A particle of mass m in the figure below slides down a frictionless surface through height h and collides with a uniform vertical rod (of mass M and length d), sticking to it. The rod pivots about point O through an angle θ when it momentarily stops. Find θ in terms of m, M, g, h, d and various constants. (Please type answer no
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A particle of mass m in the figure below slides down a frictionless surface through height h and collides with a uniform vertical rod (of mass M and length d), sticking to it. The rod pivots about point O through an angle θ when it momentarily stops. Find θ in terms of m, M, g, h, d and various constants.
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- A uniform plate of height 0.740 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: y= 1.100x^2 . Find the distance from the rounded tip of the plate to the center of mass. Needs Complete typed solution with 100 % accuracy.Chapter 11, Problem 066 GO Your answer is partially correct. Try again. In the figure, a small 0.109 kg block slides down a frictionless surface through height h = 1.46 m and then sticks to a uniform vertical rod of mass M = 0.218 kg and length d = 1.55 m. The rod pivots about point O through angle e before momentarily stopping. Find 0. Number Units (degrees) em the tolerance is +/-2% em Click if you would like to Show Work for this question: Open Show WorkA projectile of mass m moves to the right with a speed vi (see figure below). The projectile strikes and sticks to the end of a stationary rod of mass M, length d, pivoted about a frictionless axle perpendicular to the page through O. We wish to find the fractional change of kinetic energy in the system due to the collision. The moment of inertia of a rod is I = Md² and the moment of inertia of a particle is I = mr². 12 a. What is w, the angular speed of the system after the collision. mdvi Md²+1md² 12 b. What is the kinetic energy before the collision? Answer: KE¡ = m(v¡)². c. What is the kinetic energy after the collision? (0₁)² Answer: W= 1 m²d² Answer: KE₁ = = =+M₁2² +² md² 12 4 m O d M O 3