Desperate! Circular motion - Acceleration and Velocity Analysis

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SUMMARY

This discussion focuses on analyzing circular motion, specifically calculating velocity (v) and tangential acceleration (dv/dt) for particles moving in a circle with a radius of 5m under varying acceleration conditions. The participants utilize the centripetal acceleration formula, a = mv²/r, and discuss the relationship between centripetal and tangential acceleration to find the required angles. Key calculations include determining v for given accelerations of 20 m/s², 30 m/s², and 50 m/s², resulting in velocities of 10 m/s and approximately 15.8 m/s, respectively.

PREREQUISITES
  • Centripetal acceleration formula (a = mv²/r)
  • Understanding of tangential acceleration (at = rθ')
  • Basic vector addition principles in physics
  • Knowledge of angular motion and its derivatives
NEXT STEPS
  • Study the relationship between centripetal and tangential acceleration in circular motion.
  • Learn how to derive angular velocity and its impact on linear velocity.
  • Explore the concept of angular acceleration and its calculation.
  • Investigate the effects of varying radius on circular motion dynamics.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to clarify concepts related to acceleration and velocity in circular paths.

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\theta'=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\theta'=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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