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Phy 121
Your 'cq_1_25.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_25.1_labelMessages.txt **
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: ->->->->->->->->->->->-> :
aR = v^2 / r
aR = (.30 m/s) ^2 / 20 cm = .9 m/s / .20 m = .45 m/s^2
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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: ->->->->->->->->->->->-> :
Fr = m * (v^2 / r)
Fr = .110 kg * ((.30 m/s)^2 / .2 m)
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Your units at this point are incorrect. (m/s)^2 is m^2 / s^2, so you would get
.110 kg * .09 m^2/s^2 / .2 m.
These units would give you kg m/s^2, which would yield N. The units you get below, kg * s, will not give you units of N.
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= .110 kg * .45 s
= .0495 N
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Except for the units, this is correct.
However having obtained the centripetal acceleration .45 m/s^2, you could use the mass with this result to directly get the result as
F_cent = .110 kg * .45 m/s^2 = .0495 N.
The formula m v^2 / r is fine, but in this context where you already know mass and acceleration it is redundant.
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15 min
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&#Good responses. See my notes and let me know if you have questions. &#