how to Deal with angular momentum of rigid bodies?
I'm taking dynamics of rigid bodies, I'm having some trouble with impulse and momentum. Basically ,I know that angular momentum abt. a point ,is the linear momentum multilied by the moment arm . BUT ,I'm not feeling at all comfortable appliying it For example , When is H abt. a pt. p = (moment of intertia abt. p )(w) and when is it m*V*d ,and when is it Ig*w +mVg *(d) ,i'm very confused about the whole thing and when to apply what so any clarification on the subject would be very helpfull , preferably I would like the most general case stated and explained,than when to cancel out terms and why to get to more specific cases ..... ANY clarification/comment can be helpfull Thanks
electron12012-06-06T13:53:11Z
Favorite Answer
Angular momentum = linear momentum ÷ length of moment arm.
Let’s start at the beginning. When an object is rotating, the molecules are moving in a circular pattern. When an object is moving in a straight line, the molecules are moving in a straight line. As the object rotates one complete circle, the molecules move a distance which is equal to the circumference of the circle. As the object rotates one complete circle, the molecules on the outside of the object move a greater distance that the molecules which are closer the center of the object.
Go to the website below to see a picture of concentric circles. https://legal.entrepreneur.com/wp-content/uploads/2010/09/concentric-circles.gif
The picture represents the distribution of the mass of solid cylindrical disc. The momentum of inertia describes the distribution of the mass with respect to the axis of rotation. For a solid cylindrical disc, I = ½ * mass * radius^2
Go to the website below to see what 1 radian actually means. http://lamar.colostate.edu/~hillger/faq-images/faq-radian.gif
As the disc rotates an angle of 1 radian, a particle on the outer edge of the circle moves a linear distance, which is equal to the radius of the circle. As the disc rotates one complete circle, a particle on the outer edge of the circle moves a linear distance, which is equal to the circumference of the circle. SO 360˚ = 2 * π radians and Circumference = 2 * π * (radius) meters 2 * π * r ÷ 2 * π = r Linear distance ÷ (number of radians) = radius d ÷ θ = r So d = θ * r and θ = d ÷ r
Example #1: The radius of the disc is 3 meters and the mass is 5 kg. When the disc has rotates an angle of 1 complete circle, how many meters, has a particle on the outer edge of the disc moved? Linear distance, d = 2 * π * 3 ≈ 18.85 meters θ = 2 * π = 6.283 d ÷ θ = 18.85 ÷ 6.283 = 3 = radius
Example #2: The same disc rotates 4 times in 0.6 seconds. What is the linear velocity of a particle on the outer edge of the disc? What is the angular velocity of the disc?
Linear velocity = (number of meters) ÷ time = 4 * 18.85 ÷ 0.6 = 125.7 m/s Angular velocity = (number of radians) ÷ time As the disc rotates 1 time, it rotates an angle of 2 * π radians. As the disc rotates 4 time, it rotates an angle of 8 * π radians. Angular velocity = 8 * π radians ÷ 0.6 seconds = 41.9 radians/second
Let’s compare the linear velocity to the angular velocity. Linear velocity ÷ angular velocity = 125.7 ÷ 41.9 = 3 The radius of the circle is 3 meters. Linear velocity ÷ angular velocity = radius v ÷ ω = r So v = ω * r and ω = v ÷ r
In each case above, the linear measurement is equal to the angular measurement times the radius. OR In each case above, the angular measurement is equal to the linear measurement divided by the radius
Send me a problem that deals with these concepts. I will show you how to solve it.
in case you concentration on that in an elliptical orbit, the gap between the orbiting body and the critical body is continuously replacing, that's the position gravitation will strengthen and reduces tangential or on the spot p.c.. at the same time as an orbiting body is shifting in route of the perihelion/perigee factor, it is fairly falling in route of the middle of mass or barycenter. this supplies your gravitational ingredient that will strengthen velocity. it is not a torque as that factor period is often defined, it is a stress or area of stress utilized at a distance from the middle of mass. at the same time as it is real that gravitational stress is continuously radial and directed in route of the middle of mass, this does no longer necessisarily correspond to the route of shuttle of the item being acted upon. in case you word a stress by the middle of mass to a shifting body, you introduce an acceleration that adjustments rapidly-line action right into a curve. there is your ellipse. Draw an elliptical orbit, and at countless factors attraction to those vectors for the orbiting body (loose body diagram): a) gravitational stress b) on the spot velocity c) radial and tangential acceleration that can make it clearer. And convinced, Kepler's regulation does word, yet do not forget that that's only a quantitative description of elliptical / planetary action. Kepler knew no longer something about gravitation, or the motives of the elliptical action we now comprehend as orbits.