Question

A rigid body, with its center of mass located at the position $LaTeX: {\bf r}_{\scriptscriptstyle G} = -2.76\hat{\bf I} +0.96\hat{\bf J}$ r G = − 2.76 I ^ + 0.96 J ^ (in), and rotating about the origin A, has a mass of m = 8.24lbm and a moment of inertia about its center of mass of I G = 11.3290 lbf in s2. At the instant shown, it has an angular velocity of $LaTeX: \dot\theta = -0.21$ θ ˙ = − 0.21 rad/s. In addition to the gravitational force acting in the $LaTeX: -\hat{\bf J}$ − J ^ direction, an external force, $LaTeX: {\bf F} = 3.13\hat{\bf I} -3.37\hat{\bf J}$ F = 3.13 I ^ − 3.37 J ^ (lb), is applied to the body at the position $LaTeX: {\bf r}_i = -2.88\hat{\bf I} -9.96\hat{\bf J}$ r i = − 2.88 I ^ − 9.96 J ^ (in).

What is the acceleration of the center of mass in the x direction (in units of in/s/s)?

Answer 1

M = 8.24lbm la- o mg = -8.24 x 82.175 - Apply De Alberts frnciples, - 3.37) F = 3.18 ETA = I4 = -2.8875 2.983 fa = 11-329 lbf in i i que Pepit mllam) € 11.324 + 8.24 2.76 +0.46 L = 81.69 Hof ins ETA = mg x fg + F XF a position of Where is a force Where acting on body 3.13 - 3.37 -8.24 -2.88 -9.96 0 1-2.76 o 0.96 - 22.74 j + (-39.95 E)

az -22.747 - 39.95€ 81-69 ē = -0.278 j -0.49R ā= ăxto I -0.278 096 no.49 0 1-2-76 á = 0.4704 T +(1.3524) + K P-0-7673) - Acceleration of Centre of mass in a-disection Aa=0.470 A m/ see