How Long Does It Take for the Cart to Complete One Rotation?

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Homework Statement


An energetic father places his 20 kg child on a 5.0 kg cart to which a 2.0 m long rope is attached. He then holds the end of the rope and spins the cart and child around in a circle, keeping the rope parallel to the ground. If the tension in the rope is 100 N, how much time does it take for the cart to make one rotation?




Homework Equations


F=ma=m(w^2)r where w is omega then w^2= (2pi/T)^2)= 4 * pi^2/ T^2

since w for one rotation in the circle is equal to 2pi radians/T then F=m * ((2pi/T)^2) * r /r. The r's in the equation cancel and you are left with F= 25 * 4 * pi^2/T^2

So 100T^2=987 T=3.1 seconds but the answer is 4.4 seconds. what am I doing wrong?


The Attempt at a Solution



since w for one rotation in the circle is equal to 2pi radians/T then F=m * ((2pi/T)^2) * r /r. The r's in the equation cancel and you are left with F= 25 * 4 * pi^2/T^2

So 100T^2=987 T=3.1 seconds but the answer is 4.4 seconds. what am I doing wrong?

Thank you kindly for your help.
 
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sona1177 said:
F=ma=m(w^2)r where w is omega then w^2= (2pi/T)^2)= 4 * pi^2/ T^2

since w for one rotation in the circle is equal to 2pi radians/T then F=m * ((2pi/T)^2) * r /r. The r's in the equation cancel ...
Why are you dividing this by r?
 
dulrich said:
Why are you dividing this by r?

Because I was being careless and mixing F=mw^2r with F=mv^2/r and thinking the equation was F=mw^2r/r. Sorry that was an ignorant error on my part. Thanks for asking that question though, it helped me realize that I was mixing the two equations up!
 
No apologies necessary. :smile: I'm glad the nudge worked for you to be able to answer your own question!