Answer 7_1_11

(a)  Find the angular velocity at the start:

w1 = 2 x p x (60 ÷ 60) = 2p = 6.28 rad/s

(b) Angular acceleration:  w2 = w1 + at 

0 = 6.28 rad/s + 40 s x a

a =  6.28 rad/s ÷ 40 s = -0.157 rad/s²

(c) The torque is constant, so after 20 s the rate of turning is now 30 rpm = p rad/s.

Now use the formula:

q =(w1 + w2)t/2

q  = (6.28 rad/s + 3.14 rad/s) x 20 s = 94.2 rad

                                     2

(d) Use T = Ia

T = 10 000 kg m^-2 x 0.157 rad/s² = 1570 N m

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