Angular momentum. To finish off our comparison of translational (straight-line) and rotational motion, let's consider the rotational equivalent of momentum, which is angular momentum. For straight-line motion, momentum is given by p = mv. Momentum is a vector, pointing in the same direction as the velocity. Plane Motion of Rigid Bodies: Energy and Momentum Methods
A small ball of mass m is projected horizontally along the tangent with a speed v from a point C. The centre O, lowest point A and point C lie in same vertical plane. Column I Column II (P) Angular momentum about O (1) mvR (Q) Angular momentum about C (2) 2mvR cos 2 ⎛⎞θ ⎜⎟ ⎝⎠ θ A O C (R) Angular momentum about A (3) mvR sinθ

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If we take Conservation of Angular Momentum into account for a baseball or softball swing, you realize that: The further away the bathead and/or the hands and arms are from the body during the swing, the slower the rotation of the body, the slower the turn, resulting in much slower bat speed. For instance, for a golf ball you’re whirling around on a string, the moment of inertia depends on the radius of the circle the ball is spinning in: I = mr 2. Here, r is the radius of the circle, from the center of rotation to the point at which all the mass of the golf ball is concentrated.

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When a collision is elastic and no external torque acts on a system, angular momentum is conserved. I found this example and checked the results: A ball (m = 1 Kg , v = p =+22 m/s, Lm = +11, Ke = 242 J) hits the tip of a rod (M = 10Kg , length = 1m, I = 10 ∗ 1 2 / 12 = 5/6 ) in an elastic collision. If the rod is pivoted, the ball bounces back with v, p = -11.846 m/s , L = -5.923, (Ke = 70.16) and the rod rotates with ω = 20.3 , L is conserved : Lr = (20.3 *5 /6) = 16.923 and Ke = 70.166 ...

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The angular momentum of a rigid object rotating about an axis is the sum of the angular momenta of all its parts. It is a measure of an object's rotational motion about this axis. The angular momentum L about one of the object's symmetry axis is the product of the object's moment of inertia I times its angular velocity ω about this symmetry axis.

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We can determine the average acceleration experienced by the ball from Newton's second law: F=m a. Using the numbers above, we obtain an average acceleration of 127,145m/s 2 , or 12,740 times the acceleration due to gravity! The angular momentum also decreases to zero during the swing. We can't simply say it was given to the apparatus, for the pivot is sufficiently frictionless that it provides insufficient torque. The angular momentum is decreased by the torque of gravity on the pendulum, and the earth gains the angular momentum the pendulum loses.

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It has angular velocity $ω$ right before hitting a ball (mass $M_2$) at its lowest position. As in this picture: The rod comes to full rest after the collision. To solve for the speed of the ball, you can use angular momentum conservation: $$I\omega = mvD$$ However, by dividing through by $D$, you have units of linear momentum. This muscle in a professional boxer exerts a force of. 2.00 × 10 3 N 2.00 × 10 3 N. size 12 {2 "." "00" times "10" rSup { size 8 {3} } `N} {} with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of. 120 rad/s 2 120 rad/s 2.

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