Phy 201
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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: ->->->->->->->->->->->-> :
aCent = v^2 / r
aCent = (.3 m/s)^2 / .2 m
aCent = .09 m^2/s^2 / .2 m
aCent = .45m/s/s
Good.
atan = r * pi
atan stands for the 'arc-tangent' function; this function must be applied to an angle in order to have a meaning.
It's possible that you are confusing this with arc distance around the circle, which would be 2 pi r or about 1.26 meters.
atan = .2 m * pi
atan = .63 m/s/s
.2 m * pi is about 1.26 meters, half the distance around the circle; this product cannot give you units of m/s/s.
Magnitude = sqrt((.45 m/s/s)^2 + (.63 m/s/s)^2)
Magnitude = sqrt((.202 m^2/s^4) + (.397 m^2/s^4))
Magnitude = sqrt(.594 m^2/s^4)
Magnitude = .774 m/s/s
Direction = theta = tan^-1 (atan / aCent)
Direction = theta = tan^-1 (.63 m/s/s / .45 m/s/s)
Direction = 54.46 degrees
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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 = Gravitational Force
Fg = .110 kg * 9.8 m/s/s
Fg = 1.078 Newtons, so the centripetal force would have to be 1.078 N, in the positive direction.
Gravity is not assumed in this problem, nor need it be assumed. In any case, while all or part of the centripetal acceleration at certain points in certain situations might be gravitational in nature, gravity isn't relevant to the present question.
#$&*
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30 minutes
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Solution
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