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What have I done wrong? (torque and angular momentum)
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[QUOTE="PhyIsOhSoHard, post: 4684082, member: 408052"] [b][SOLVED] What have I done wrong? (torque and angular momentum)[/b] [h2]Homework Statement [/h2] [ATTACH=full]167983[/ATTACH] A billiardball is hit from rest by the cue at height "h" from the table with a force F in the time interval Δt. The mass M and radius R of the ball is known as well as the moment of inertia which is [itex]I=\frac{2}{5}MR^2[/itex] Find an expression for the height "h" at which the billiardball will roll without slipping when it is hit. [h2]Homework Equations[/h2] Condition for roll with no slipping: [itex]v_{CM}=R\omega[/itex] [h2]The Attempt at a Solution[/h2] I start by finding an expression for [itex]v_{cm}[/itex]. Center of mass differentiated by Δt gives: [itex]v_{CM}=\frac{Mv}{M}=v[/itex] Newton's 2nd law: [itex]F=M\frac{v}{Δt}[/itex] Isolating velocity gives: [itex]v=MFΔt[/itex] Since the velocity is equal to the velocity of the center of mass: [itex]v_{CM}=MFΔt[/itex] Now I find an expression for the angular velocity. The net torque is given by: [itex]∑τ=Iα[/itex] The only force is the force F from the cue which gives the torque [itex]τ=F(h-R)[/itex] where (h-R) is the perpendicular length from the force F to the center of mass of the ball. [itex]F(h-R)=I\frac{\omega}{Δt}[/itex] The angular velocity is: [itex]\omega=\frac{F(h-R)Δt}{I}[/itex] Now I insert the velocity of CM and the angular velocity into the rolling without slip equation: [itex]MFΔt=R\frac{F(h-R)Δt}{I}[/itex] And I end up with: [itex]h=R(2/5M^2+1)[/itex] But my expression for the height has the mass squared in it. What did I do wrong? [/QUOTE]
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What have I done wrong? (torque and angular momentum)
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