Question

A ball hangs on a string and is rotated in a circle(m=.275, r=.850). If the String breaks when its tension exceeds 22.5N, What is the maximum speed the ball can have at the bottom before it breaks?

Answer 1

Vertical or horizontal circle. Like a water wheel on an old corn mill or like a record on a turn table. (These are probably terrible analogies for the younger crowd.)

Answer 2

Vertical circles like the water wheel

Answer 3

Since we only care about the case when the mass is at the bottom of the circle, this simplifies things.

Draw a FBD. We assume \(\omega\) (angular velocity) is constant.

Let's balance the forces. \[\sum F = ma = 0 = T - mg\]T is equal to the centripetal force. \[T = {mv^2 \over r}\]

With me?

Answer 4

Yes I am with you so far

Answer 5

Hmm. We are over defined.

Answer 6

So I would set that up as

\[22.5N = \frac{mv^2}{r}-mg\]

When I do that I get a different answer than the book says

Answer 7

That's because I did it wrong.

T doesn't equal the centripetal force. \[\sum F_r = ma_c = T - mg\]

Now, \[ma_c = {mv^2 \over r}\]

This yields, \[{mv^2 \over r} = 22.5 - mg\]

Answer 8

Awesome bro Thanks a ton.

Answer 9

I attempted to set it up the same way you did the first time.

Answer 10

Yeah. Sorry about the misstep.

Hopefully you're clear on the "why" behind the proper set-up.

Answer 11

If not, let me know and I'll lay it out more fundamentally.

Answer 12

No i'm with you bur for whatever reason I wanted to set ma=22.5 instead of setting tension to 22.5. Thanks for the help

Answer 13

Great. Good luck.