Chapters 11 – 14
Andre Mallqui
5
th
Period
11.1 Rolling Motion of a Rigid Object
 center of mass moves a linear distance so
speed is
=
=
=
vmc
dsdt R dθdt Rω
 center of mass for rolling motion finds
acceleration is
=
=
acm dvcmdt
=
Rdωdt Rα
 Kinetic energy is
=
K
12Ipω2
 Inertia is
=
+
K
12ICMω2
MR2ω2

kinetic energy is sum of rolling and
translation kinetic energy
Ex 11.1 Sphere Rolling Down an Incline
The potential energy is mgh and kinetic is
zero so the change can find the acceleration, all
acceleration and velocity of spheres are the
same
Ex 11.2 Another Look at the Rolling Sphere
A FBD from the center of mass of the sphere
can be used with torque to find the same
acceleration
11.2 The Vector Product and Torque
 origin is inertial frame, Newton’s law is true
and vector product of r and F is used to find
torque
 this is not commutative but is distributive;
preserve order of multiplication
Ex 11.3 The Cross product
A = 2i + 3j and B = 1i + 2j, cross product is7k
for AxB and –BxA so they are equal
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 Spring '10
 Dude
 Angular Momentum, Force, Kinetic Energy, Special Relativity, Andre Mallqui

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