will it fall off?

[methodsboy]

In a loop the loop novelty ball game.....eh fcuk it (im going straight to the point).... Ball has mass 50g. It must go fast enough, or it will lose contact with the track.
- v= 1.5 m/s. radius = 0.25m
Will it fall off?

Show working etc.

Cheers

[vce08]

centripetal force formula

Fcentripetal = mv^2/r
If this force isn't bigger than the mg force, then it will fall.

[Mao]

:. it will fall off.

[vce08]

:. it will fall off.

Mao are you teaching these kids that g = 10 and not 9.8!!?!??!!?

[shinny]

Isn't that how it is in the wonderful world of VCAA's VCE physics-lite edition?

[methodsboy]

:. it will fall off.

can this formula also be used: (mg+N) = mv^2/r ?          where N is the frictional force.

[MattPritchard]

Mao are you teaching these kids that g = 10 and not 9.8!!?!??!!?

In the new study design for VCE Physics, Acceleration due to Gravity is assumed to be at 10ms-1 down instead of 9.8ms-1 down unless otherwise stated that "For the following questions assume gravity is 9.8"
In the exam we must use 10 unless as previously mentioned it is stated otherwise.

[methodsboy]

:. it will fall off.

what if : v^2/r = g?
would that mean it would also fall off?

[methodsboy]

centripetal force formula

Fcentripetal = mv^2/r
If this force isn't bigger than the mg force, then it will fall.

what if mv^2/r= mg?
does that also mean that it would fall off?

[Mao]

I don't think so [not 100% sure, but fairly sure], as gravity supplies all of the centripetal force, that will just mean the passengers would feel momentarily weightless as the normal reaction force would be zero.

[methodsboy]

I don't think so [not 100% sure, but fairly sure], as gravity supplies all of the centripetal force, that will just mean the passengers would feel momentarily weightless as the normal reaction force would be zero.

Okays