Question

a 3.2m long rope can with stand 300N before breaking. If a mass of 18kg is attached to end of rope and rotated in a vertical circle. what is max speed at bottom position before breaking?

I thought that you would just use Fnet = m(v^2/R) and then solve for v, but when i do that the answer is not one of the multiple choice answers in my book. Not sure what i am missing.

Answer 1

the equation i used was \[F _{net} = m(v ^{2}/R)\]
Is there a way to put equations into the original post. i don't see the equation button when originally asking a question

Answer 2

i said
300 = 18(v^2/3.2)
divide by 18
16.67 = v^2 / 3.2
multiply both sides by 3.2
53.33 = v^2
and then square root both sides
v = 7.3
but that is not one of the answers so i dont understand what im not getting

Answer 3

You worked out the centrifugal force but did not take into account the force of gravity acting on the mass, hence the rope, at the bottom of the "vertical" circle.
You need to add the additional force, in Newtons, for an 18 kg mass when accelerated at 9.8 m/s^2 (gravity).
You will need to start with this force as if the rope were just hanging with the 18 kg mass on it (regardless of the length of rope) and work backwards with the maximum total centrifugal force being equal to 300 N minus the force from gravity.

Answer 4

Alright i took your advice and all i had to do was multiply 18kg * 9.8m/s and then i subtracted that from 300N and worked it like before and it gave me an answer that was part of the multiple choice. Also as a trivial side note. My physics teacher made a HUGE point to the class that it is NOT called centrifugal it is centripetal force. Said that was actually called radial acceleration and that when labeling a free body diagram centrifugal is never used only \[a_{rad}\] which is centripetal force. But thanks for the help.

Answer 5

You're right not to use the term centrifugal. It is quite confusing.
Glad you got an answer.