A motorized wheel is spinning with a rotational velocity of ub = 1.33 rad/s. At

Question

A motorized wheel is spinning with a rotational velocity of ub = 1.33 rad/s. At t = 0 s, the operator switches the wheel to a higher speed setting. The rotational velocity of the wheel at all subsequent times is given by where c = 1.5574 and b = 5.43 s-2 At what time, tamax, is the rotational acceleration a maximum? Number ms amax What is the maximum tangential acceleration, at, of a point on the wheel a distance R = 1.17 m from its center? Number dl m/s What is the magnitude of the total acceleration of this point? Number a= m/ s

Explanation / Answer

1) a=dw/dt

=w0*2bt/(c+bt2)2da/dt=0

c=3bt2tmax

tmax=sqroot(c/3b)

=sqroot(1.5574/35.43)

=0.043957s

2)Tangential acceleration=(radius of rotation)(rotational acceleration)

for this we need to calculate rotational acceleation at t max i.e at t=0.0439s

at=(1.17)(0.634 sec2(1.5678))

3)totatl acceleration =ar+at here,ar=0.623sec2(1.5678)       

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