The 20-kg wheel shown has a moment of inertia of I_G = 0.349 kg-m^2. Assuming th
ID: 1561397 • Letter: T
Question
The 20-kg wheel shown has a moment of inertia of I_G = 0.349 kg-m^2. Assuming that the wheel does not slip or rebound, determine the minimum velocity v_G it must have to just roll over the obstruction at A. Assume the obstruction is 0.07 m tall. Which of the following equations represents the conservation of energy of the wheel from after the impact to when it is on top of the obstruction? 1/2(20)(upsilon_G)_2^2 + 1/2 (0.349)omega_2^2 = (20) (0.07) 1/2 (20)(upsilon_G)_2^2 + (0.349)omega_2^2 = (20) (9.81)(0.07) 1/2 (20)((upsilon_G)_2^2 + 1/2(0.49)omega_2^2 = (20) (9.81)(0.07) 1/2 (20) ((upsilon_G)_2 + 1/2 (0.349)omega_2 = (20) (9.81) (0.07) (20) ((upsilon_G)_2 + (0.349)omega-2 = (20) (2.81)Explanation / Answer
conserve energy
0.5 mv^2 + 0.5 I w^2 = mgh {w = v/r }
0.5*20*v^2 + 0.5*0.349*(v/0.2)^2 = 20*9.81*0.07
v = 0.9778 m/s
5.)option D is correct
0.5*20*V^2 + 0.5*0.349*w^2 = 0.07*20*9.81
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