Circular Motion with constant angular acceleration

In summary, the object travels a circular path with a radius of 2πR, and has a constant angular acceleration of α. At time t = 0, its coordinates are r(0) = r(T), and at time t = T, its coordinates are r(T) = r(0). The magnitude of r(t) is R, and the object returns to its original position at time t = T when its coordinates are r(T) = r(0).
  • #1
anhchangdeptra
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0

Homework Statement



An object travels counterclockwise on a circular path with radius R and constant angular
acceleration α , so that

vector r(t) = R cos(αt^2/2) i^+ R sin(αt^2/2) j^


Homework Equations



b. Find the time T when the object made a single revolution and returned to its
original position. Evaluate vectors r, v, and a at both t = 0 and t = T.
c. Show by computation that at t = T, the acceleration vector is the sum of
a part parallel to the velocity vector with magnitude dv/dt , and a part perpendicular to the
velocity vector with magnitude v^2/R

The Attempt at a Solution



I am calculating based on the fact that the object will travel a distance of 2πR at the time it made a revolution, but it doesn't work !
 
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  • #2
A revolution brings the object where it was originally. Express that mathematically.
 
  • #3
voko said:
A revolution brings the object where it was originally. Express that mathematically.

I am sorry I really don't know how to express that mathematically. I have just calculate its speed to be Rαt but I can not make an equation because the object has an increasing acceleration (because its angular accel is constant). This is quite new to me.
 
  • #4
Well, take your equation for r(t) from the OP.

What are the coordinates for the object at time t = 0?

At time t = T, you will have these same coordinates. Knowing that sine and cosine are periodic functions, use this fact to figure out what T must be to return the object to its original position.
 
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  • #5
r(t) that you were given is the position of the object. As one revolution brings the object where it started from, you should have r(0) = r(T).
 
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  • #6
Oh thank you SteamKing and Voko I know how to do it now. My problem is that I was stuck with the idea that the magnitude of r(t) is always R so I thought I must use another equation rather than r(t). Thanks a ton!
 

1. What is circular motion with constant angular acceleration?

Circular motion with constant angular acceleration is a type of motion where an object moves in a circular path with a constant change in its angular velocity. This means that the object's speed or direction of motion is changing at a constant rate as it moves along the circular path.

2. How is angular acceleration different from linear acceleration?

Angular acceleration is the rate of change of angular velocity, while linear acceleration is the rate of change of linear velocity. In circular motion, angular acceleration is caused by a change in the object's direction, while linear acceleration is caused by a change in its speed.

3. What is the formula for calculating angular acceleration?

The formula for calculating angular acceleration is α = (ωf - ωi) / t, where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time taken for the change in angular velocity to occur.

4. How does the direction of angular acceleration affect circular motion?

The direction of angular acceleration determines the direction of the change in angular velocity. If the angular acceleration is in the same direction as the object's motion, the object's speed will increase. If the angular acceleration is in the opposite direction, the object's speed will decrease.

5. How does centripetal acceleration relate to circular motion with constant angular acceleration?

Centripetal acceleration is the acceleration directed towards the center of a circular path. In circular motion with constant angular acceleration, the centripetal acceleration is caused by a change in the direction of the object's velocity, and it is always perpendicular to the object's velocity.

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