#$&*
Phy 121
Your 'cq_1_25.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_25.1_labelMessages.txt **
A steel ball of mass 110 grams moves with a speed of 30 cm / second around a circle of radius 20 cm.
• What are the magnitude and direction of the centripetal acceleration of the ball?
answer/question/discussion: ->->->->->->->->->->->-> :
To find the centripetal acceleration, we use the equation for centripetal acceleration, which is aR = v^2 / r. Thus we would plug in aR = (.30 m/s) ^2 / 20 cm = .9 m/s / .20 m = .45 m/s^2, which is equal to the magnitude. Direction must be toward the center of the circle.
#$&*
• What is the magnitude and direction of the centripetal force required to keep it moving around this circle?
answer/question/discussion: ->->->->->->->->->->->-> :
In order to find the magnitude of the centripetal force that would keep the ball going, we need to use the equation Fr = m * (v^2 / r) to get Fr = .110 kg * ((.30 m/s)^2 / .2 m) = .110 kg * .09 m/s / .2 m = .110 kg * .45 s = .0495 N. This is true because F = ma, so Fr = maR, so that Fr = m * (v^2/r). The circular motion is indicative of a v = constant, so that aR or acceleration is directed toward the center of the circle at any moment. Thus the net force must also be directed at the center of the circle.
#$&*
&#Very good responses. Let me know if you have questions. &#