Posted by Tammy on Wednesday, October 31, 2007 at 8:15pm.
A .5kg ball on the end of a string is revolving uniformly in ahorizontal circle of radius 1m. The ball makes 2 revolustion in a second.
a) determine the linear (tangential) speed of the ball.
For this part of the question would I do this:
v=d/t
=1m/.5s
=2m/s
b)determine the ball's centripetal acceleration
a_c=v^2/r
=2m/s^2/1
=4m/s^2
c)determine the force a person must exert on the opposite end of the string.
F_c=mv^2/r
=.5kg(2m/s)^2/1
=2N
Is this correct? The only thing i'm not too sure about is part a). Thanks in advance for your help.

physics  bobpursley, Wednesday, October 31, 2007 at 8:28pm
The distance around the circle is 2PI*radius, v= distance/time You have distance equal to radius.
The centripetal force is indeed v^2/r, your numbers are of course wrong.
part A messes up part b, c.
Answer This Question
Related Questions
 physics  A ball on the end of a string is revolving at a uniform rate in a ...
 physics  A ball on the end of a string is revolving at a uniform rate in a ...
 physics  An athlete swings a 4.30 kg ball horizontally on the end of a rope. ...
 physics  a ball of weight 200 grams,is tied to the end of a cord and whirled in...
 Physics  An athlete swings a 3.46 kg ball horizontally on the end of a rope. ...
 Physics  An athlete swings a 2.01 kg ball horizontally on the end of a rope. ...
 Physics  An athlete swings a 3.06 kg ball horizontally on the end of a rope. ...
 physics  An athlete swings a 3.06 kg ball horizontally on the end of a rope. ...
 Physics  A ball is fastened to one end of a 30cm string, and the other end is ...
 Physics  A ball is fastened to one en of a 28 cm string, and the other end is ...
More Related Questions