.

This gives me a value of 11.33ms-2 which surely cant be right since the overall resultant acceleration was found to be 11.33ms-2?!

If I do the same for w(t), I get aθ = 2.23ms-2 which seems to concur with your explanation of decreasing tangential acceleration.

If so, how am I to calculate the minimum time if the acceleration in the tangential direction is constantly changing?

Another thing I am confused about, is the fact that the OP said uniform acceleration. I thought that this means uniform acceleration in the tangential direction. Surely if you were in a car and put your foot down on the gas, and left it in one position, you would accelerate uniformly in the tangential direction. Does this mean, that in order to keep the side force below 6800N, you would need to ease off on the gas as your angular velocity increases?

8. Nov 5, 2009

### Staff: Mentor

I don't see where the OP problem statement says uniform acceleration... It says constant radius, and asks for the minimum possible time to accelerate, but that's about it (unless I'm missing something).

As for the calculation, I'm guessing it will be an integral equation, since the accelerations are not constant.

EDIT -- And yes, you would be lifting a bit on the accelerator as you got going faster around the circle...

9. Nov 5, 2009

### paul_harris77

I am assuming that it means uniform tangential acceleration, since that it the direction of the velocity increase. Surely this means that the side force cannot be constant at 6800N, rather it is an upper bound, since radial acceleration HAS to increase?

10. Nov 5, 2009

### Staff: Mentor

I'm interpreting it as the magnitude of the total acceleration, which is the vector sum of the radial and tangential accelerations. You have the radial acceleration vector pointing in toward the center of the circle, and the tangential acceleration vector pointing in the direction that the car is moving (tangent to the circular motion). The vector sum is going to point mostly toward the center of the circle, but a bit forward of that with respect to the car. The magnitude of that vector sum is constrained by the max acceleration specified. So as the car moves faster and the radial component increases, the tangential component must decrease in order to keep the magnitude of the vector sum at that maxumum limit.