Answer:
Explanation:
20° from the normal = 110° from parallel
1a) τ = (200sin90)[6] + (75sin110)[3] - (100sin110)[3]
τ = 1,129.52305... = 1100 N•m CCW
1b) τ = 200(6)(sin90) + 75(-3)(sin(360-110)) + 100(3)(sin(270 + 20)
τ = 1,129.52305... = 1100 N•m CCW
1c) My directions agree, both are positive z by right hand rule.
1d) Moment of inertia for a thin rod about an axis perpendicular to its center is
I = (1/12)mL²
τ = Iα
α = τ/I = 1129.523 / ((1/12)(200)(12²)) = 0.4706 rad/s²
θ = ½αt²
θ = ½(0.4706)(2•60)² = 3,388.56916... radians
θ = 3400 radians
at which time is is spinning about 9 revolutions per second
