Complete Question:
The 30-in. slender bar weighs 20 lb and is mounted on a vertical shaft at O. If a torque, M = 100 lb-in. is applied to the bar through its shaft, calculate the horizontal force R on the bearing as the bar starts to rotate.
The attached diagram completes the question
Answer:
α = 23.02 rad/s²
Explanation:
From the Free Body Diagram attached to this solution:
[tex]I_{0} = I_{c} + md^{2}[/tex]...............(1)
[tex]I_{c} = \frac{ml^{2} }{12} \\W = mg, m = W/g[/tex]
Substitute Ic and m into equation (1)
[tex]I_{0} =\frac{ml^{2} }{12} + md^{2}\\I_{0} =\frac{Wl^{2} }{12g} + \frac{Wd^{2}}{g}[/tex]
W = 20 lb
l = 30 in = 30/12 ft = 2.5 ft
g = 32.2 ft/s²
d = 3 in = 3/12 ft = 0.25 ft
[tex]I_{0} =\frac{20 * 2.5^{2} }{12*32.2} + \frac{20*0.25^{2}}{32.2}\\I_{0} = 0.362 lb-ft-s^2[/tex]
Taking moment about O
[tex]\sum M_{0} = I_{0} \alpha[/tex]
M = 100 lb-in = 100/12 lb - ft = 8.33 lb - ft
8.33 = 0.362 α
α = 8.33/0.362
α = 23.02 rad/s²
