Phy 121
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A steel ball of mass 110 grams moves with a speed of 30 cm / second around a circle of radius 20 cm.
• What are the magnitude and direction of the centripetal acceleration of the ball?
answer/question/discussion: ->->->->->->->->->->->-> :
Centripetal acceleration is determined by v^2/r. To solve for our centripetal acceleration, we use 30 cm/s^2 / 20cm. = 900cm^2/s^2 / 20cm. This gives us an acceleration of 45cm/s^2. I think we still need to take the square root of this. If we did so it gives us 6.7 m/s^2. I solved for the centripetal acceleration, but I am not sure how to solve for the direction of the centripetal acceleration.
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• What is the magnitude and direction of the centripetal force required to keep it moving around this circle?
answer/question/discussion: ->->->->->->->->->->->-> :
Centripetal force is equal to mass * centripetal acceleration. So Fcent = 110 grams* 6.7cm/s^2. I also think the mass of the object needs to be converted over to kilograms. 110 grams would be equal to .011kg. So our problem now reads as Fcent= .11kg * 6.7 cm/s^2. We also need to convert our cm over to m. This gives is .67m/s^2. To solve for the Fcent we now have .011kg * .67m/s^2 which gives us .00737N. This seems like a really small amount though and I am not sure if it is correct or not.
I am not quite sure how to determine the direction of the centripetal force for this situation.
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20 minutes
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You didn't need to take that square root, and you need to understand the direction of the centripetal force. However most of your steps are correct, so you should easily understand the solution given in the discussions. Let me know if not.
&#Please compare your solutions with the expanded discussion at the link
Solution
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