What is the magnitude of total acceleration for a particle on a circular path?

In summary, the problem involves a particle moving along a circular path with a radius of 2.0 m. At an instant when the speed is 3.0 m/s and changing at a rate of 5.0 m/s^2, the magnitude of the total acceleration of the particle is 6.7 m/s^2, with a component towards its center. This was found by using the equation a=sqrt(at^2+ar^2) and plugging in the values for at (5 m/s^2) and ar (4.5 m/s^2).
  • #1
rasikan
21
0

Homework Statement


A particle moves along a circular path having a radius of 2.0 m. At an instant when the speed of the particle is equal to 3.0 m/s and changing at the rate of 5.0 m/s2, what is the magnitude of the total acceleration of the particle?



Homework Equations


magitude= V^2/R




The Attempt at a Solution



But i don't know how to slove with this equation
 
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  • #2
Acceleration is a vector. There will be a component tangent to the circle and a normal component. Which one is given to you in the problem statement and which do you have to find using your equation?
 
  • #3
Dick,
v=3m/s
a=5 m/s^2

I guess I can find the magnitude when v=3 m/s . is that
right?


Dick said:
Acceleration is a vector. There will be a component tangent to the circle and a normal component. Which one is given to you in the problem statement and which do you have to find using your equation?
 
  • #4
Ok, so what are the two components of the acceleration.
 
  • #5
the two components are ax and ay
 
  • #6
The components are normal to the circle and tangent to the circle, so x,y coordinates are probably not appropriate. I would say the tangent component is 5 m/s^2. Why would I say that?
 
  • #7
coz that is a constant acceleration?
 
  • #8
No, because they told me. If an object is moving on a circle and it's speed is changing at 5 m/s^2, in what direction is that acceleration pointed?
 
  • #9
this is my calculation

a=Sq root of (at)^2+(ar)^2

at= 5m/s^2
ar=v^2/r= 9/2=4.5
a=sq root(5^2+4.5^2)
=6.7 m/s^2

is this is right??
 
  • #10
towards its center
Dick said:
No, because they told me. If an object is moving on a circle and it's speed is changing at 5 m/s^2, in what direction is that acceleration pointed?
 
  • #11
rasikan said:
towards its center

Your calculation is fine. What does "towards its center" mean? You labelled it as at in your calculation?
 
  • #12
u mean my answer is right??
 
  • #13
Yeah. Does that surprise you? If you're clear on the two components of the acceleration then I think you can go on to the next problem...
 
  • #14
yaa i got it now dick thanks for you help
 

What is magnitude of acceleration?

The magnitude of acceleration is the measure of how much an object's velocity changes over a specific period of time. It is usually represented by the symbol "a" and measured in meters per second squared (m/s²).

How is magnitude of acceleration calculated?

The magnitude of acceleration can be calculated by dividing the change in an object's velocity by the time it took for that change to occur. This can be represented by the equation a = Δv/Δt, where Δv is the change in velocity and Δt is the change in time.

What does a positive magnitude of acceleration indicate?

A positive magnitude of acceleration indicates that an object is speeding up, or increasing its velocity. This could be due to a number of factors, such as a force acting on the object or an incline causing it to move downhill.

What does a negative magnitude of acceleration indicate?

A negative magnitude of acceleration indicates that an object is slowing down, or decreasing its velocity. This could be due to factors such as friction or a force acting in the opposite direction of motion.

Why is magnitude of acceleration important in physics?

Magnitude of acceleration is important in physics because it helps us understand how objects move and how forces affect their motion. It is a fundamental concept in studying motion, and is crucial in fields such as mechanics, kinematics, and dynamics.

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