I need an answer please At t = 2.80 s a point on the rim of a 0.185-m-radius whe

Question

I need an answer please

At t = 2.80 s a point on the rim of a 0.185-m-radius wheel has a tangential speed of 45.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 14.5 m/s2. (a) Calculate the wheel's constant angular acceleration. rad/s2 (b) Calculate the angular velocities at t = 2.80 s and t = 0. 2.80 s = rad/s 0 = rad/s (c) Through what angle did the wheel turn between t = 0 and t = 2.80 s? rad (d) At what time will the radial acceleration equal g? s after t = 2.80 s

Explanation / Answer

Part A)

alpha = a/r

alpha = 14.5/.185

alpha = -78.4 rad/s2 (Negative is sign convention since it is slowing down - ignore it if needed)

Part B)

At 2.8...

w = v/r

w = 45/.185 = 243.2 rad/s

At t = 0

wf = wo + (alpha)t

243.2 = wo + (-78.4)(2.8)

wo = 681 rad/s

Part C)

wf2 = wo2 + 2(alpha)theta

243.22 = 6812 + 2(-78.4)(theta)

theta = 2580 rad

Part D) You will need to re-type this in a comment. It makes no sense how you have it typed due to Cheggs reformatting issue. (It a good idea to always proof read your post after you submit it for clarification if needed)

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