# Circular Motion -- Finding Centripital Acceleration, Speed, etc.

1. Sep 16, 2014

### beniamin

1. The problem statement, all variables and given/known data
An object is constrained to move in a circular path of radius r = 3.1 m. At the instant diagrammed in the Figure 1, the directions of the velocity and acceleration are indicated by the v and a, respectively.The magnitude of the acceleration is a = 11.2 m/s^2 and the angle shown is θ = 32.

2. Relevant equations

a) What is the centripetal acceleration of the object?
b) What is the speed of the object?
c)What is the tangential acceleration of the object?
d) Is the speed increasing or decreasing? Explain.

3. The attempt at a solution
a = ac + at

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Last edited: Sep 16, 2014
2. Sep 16, 2014

### DEvens

So, how do you work out the components of the acceleration? You have the magnitude and the angle relative to the radius.

Once you have the radial acceleration, how do you then get the velocity? What does the fact that the object is constrained to move in a circle mean about the velocity?

Given that the velocity is towards the upper left, and given the answer to c, what then will be the answer to d?

3. Sep 16, 2014

### tjmiller88

Starting first with part (a), draw a vector showing the centripetal acceleration (ac) on your diagram. Ask yourself what centripetal acceleration is, and look up the formula.

Get this down first and you'll be able to solve part (a) and part (b).

Then worry about part (c) and (d) after you get to that point.

4. Sep 16, 2014

### beniamin

Thank you guys Ac is a cos 32
Velocity I get it from ac= v'2 / r
At is a sin 32
d) increasing since centripetal and tangential acceleration are positive

5. Sep 16, 2014

### Orodruin

Staff Emeritus
How did you define positive tangential acceleration? Did you define positive velocity in the same way? What is the speed in terms of the velocity?

6. Sep 16, 2014

### tjmiller88

You've got the formulas correct, but your answer to part (d) is not correct based on your initial drawing of the problem.

Go back and see if you can draw the at and ac vector components on your circle. Then compare the direction of at and v. Are they going in the same direction or different?

7. Sep 16, 2014

### beniamin

I see what you guys are saying. That means that it is decreasing since based on my drawing the tangential acceleration is in the opposite way of velocity.

8. Sep 16, 2014

### Orodruin

Staff Emeritus
Spot on. More generally, the square of the speed is the same as the square of the velocity and the square of the speed will be increasing/decreasing if its square is since it is a positive quantity. If we call the speed $v$ and the velocity $\vec v$, we obtain
$$\frac{dv^2}{dt} = \frac{d\vec v^2}{dt} = 2 \vec v \cdot \frac{d\vec v}{dt} = 2 \vec v \cdot \vec a.$$
Thus, the speed is increasing if the scalar product between the velocity and acceleration is positive and decreasing if it is negative. (This is equivalent to the acceleration projected onto the velocity direction being positive/negative.)