#$&*
Phy 201
Your 'cq_1_25.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A steel ball of mass 110 grams moves with a speed of 30 cm / second around a circle of radius 20 cm.
• What are the magnitude and direction of the centripetal acceleration of the ball?
answer/question/discussion: ->->->->->->->->->->->-> :
We can first find the Force by multiplying m*a to get: 3300N.
The mass is 110 grams. You don't know the acceleration.
The units of your apparent calculation would be g cm / second, which is not a unit for force. You appear to have calculated the momentum.
We can then find the x by: 3300cos(20)=3101
Y= 3300sin(20)= 1129
The magnitude is: sqrt( 3301^2+1129^2)= 3300
The direction is: artan(1129/3101)= 20degrees, we then add 180 to get 190 degrees
#$&*
• What is the magnitude and direction of the centripetal force required to keep it moving around this circle?
answer/question/discussion: ->->->->->->->->->->->-> :
The same magnitude and direction found in the first question should allow the ball to keep moving.
#$&*
** **
10 minutes
** **
You do need to submit a revision on this one.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#