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Unformatted text preview: Example : The drawing shows a device that can be used to measure the speed of a bullet. The device consists of two rotating discs separated by a distance d = 0.850 m, and rotating with an angular speed of 95.0 rad/s. The bullet first passes thru the left disk and then thru the right disk. It is found that the angular displacement between the two bullet holes is θ = 0.240 rad. What is the speed of the bullet? Projection of bullet θ hole thru first disk. The speed of the bullet is given by: t d v = Thus, we need to find the time. We are given ω and Δθ . From this we can find the time: t Δ Δ = θ ω 50 5 ω θ Δ = t Thus, θ ω Δ = d v m/s 336 240 . ) 850 )(. 95 ( = = Example : A child hunting for his favorite wooden horsey is running on the ground around the edge of a stationary merrygoround. The child has a constant angular speed of 0.250 rad/s. His horsey is sitting on the outer edge of the merrygoround. At the instant the child spots the horsey, onequarter of a turn away, the merrygoround begins to move (in the same direction the child is running) with a constant angular acceleration of 0.010 rad/s 2 . What is the shortest time it takes for the child to catch up ith the horse? with the horse? When the child spots the horse ¼ of a turn away, the MGR starts to accelerate. ¼ turn is = π /2 rad. he condition we want is for the child to reach the horse α = 0.01 rad/s 2 The condition we want is for the child to reach the horse....
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 Fall '08
 SPRUNGER
 Physics, Rotation, tangential acceleration

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