Question

M Review The reel has a weight of 168 lb and a radius of gyration about its center of gravity of kc = 1.25 ft. (Figure 1) Part A If it is subjected to a torque of M = 25 lb. ft. and starts from rest when the torque is applied, determine its angular velocity in 3 seconds. The coefficient of kinetic friction between the reel and the horizontal plane is = 0.15 Express your answer to three significant figures and include the appropriate units. Figure < 1 of 1 > - LÀ 6 + ? rad w= 23 115ft M = 25 lb. ft JE Submit Previous Answers Request Answer X Incorrect; Try Again; 9 attempts remaining

Answer 1

Given that,

Coefficient of kinetic friction _{, $\mu$ k} = 0.15

Center of gravity, k_{G =}1.25  

M = 25 lb

r₁ = 1.5 ft

r₂ = 1 ft

t = 3 sec

F = 168

Auswers -1-5fd Using principle of angular impulse and momentum about G, IGW+& Gdt IG Wa o+MH-Confcri-r))t = (mxk6) W₂ + (mxrJwz 25*3-(0.15 x168 *(1-6-13) 2 . 162XJ2 32-9 75-(37.8) = (8.1521) W2 + (3.2173)W, 37.2 = (1 3. 3694) W2 W₂ = 37.2 13.3694 Wz = 2.78 road sec