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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.
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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: ->->->->->->->->->->->-> :
centA = (.3 m/s)^2 / .2 m = 1.5 m/s^2
@& I suspect you didn't square the 3. I believe that calculation gives you .45 m/s^2.*@
centF = 1.5 m/s^2 * .11 = .165 N
circumference = 2 * pi (.2 m) = 1.26 m
@& The direction of the force is toward the cente of the circle.
1.5 m/s^2 is an acceleration, not an angle. The cosine of an acceleration doesn't have any meaning.*@
x comp = 1.26 m * cos(1.5 m/s^2) = 1.26 m/s^2
y comp = 1.26 m * sin(1.5 m/s^2) = .03 m/s^2
magnitude = sqrt( (1.26 m/s^2)^2 + (.03 m/s^2)^2 ) = 1.26 m/s^2
ang = tan^-1(.03 m/s^2 / 1.26 m/s^2) = 1.36 deg
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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: ->->->->->->->->->->->-> :
x comp = 1.26 m * cos(.165 N) = 1.26 N
y comp = 1.26 m * sin(.165 N) = .004 N
magnitude = sqrt( (1.26 N)^2 + (.004 N)^2 ) = 1.26 N
ang = tan^-1(.004 N / 1.26 N) = .18 deg
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15 mins
@& `c251
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