ch10-p025

# ch10-p025 - Eq. 10-23) places an upper limit of the rate of...

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which leads to t = 20 s. The second half of the motion takes the same amount of time (the process is essentially the reverse of the first); the total time is therefore 40 s. (b) Considering the first half of the motion again, Eq. 10-11 leads to ω = o + α t ¡ = 40 rad/s 20 s = 2.0 rad/s 2 . 25. (a) The upper limit for centripetal acceleration (same as the radial acceleration – see
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Unformatted text preview: Eq. 10-23) places an upper limit of the rate of spin (the angular velocity ) by considering a point at the rim ( r = 0.25 m). Thus, max = a/r = 40 rad/s. Now we apply Eq. 10-15 to first half of the motion (where o = 0): o = 1 2 ( o + ) t 400 rad = 1 2 (0 + 40 rad/s) t...
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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