- #1

a.mlw.walker

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- 0

so i am looking at the issue of a ball, spinning in a circle, and the surface the ball is on, is slanted, towards the centre

- same as spinning a piece of string with a ball on around your head i think - don't worry about that...

I have attached 2 pictures of ball on the surface at one time, tilted at an angle theta.

In the side on image i have labeled the horizontal component Rsin[theta],

and the component = to the force of gravity = Rcos[theta].

thefore R = mg/cos[theta]

therefore horiztonal component Rsin[theta] = mgsin[theta]/cos[theta] or mgtan[theta]

centrepetal force is equal to horizontal component, it is not a new force of its own, so

mv^2/r = mgtan[theta]

[theta] is the angle of the slope.

so the bit i am confused about is whether i can use SUVAT equations here to work out where the ball will end.

for instance if i had taken times, of the ball so i knew the speeds it was traveling at for two separate but consecutive revolutions, (2*PI/t) then i could calculate the deceleration from

dv/dt = a

So if i know these two speeds for two revolutions, and work out the deceleration, can i work out how many radians the ball will now go through until it drops off the slope?

I think that the speed it drops at will always be the same, because the horizontal component, Rsin[theta] less than mg, the ball will begin to fall.

Yeah so basically how do i work out where the ball will be (in radians) when the ball begins to fall from the track.

Thanks

- same as spinning a piece of string with a ball on around your head i think - don't worry about that...

I have attached 2 pictures of ball on the surface at one time, tilted at an angle theta.

In the side on image i have labeled the horizontal component Rsin[theta],

and the component = to the force of gravity = Rcos[theta].

thefore R = mg/cos[theta]

therefore horiztonal component Rsin[theta] = mgsin[theta]/cos[theta] or mgtan[theta]

centrepetal force is equal to horizontal component, it is not a new force of its own, so

mv^2/r = mgtan[theta]

[theta] is the angle of the slope.

so the bit i am confused about is whether i can use SUVAT equations here to work out where the ball will end.

for instance if i had taken times, of the ball so i knew the speeds it was traveling at for two separate but consecutive revolutions, (2*PI/t) then i could calculate the deceleration from

dv/dt = a

So if i know these two speeds for two revolutions, and work out the deceleration, can i work out how many radians the ball will now go through until it drops off the slope?

I think that the speed it drops at will always be the same, because the horizontal component, Rsin[theta] less than mg, the ball will begin to fall.

Yeah so basically how do i work out where the ball will be (in radians) when the ball begins to fall from the track.

Thanks