The uniform slender bar ABC weighs 4.3 lb and is initially at rest with end A bearing against the stop in the horizontal guide. When a constant couple M = 62.1 lb-in. is applied to end C, the bar rotates causing end A to strike the side of the vertical guide with a velocity of 11.6 ft/sec. Calculate the loss of energy ΔE (enter a positive number) due to friction in the guides and rollers. The mass of the rollers may be neglected.

Respuesta :

Answer:

the energy loss due to friction = 0.2344 ft-lb

Explanation:

The principle of conservation of energy is used to calculate the loss of energy  due to friction.

The difference between the gain in potential (Mθ) due to the restoring moment M = 62.1 lb-in and loss in potential energy due to weight (mgh) and kinetic energy as a result of restoration [tex](\frac{1}{2}*I_{AC}* \omega ^2)[/tex] of AC adds up to the energy loss ( ΔE).

Expressing the formula; we have:

ΔE = [tex]M \theta - mgh - \frac{1}{2}*I_{AC}* \omega^2[/tex]

ΔE = [tex]M \theta - mgh - \frac{1}{2}*\frac{mb^2}{12}* (\frac{v}{r})^2[/tex]

From the diagram below; the weight of the slender = W

acceleration due to gravity = g

length of the slender bar = r

the couple  = M

the vertical guide velocity = v

and the angle = θ

ΔE =  [tex]\frac{62.1}{12}*\frac{\pi}{4}-4.3*\frac{8}{12} *(1-\frac{1}{\sqrt{2}})- [\frac{1}{2}*(\frac{1}{12})*\frac{4.3}{32.2}*(\frac{16}{12})^2*(\frac{11.6}{\frac{8}{12}})^2][/tex]

ΔE = 4.064 - 0.8396 - 2.99

ΔE = 0.2344 ft-lb

Thus, the energy loss due to friction = 0.2344 ft-lb

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