A fan blade rotates with angular velocity given by&nbsp;ωz(t)=γ−βt2, where&nbsp;γ=5.00&nbsp;rads&nbsp;and&nbsp;β=0.800&nbsp;rads3. (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration&nbsp;az&nbsp;at&nbsp;t=3.00&nbsp;s&nbsp;at t = 3.00 s and the average angular acceleration&nbsp;aav−z&nbsp;for the time interval t = 0 to t = 3.00 s. How do these two quantities compare? If they are different, why?

Question

A fan blade rotates with angular velocity given by&amp;nbsp;ωz(t)=γ−βt2, where&amp;nbsp;γ=5.00&amp;nbsp;rads&amp;nbsp;and&amp;nbsp;β=0.800&amp;nbsp;rads3. (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration&amp;nbsp;az&amp;nbsp;at&amp;nbsp;t=3.00&amp;nbsp;s&amp;nbsp;at t = 3.00 s and the average angular acceleration&amp;nbsp;aav−z&amp;nbsp;for the time interval t = 0 to t = 3.00 s. How do these two quantities compare? If they are different, why?

mauricio0815sh · Accepted Answer

a) From equation and the given angular velocity ωz(t) we can find the instantenious angular acceleration az in some time t:  az(t)=dωz(t)dt=−2β  az(t)=−2βt−(1.6⋅rads3)⋅  b)From the result of a) we have for the instantenious angular acceleration az at t=3 s:  az(t)=−(1.6⋅rads3)⋅3 s  az(t)=−4.8rads2  Frome equation we can find the average angular acceleration aav−z  aav−z=ωz(t1)−ωz(t0)t1−t0=γ−βt12−(γ−βt02)t1−t0=βt12−t02t1−t0=−0.8rads3(3 s)23 s  aav−z=−2.4⋅rads2  The average angular acceleration aav−z and the instantaneous acceleration az are different, and the instanteneous angular acceleration is twice the size of the average angular acceleration.  They are different because the acceleration is not constant, az(t) is changing linear with time.

Bukvald5z · Accepted Answer

Very good answer

A fan blade rotates with angular velocity given by \omega

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