At a time t = 2.70 s , a point on the rim of a wheel with a radius of 0.220 m ha

Question

At a time t = 2.70s , a point on the rim of a wheel with a radius of 0.220m  has a tangential speed of 50.0m/s  as the wheel slows down with a tangential acceleration of constant magnitude 10.7m/s2

a)Calculate the wheel's constant angular acceleration.

b)Calculate the angular velocity at t = 2.70s

c)
Calculate the angular velocity at t=0.

d)
Through what angle did the wheel turn between t=0 and t = 2.70s ?

e)
Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81m/s2 ?

Explanation / Answer

vo = 50 m/s

a_tan = -10.7 m/s^2

a) alfa = a_tan/r = -10.7/0.22 = -48.63 rad/s^2

b)

at t = 2.7s

v = vo + a_tan*t = 50 - 10.7*2.7 = 21.11 m/s

w = v/r = 21.11/0.22 = 95.95 rad/s

c) at t = 0

wo = vo/r = 50/0.22 = 227.27 rad/s

d)

w^2 - wo^2 = 2*alfa*theta

theta = w^2-wo^2/(2*alfa)

    = (95.95^2-227.27^2) / (2*(-48.63))

    = 51558.23 radians

    = 8209 revolutions

e)


a_rad = v^2/r

9.8 = v^2/0.22

==> v = sqrt(9.8*0.22) = 1.468 m/s


v = vo + a_tan*t

t = (v-vo)/a_tan

= (1.468-50)/(-10.7)

= 4.54 s

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