Desperate! Circular motion - Acceleration and Velocity Analysis

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Homework Help Overview

The discussion revolves around a problem involving circular motion, specifically analyzing acceleration and velocity of a particle moving in circles of radius 5m. The problem presents three scenarios with varying acceleration vectors and asks for the values of velocity and the rate of change of velocity at specific times.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of centripetal acceleration and the need to calculate tangential acceleration. There are questions regarding how to derive tangential acceleration from given formulas and how to combine different acceleration components to find angles.

Discussion Status

Some participants have provided guidance on using the centripetal acceleration formula and the relationship between centripetal and tangential acceleration. There is an ongoing exploration of how to calculate the required values, with some participants expressing confusion about specific calculations and concepts.

Contextual Notes

Participants are working under time constraints and express uncertainty about how to proceed with calculations, particularly regarding the tangential acceleration and its relationship to the overall acceleration vector.

nns91
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Desperate ! Circular motion

Homework Statement



Particles travels counterclockwise in circles of radius 5m. The acceleration vectors are indicated at three specific times. Find the values of v and dv/dt at each time.

a. a=20 m/s^2 . vector a points toward the center of the circle, vector v tangent to the circle

b. a=30 m/s^2. vector a points toward the center of the circle at a 30 degree angle in the direction of motion ( so imagine the case a, vector a point directly to the center, this case vector a and the vector a in part a creates a 30 degree)

c. a= 50 m/s^2. vector a creates a 45 degree angle clockwise with the radius.

Homework Equations



a=mv^2/s

The Attempt at a Solution



I know that it's confusing. Please give some input. if you guys are unclear I can explain more.
 
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Hi nns91! :smile:

(have a theta: θ :wink:)

I assume you know how to do a., using centripetal acceleration.

ok, for the others, you have the usual centripetal acceleration formula, and you also need to calculate the tangential acceleration (for constant r, that's at = rθ'' = v').

Acceleration is a vector, so you add accelerations just like velocities … add the centripetal and tangential accelerations, to get the required angle. :wink:
 


Thanks.

I know acentripetal=mv2/2

The tangential part confuses me the most. How can I calculate it from that formula at=r[tex]\theta[/tex]'=v' ??

What do you mean by to get the require angle ?
 
nns91 said:
Thanks.

I know acentripetal=mv2/2

The tangential part confuses me the most. How can I calculate it from that formula at=r[tex]\theta[/tex]'=v' ??

What do you mean by to get the require angle ?

Hi nns91! :smile:

(why didn't you use that θ i gave you? :rolleyes:)

at = dv/dt, which you get from the question.

And if you want the total acceleration to be at 45º, for example, then obviously you need the centipetal acceleration to be equal to the tangential acceleration … just like adding velocities! :wink:
 


But I don't have time ? so how can I get dv/dt ?
 
nns91 said:
But I don't have time ? so how can I get dv/dt ?

Well, for 45º, for example, you need v2/r = dv/dt, so v = … ? :smile:
 


How about a 30 degree angle ?
 
nns91 said:
How about a 30 degree angle ?

Oi! Do it yourself!
 


I mean I still have not fully understand.

So they give a=20 m/s^2 in part a.

Thus: v=sqrt(ar)= sqrt(20*5)= 10 m/s

Is that right ??

I still don't understand the calculation of acceleration
 
  • #10
nns91 said:
So they give a=20 m/s^2 in part a.

Thus: v=sqrt(ar)= sqrt(20*5)= 10 m/s

Is that right ??

That's right. :smile:

Now try c.
 
  • #11


c.

v= sqrt(a*r)=sqrt(50.5)= 15.8 m/s ??

So I got v.

How then can I calculate dv/dt ?
 

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