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August 21, 2014

August 21, 2014

Posted by **Sarah** on Monday, June 8, 2009 at 12:49pm.

- Physics -
**MathMate**, Monday, June 8, 2009 at 1:04pm(a), (b)

acceleration due to circular motion

= rω^{2}towards the centre

It is necessary to add/subtract the acceleration due to gravity at different points on the wheel

(c)

tangential speed, v= rω

v and r are know, so ω can be determined.

- Physics -
**Sarah**, Monday, June 8, 2009 at 1:12pmFor C i got .5 or 1/2 v=rw

7.0=14.0 w is that correct

im still a little confused on a and b

- Physics -
**bobpursley**, Monday, June 8, 2009 at 1:36pmat the top, net acceleration=centacc-9.8

at the bottom, add them to get net acceleration

centripetal acceleration=v^2/r r, v given.

distanceAround/time=v

solve for time.

- Physics -
**MathMate**, Monday, June 8, 2009 at 1:42pmThat's correct. The unit is radians / s.

For a and b, the acceleration I was referring to corresponds to the centripetal force required to keep the passenger in place.

Thus at the lowest point, the weight of the passenger adds to this centripetal force, and so the acceleration due to gravity, g, is additve to this acceleration.

On the other hand, at the highest point, the acceleration due to gravity tends to reduce the centripetal force required to keep the passenger in its place, so g must be subtracted from the centripetal acceleration that you calculate from the formula a=ω^{2}r.

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