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Equations of Motion: The mass moment of inertia of the wheel about the z axis is. Referring to the free-body diagram of

the wheel shown in Fig. a, we have

Ans.

a Ans.+©MG = IGa; -150(0.25) = -1.6875a a = 22.22 rad>s2

+ T©Fx = m(aG)x ; 150 = 75aG aG = 2m>s2

(IG)z = mkz 2 = 75 A0.152 B = 1.6875 kg # m2

*17–96. The 75-kg wheel has a radius of gyration about thez axis of . If the belt of negligible mass issubjected to a force of , determine the accelerationof the mass center and the angular acceleration of the wheel.The surface is smooth and the wheel is free to slide.

P = 150 Nkz = 150 mm

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P ! 150 N

G

z

yx

250 mm

a

Assume the wheel does not slip.

Solving:

Ans.

OKFmax = 0.2(29.34) = 5.87 lb 7 1.17 lb

a = 4.35 rad>s2

aG = 5.44 ft>s2

N = 29.34 lb

F = 1.17 lb

aG = (1.25)a

+©MG = IG a; F(1.25) = c a 3032.2b (0.6)2 da

+a©Fy = m(aG)y ; N - 30 cos 12° = 0

+b©Fx = m(aG)x ; 30 sin 12° - F = a 3032.2baG

•17–97. The wheel has a weight of 30 lb and a radius ofgyration of If the coefficients of static andkinetic friction between the wheel and the plane are

and determine the wheel’s angularacceleration as it rolls down the incline. Set u = 12°.

mk = 0.15,ms = 0.2

kG = 0.6 ft.

1.25 ft

G

u

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