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To get a billiard ball to roll without sliding from the start, the cue must hit the ball not at the center (that is, a height above the table equal to the ball's radius, r) but at exactly at a height of 2r/5 above the centre. Prove the result.
For a ball to roll without slipping, it must have tangential velocity twice that of its velocity of the centre of mass.
so tangengial speed = ωr
at the point where the wheel contact the surface must be at rest
vcm - ωr = 0
vcm = ωr
L=Iω = Ip/mr
p = Lmr/I
so for conversation of momentum
mvcmi = mvcmf + Lmr/I
Δvcm = Lr/I
I for a sphere = (2/5)mr^2
where r is where the ball needs to hit
Δvcm = L/(2/5)mr
(2/5)mΔvcm/L = r
(2/5)Δp/L = r
(2/5)r = r
these last few steps don't seem right?
are they? am i even correct up to there?
thanks
For a ball to roll without slipping, it must have tangential velocity twice that of its velocity of the centre of mass.
so tangengial speed = ωr
at the point where the wheel contact the surface must be at rest
vcm - ωr = 0
vcm = ωr
L=Iω = Ip/mr
p = Lmr/I
so for conversation of momentum
mvcmi = mvcmf + Lmr/I
Δvcm = Lr/I
I for a sphere = (2/5)mr^2
where r is where the ball needs to hit
Δvcm = L/(2/5)mr
(2/5)mΔvcm/L = r
(2/5)Δp/L = r
(2/5)r = r
these last few steps don't seem right?
are they? am i even correct up to there?
thanks