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course Phy 121
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:
m = 110 g * 0.001 kg/ 1 g = 0.11 kg
v = 30 cm/s * 0.01 m/ 1 cm = 0.3 m/s
r = 20 cm * 0.01 m/ 1cm = 0.2 m
aCent = v^2/r
aCent = (0.3 m/s)^2/0.2 m
aCent = 0.09 m^2/s^2/0.2 m
aCent = 0.45 m/s^2
F = m*a
F = 0.11 kg * 0.45 m/s^2
F = 0.0495 kg m/s^2 or 0.0495 N
Direction ????
• What is the magnitude and direction of the centripetal force required to keep it moving around this circle?
answer/question/discussion:
Fgrav = m * v^2 /r
Fgrav = 0.11 kg * (0.3 m/s)^2/(0.2 m)
Fgrav = 0.11 kg * (0.09 m^2/s^2)/(0.2 m)
Fgrav = 0.11 kg * 0.45 m/s^
Fgrav = 0.24 kg m/s^2 or 0.24 N
Direction ????
Good.
The direction is toward the center of the circle.
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