- #1
pff
- 17
- 0
hi, i was given
[itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] = 2[itex]\tau[/itex]Δ[itex]\theta[/itex]
for working out kinetic energy transferred to a ball in a tennis ball launcher type scenario, where two wheels are counterrotating and the ball goes between them.
Is this right? is the speed of the wheels not taken into account?
If the drive motor was geared down to give the wheels outrageous torque at an incredibly low rpm then surely the ball would just fall out of the end with no kinetic energy at all.
[itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] = 2[itex]\tau[/itex]Δ[itex]\theta[/itex]
for working out kinetic energy transferred to a ball in a tennis ball launcher type scenario, where two wheels are counterrotating and the ball goes between them.
Is this right? is the speed of the wheels not taken into account?
If the drive motor was geared down to give the wheels outrageous torque at an incredibly low rpm then surely the ball would just fall out of the end with no kinetic energy at all.