Ball in Vertical Circle on End of String, Tangential and Radial Acceleration

AI Thread Summary
The discussion revolves around a physics problem involving a ball swinging in a vertical circle at the end of a 1.30 m rope, with a given total acceleration vector. The main challenge is determining the radial acceleration by first calculating the tangential acceleration, which is initially assumed to be zero. However, the tangential acceleration cannot be zero since the ball's velocity changes direction as it moves along the circular path. The confusion arises from the interpretation of the coordinate system, particularly the orientation of the radial direction relative to the angle of 36.1° past the lowest point. Understanding these concepts is crucial for accurately solving for both radial acceleration and the ball's speed and velocity.
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Homework Statement



A ball swings in a vertical circle at the end of a rope 1.30 m long. When the ball is 36.1° past the lowest point on its way up, its total acceleration is (-22.5 i + 20.2 j) m/s2.

(a) Determine the magnitude of its radial acceleration.

(b) Determine the speed and velocity of the ball.

Homework Equations



A = v^2/r
Ar = Atotal - Atangential

The Attempt at a Solution



I have no idea where to go with this. I am stuck on the first part. My issue is in figuring out the tangential acceleration so that I can subtract that from the total acceleration to find the radial/centripetal acceleration. My thought is that tangential acceleration should be zero, because the ball is swinging on the end of a string, and its speed should not be changing, only its velocity.
 
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You have a=(-22.5 i + 20.2 j) m/s2. The acceleration vector is made up of the sum of the radial and tangential acceleration. Using your coordinate system, the j vector points up or towards the center of rotation right? So the radial acceleration is?
 
rock.freak667 said:
You have a=(-22.5 i + 20.2 j) m/s2. the j vector points up or towards the center of rotation right?

no, up is not towards the center of rotation. it is 36.1 degrees past the bottom
 

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