Hitting a billiard ball h above centerline

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SUMMARY

The discussion focuses on the physics of a billiard ball struck by a cue held at a height h above the centerline. The ball, initially at rest, is propelled with an initial speed v0 and ultimately reaches a final speed of 9vo/7 due to the applied impulse. The key conclusion derived from the problem is that the height h can be expressed as h = 4R/5, where R represents the radius of the billiard ball. The equations used include torque (FRsin(θ) = Iα) and impulse (a = v0/Δt), which are critical for solving the problem.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with impulse and momentum equations
  • Knowledge of angular velocity and its relation to linear velocity
  • Basic principles of billiard ball physics
NEXT STEPS
  • Study the relationship between torque and angular acceleration in rotational motion
  • Learn about the conservation of angular momentum in billiard ball collisions
  • Explore the effects of impulse on rotational motion in rigid bodies
  • Investigate the mathematical derivation of motion equations for spherical objects
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This discussion is beneficial for physics students, educators, and anyone interested in the mechanics of billiard games, particularly those studying rotational dynamics and impulse effects in sports physics.

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Homework Statement



A billiard ball, initially at rest, is given a sharp impulse by a cue. the cue is held horizontally a distance h above the center line. the ball leaves the cue with a speed vo, and because of its "forward English", eventually acquires a final speed of 9vo/7. Show that h=4R/5, where R is the radius of the ball.

Homework Equations


The Attempt at a Solution


I've tried using FRsin(θ)=Iα and coming up with h = (2/5)R^2(α/a). Then using the impulse equation a = v0/Δt and α= ω0/t i got h=(2/5)R^2(ω0/v0). The thing is I keep getting wrong and contradictory answers about ω0.
 
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