11 of 24 Sapling Learning A motorized wheel is spinning with a rotational veloci

Question

11 of 24 Sapling Learning A motorized wheel is spinning with a rotational velocity of ab -1.33 radis. 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= 5.21 s At what time, ma, is the rotational acceleration a maximum? ms what is the maximum tangential acceleration, a, of a point on the wheel a distance R = 1.19 m from its center? What is the magnitude of the total acceleration of this point? ? Hint

Explanation / Answer

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given angular velocity W (t) = Wo tan^-1((c + bt^2)

W(t) = Wo tan-1(1.5574 + 5.21 t^2)

now take time derivative on both sides

dW/dt = (t) = 10.42 a Wo /((5.21 t^2 + 1.5574)^2 + 1)

taking derivative again

(t) = 0

d/dt (10.42 t Wo /((5.21 t^2 + 1.5574)^2 + 1)) = 0

tmax = 0.358 sec

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maximum tangential acceleration is given by

a* tmax = R * (t)

at = 1.19* (10.42 t *1.33 /((5.21t^2 + 1.5574)^2 + 1))

at t = 0.358

at = 1.19 *(10.42* (0.358) (1.33) /((5.21 (0.384^)2 + 1.5574)^2 + 1))

at   = 0.921 m/s^2

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radial acleration a = ar = R W^2

here

W = wo *tan^-1(C + btmax^2)

Wmax = 1.33tan^-1( 1.5574 + 5.21*0.384^2)

Wmax = 1.54 rad/s

ar = 1.19 * 1.54^2

ar = 2.822 m/s^2

the magnitude of total acleration a^2 = ar^2 + at^2

a^2 = 0.921^2 + 2.822^2

atotal = 2.96 m/s^2

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