At t = 2.95 s a point on the rimof a 0.195-m-radius wheel has a tangentialspeed

Question

At t = 2.95 s a point on the rimof a 0.195-m-radius wheel has a tangentialspeed of 48.5 m/s as the wheel slows downwith a tangential acceleration of constant magnitude 13.0 m/s2. (a) Calculate the wheel's constant angularacceleration.
1 rad/s2

(b) Calculate the angular velocities at t = 2.95 s andt = 0. 2.95 s = 2 rad/s 0 = 3 rad/s
(c) Through what angle did the wheel turn between t = 0and t = 2.95 s?
4 rad

(d) At what time will the radial acceleration equal g?
5 s after t = 2.95 s
(a) Calculate the wheel's constant angularacceleration.
1 rad/s2

(b) Calculate the angular velocities at t = 2.95 s andt = 0. 2.95 s = 2 rad/s 0 = 3 rad/s
(c) Through what angle did the wheel turn between t = 0and t = 2.95 s?
4 rad

(d) At what time will the radial acceleration equal g?
5 s after t = 2.95 s 2.95 s = 2 rad/s 0 = 3 rad/s

Explanation / Answer

(a)   angular acc = tang acc / r =13.0 / 0.195 =    66.67rad/s2 . (note:   it might want   -66.67   because it says it is slowing down) . . (b)    at t=2.95      = v/r = 48.5 / 0.195 =   248.72 rad/s .        at t =0        = 248.72 +   2.95 * 66.67 =    445.38 rad/s . (c) average speed = (248.72 + 445.38) /2 = 347.05 .      angular dist = avg speed* time = 347.05 * 2.95 =    1023.8radians . (d) radial acc = 2r .          9.80 = 2 * 0.195 .     = 7.0892 rad/s .     time = change in speed /acc = (445.38 - 7.0892) / 66.67 =   6.574 seconds

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