Ice Hockey Sticks Length
Question about center of mass?
A rod of mass of 1 kg and length 3.4 m is standing on frictionless ice, pointing north-south. Glue covered hockey puck of mass 0.75 kilograms slide bar at a speed of 7 m / s from 20 degrees north of east. The album hit the tip south of the bar and you Paste. Find the center of mass of the combination of rod and disk? What equation should I use?
The center of mass is the sum of each mass times the radius divided by the total mass. For this problem, the center of mass is at the helm of a point D, the north end. This section of the bar ad/3.4 a mass of kg and radius d / 2, which stretches north from the center of mass, the rest of the system should balance it. There is a Another section of the rod with mass (3.4-d) / 3.4 and radius (3.4-d) / 2 then the disk has a mass 0.75 and radius (3.4-d) Construction of the equation to the center of mass (d/3.4) * d / 2 = ((3,4-d) / 3.4) * ((3,4-d) / 2) 0.75 * (3,4-d) J 'I neglected to 1.75 by dividing each term as it is unnecessary now to solve d. This is my Home algebra multiplying both sides by 2 * 3.4 ^ 2 = d (3.4-d) ^ 2 +2 * 3.4 * 0.75 * (3.4-d) simplify d ^ 2 = (3,4-d) ^ 2 5.1 * (3.4-d) multiply (3,4-d) ^ 2 and 5.1 * ( 3,4-d) d ^ 2 = 11.56-6.8 * d + d ^ 2 * d ^ d 17.34-5,1 subtract 2 from both sides 0 = 11.56-6.8 * d + 17.34-5.1 * D * D 11.9 add both sides 11.9 * D = 11.56 17.34 divide both sides of 11.9 d = 2.43 Algebra I made a mistake in my previous post. I will correct solution here, since d = 2.43, Disc length is 0.97 m, it is an inelastic collision between the bar and disk. Conservation of momentum applies the center of mass of the rod and disc combo translation has found momentum direction is: 0.75 * 7 * cos (20) = 1.75 solve for vx * VX = 2.82 m / s in the direction north, 0.75 * 7 * sin (20) = 1.75 * v and solve to vv = 1.04 m / s So now we have the momentum shift, to calculate the amount of rotation. The first step is to determine where along the bar is the center of mass of a uniform bar is 1/3.4 kg / m, and the disc is considered a point mass located at 1.7 m from the center of the stem. The radius This disc is considered negligible. The center of mass is at the northern end, to see beyond d = 2.43 m distance from disc to the center of mass 3.4-2.43 0.97 The next step is to calculate the moment of inertia of the disk along the rod. I'll try the system into three parts: a rod length of 2.43 with a mass of 0.71 kg of food – in the center of mass I1 = 0.71 * 2.43 ^ 2 A rod with a length of 0.97 and 0.29 mass attached to the center of mass 2 = 0.97 * 0.29 ^ 2 A Mass point to 0.97 range and I3 = 0.75 mass 0.75 * 0.97 ^ 2 inertia of the combined system is I1 + I2 + I3 I = 5.19 The moment of inertia of the disc at the moment before the collision is 0.71. The angular velocity is to gain speed to travel is divided by 0.97 = 7 * cos (20) / 0.97 rad / s = 6.78 rad / s using the angular momentum conservation 0.71 * 6.78 = 5.19 * for solving ω ω ω = 0.92 rad / S 1) We have solved the first part of the point ω = 0.92 rad / s 2) 0.1 seconds after the collision to what extent has shifted to the end South (x, y). To say the crash of the tip of the South at time t = 0, x = 0, y = 0 The center of mass translational velocity of v = 2.82 m / sm = 1.04 v / s center of mass movements dy dx = 0282 m = 0104 m 0.092radians The angular displacement is 5.28 degrees or additional displacement of south because of the rotation is dx = sin (5.28) 0.97 * dy dx = 0.09 m '= 0.97 * (1-cos (5.28)) dy' = 0.0041m Thus, the total displacement is 0.09 x 0.37 m in 0282 and the total displacement is 0.0041 and 0.104 0.11 MJ
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