Question

The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, determine the velocity of the collar at the instant the rod becomes horizontal. The system is released from rest when != 45°.

Answer 1

Answer:

The velocity of the collar will be 3.076 ft/s

Explanation:

Given data

weight of the disk, Wa = 20lb

weight of rod BC, Wbc = 4lb

weight of collar, Wc = 1lb

Considering the equation of equilibrium

Vb = 1.5Wbc

Wa = 1.875 Wbc

to calculate the velocity of the collar using energy conservation equation

T1 + V1 = T2 + V2

$0+4(1.5 \sin 45)+2(3 \sin 45)=\frac{1}{2}\left(\frac{1}{2}\left(\frac{20}{32.2}\right)(0.8)\right)$              

=>$(1.875 W b c)+\frac{1}{2}\left(\frac{20}{32.2}\right)(1.5 W b c)+\frac{1}{2}\left(\frac{20}{32.2}\right)$

=>$(1.5 W b c) \frac{1}{2}\left\{\frac{1}{12}\left(\frac{4}{32.2}\right)(3)\right\}+\frac{1}{2}\left(\frac{1}{32.2}\right)$

=>$(2.598 W b c)+4(1.5 \sin 0)+2(3 \sin 0)$

Wbc = 1.18rad/sec

i.e.                                    

$V _c=2.598 \times 1.18$

= 3.076 ft/ s