Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body This system m1 m3 m4 m5 CONCEPT CENTER OF MASS Sometimes it s useful to simplify a group system of objects by replacing ALL objects with a single equivalent object m2 can be simplified replaced with a single object M When combining objects MASS of the combined object is the of the masses of all objects M CENTER OF MASS C O M of the combined object is the of all objects in the system 10 Two Obj s unequal mass Single object Two Obj s equal mass EXAMPLE Two objects m1 10kg m2 10kg are placed on the x axis m1 is placed at x 0 and m2 is placed at x 4 Calculate the center of mass for the system of objects EXAMPLE Two objects m1 10kg m2 10kg are placed on the x axis m1 is placed at x 0 and m2 is placed at x 4 Calculate the center of mass for the system of objects The C O M is NOT necessarily in the middle of objs When given different masses it is often closer to objects 10 30 10 x x Page 1 Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body PROBLEM Three weights are placed along the y axis a 1 0 kg at 2 00 m a 1 50 kg at the origin and a 7 5 kg at 1 5 m Where is the center of mass of these weights CENTER OF MASS Page 2 Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body PROBLEM Calculate the x y coordinates of the center of mass in cm for the system of objects shown below CENTER OF MASS y B 0 4 kg 0 6 kg C 8cm 8cm A 0 25 kg 0 4 kg D x Page 3 Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body However moment of inertia isn t a fixed quantity like mass it changes depending on the rotation Typically moments are given on exams or in homework but only what I d call typical moments of inertia How do we solve a problem with a non typical moment of inertia CONCEPT PARALLEL AXIS THEOREM The moment of inertia is a really important quantity to know because it acts like in rotational equations For instance we know how to find the moment of inertia of a disk rotating about its central axis What if this disk weren t rotating about its central axis but was rotating about its rim If we aren t given this new moment of inertia we can find it using the PARALLEL AXIS THEOREM The PARALLEL AXIS THEOREM will gives us the non typical moments of inertia where d is the distance between the center of mass axis and the new axis which must be PARALLEL EXAMPLE 1 A disk has a mass M and a radius R What is its moment of inertia about an axis perpendicular to the surface of the disk at the rim of the disk What about a parallel axis half way to the rim of the disk EXAMPLE 2 The moment of inertia of a thin rod of length L and mass M about an axis perpendicular to the rod at the edge of the rod is 1 3 ML2 What is the moment of inertia of the rod about a parallel axis halfway from the edge to the center of the rod Page 4 Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body MOMENT OF INERTIA Remember Motion problems do NOT depend on Mass but Energy eg K m v2 and Force F ma problems do Mass is the amount of resistance to LINEAR acceleration which we call linear In ROTATION the amount of resistance to ANGULAR acceleration depends on mass AND This combination is called and it s the rotational equivalent of You can think of it as rotational There are two types of objects Point Masses Rigid Bodies Shapes EXAMPLE A system is made of two point masses MLEFT 3 kg MRIGHT 4 kg at the ends of a 2 m long massless rod as shown Calculate the moment of inertia of the system if it spins about a perpendicular axis through the center of the rod where r is found by Table Lookup Common Moments of Inertia General Form fraction mR2 R Radius Point Masses The moment of inertia of a system of objects is the sum of each moment of inertia of the objects that make it up Center of Rod Page 5 Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body PRACTICE MOMENT OF INERTIA SIMPLE SYSTEM PRACTICE A system is made of two small masses MLEFT 3 kg MRIGHT 4 kg attached to the ends of a 5 kg 2 m long thin rod as shown Calculate the moment of inertia of the system if it spins about a perpendicular axis through the mass on the left Common Moments of Inertia Point Masses End of Rod Center of Rod EXAMPLE MOMENT OF INERTIA EARTH EXAMPLE The Earth has mass and radius 5 97 x 1024 kg and 6 37 x 106 m The radial distance between the Earth and the Sun is 1 50 x 1011 m Calculate the Moment of Inertia of the Earth as it spins around a itself treat the Earth as a solid sphere solid spheres have moment of inertia given by 2 5 mR2 b the Sun treat the Earth as a point mass Page 6 Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body PRACTICE MOMENT OF INERTIA FIND MASS PRACTICE A solid disc 4 m in diameter has a moment of inertia equal to 30 kg m2 about an axis through the disc perpendicular to its face The disc spins at a constant 120 RPM Calculate the mass of the disc EXAMPLE MOMENT OF INERTIA WITH DENSITY EXAMPLE A planet is nearly spherical with nearly continuous mass distribution with 8 x 107 m in radius and 10 000 kg m3 in density If the planet rotates around itself calculate its moment of inertia around its central axis Note VSPHERE 4 3 R3 Page 7 Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body EXAMPLE MOMENT OF INERTIA DISC WITH MASSES EXAMPLE The solid disc below has radius 4 m and mass 10 kg Two small objects are placed on top of it The object on the left has mass 2 kg and is placed half way between the disc s center and its edge The other object has mass 3 kg and is placed at the edge of the disc Calculate the system s disc masses moment of inertia around the disc s central axis Page 8 Knight Calc 5th edition Physics for Scientists and Engineers Ch 12 Rotation of a Rigid Body PRACTICE MOMENT …
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