Explanation Apply vector L vectorr vectorp mvectorr vectorv vector L

Explanation apply vector l vectorr vectorp mvectorr

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Explanation: Apply vector L vectorr × vectorp = mvectorr × vectorv vector L = bracketleftBig ( y y 0 + ( x x 0 ı bracketrightBig × m v ˆ ı = m v ( y y 0 ) ˆ k Then the magnitude of the angular momen- tum is bardbl vector L bardbl = | m v ( y y 0 ) | = | (2 kg)(7 m / s)(3 m 13 m) | = 140 kg m 2 / s 006 10.0 points A thin uniform cylindrical turntable of ra- dius 2 . 9 m and mass 28 kg rotates in a hori- zontal plane with an initial angular speed of 7 . 6 rad / s. The turntable bearing is friction- less. A clump of clay of mass 11 kg is dropped onto the turntable and sticks at a point 1 . 7 m from the point of rotation. Find the angular speed of the clay and turntable. Correct answer: 5 . 98424 rad / s. Explanation: Let : R = 2 . 9 m , x = 1 . 7 m , M = 28 kg , ω = 7 . 6 rad / s , and m = 11 kg . Basic Concepts: summationdisplay vector L = dvector τ ext dt Δ K = K f K i = Q . From conservation of the angular momentum it follows that L i = L f or I i ω = I f ω f , where I i = 1 2 M R 2 = 1 2 (28 kg) (2 . 9 m) 2 = 117 . 74 kg m 2 . Similarly, I f = I i + m x 2 = (117 . 74 kg m 2 ) + (11 kg) (1 . 7 m) 2 = 149 . 53 kg m 2 . Finally, ω f = I i I f ω = (117 . 74 kg m 2 ) (149 . 53 kg m 2 ) (7 . 6 rad / s) = 5 . 98424 rad / s . 007 10.0 points A projectile of mass m = 1 . 27 kg moves to the right with speed v 0 = 12 . 5 m / s. The pro- jectile strikes and sticks to the end of a sta- tionary rod of mass M = 8 . 65 kg and length d = 1 . 9 m that is pivoted about a frictionless axle through its center. I cm rod = 1 12 M L 2 d O O v 0 ϖ m M Find the angular speed of the system right after the collision. Correct answer: 4 . 02 rad / s. Explanation:
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kuruvila (lk5992) – HW 11 – opyrchal – (11113) 4 The initial angular momentum of the pro- jectile about the pivot O is L i = d 2 m v 0 . With the projectile stuck to the end of the rod, the rotational inertia of the projectile and the rod combined about O is, I f = m parenleftbigg d 2 parenrightbigg 2 + 1 12 M d 2 = bracketleftbigg m 4 + M 12 bracketrightbigg d 2 (1) = bracketleftbigg (1 . 27 kg) 4 + (8 . 65 kg) 12 bracketrightbigg (1 . 9 m) 2 = 3 . 75 kg m 2 . Using conservation of angular momentum, we have L i = I f ω (2) = r p = d 2 m v 0 (3) Therefore, combining Eqs. (1), (2), and (3), we have ω = d 2 m v 0 m d 2 4 + M d 2 12 · 12 12 = 6 m v 0 (3 m + M ) d = 6 (1 . 27 kg) (12 . 5 m / s) [3 (1 . 27 kg) + (8 . 65 kg)] (1 . 9 m) = 4 . 02 rad / s .
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