A question on vertical circular motion

1. Feb 13, 2010

Charismaztex

1. The problem statement, all variables and given/known data

An object of 6.0kg is whirled round in a vertical circle of radius 2.0m with a speed of 8.0m/s. If the string breaks when the tension in it exceeds 360N, calculate the maximum speed of rotation, and state where the object will be when the string breaks.

2. Relevant equations

The usual circular motion equations

3. The attempt at a solution

I calculated the max speed to be over 10m/s when the object is at the top (tension force + weight force =centripetal force, hence greater centripetal force and therefore greater maximum speed of rotation), but my sources suggest that the speed is greatest when the tension exceeds 360N at the bottom (which is tension force - weight force =centripetal force, leading to a smaller centripetal force and hence less speed.)

I am needing assistance to clarify the situation. how is it that the linear speed at the bottom is greater than the top (as my teacher's notes state). Would the object have a greater linear speed and a greater speed of rotation at the top because the centripetal force is greater there?

I may have totally confused the physics...

Charismaztex

2. Feb 13, 2010

Jebus_Chris

The tension at the top is less than the tension at the bottom because...
TOP: $$T+mg=\frac{mv^2}{r}$$
BOTTOM: $$T-mg=\frac{mv^2}{r}$$
If you slowly increase the velocity then the string will break at the bottom. If you somehow spin it so that it doesn't exceed that speed at the bottom (or anywhere else) it could break on the top which would require the maximum velocity.
You're assuming constant velocity so every point on the circle would experience the each velocity so it would break at 10m/s at the bottom.

3. Feb 13, 2010

Charismaztex

So you are saying that it wouldn't reach the speed calculated for the top (about 11m/s which I presumed was the answer instead of the correct one 10m/s) as it would have already broken at 10m/s at the bottom, which therefore would make 10m/s the max speed?

4. Feb 13, 2010

Charismaztex

So always take the values calculated at the bottom as it would always break first there (greatest tension). Is this statement correct?

Please deny or confirm, thanks (I think this is the crux of the problem.)

5. Feb 14, 2010

rock.freak667

yes if you rearrange the equations Jebus_Chris gave, you would see that

Ttop=mv2/r - mg

Tbottom=mv2/r +mg

so clearly Tbottom>Ttop, so the tension is always greatest at the bottom.

6. Feb 14, 2010

Charismaztex

Thanks to rock.freak667 and Jebus_Chris for your assistance.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}