Sect. 10.4: Rotational Kinetic Energy

Translation-Rotation Analogues & Connections Translation RotationDisplacementx VelocityvAccelerationaMass m?Kinetic Energy (K) ()mv2?CONNECTIONSs = r, v = r, at= r, ac = (v2/r) = 2r

Sect. 10.4: Rotational Kinetic EnergyTranslational motion (Chs. 7 & 8): K = ()mv2Rigid body rotation, angular velocity . Rigid means Every point has the same . Object is made of particles, masses mi. For each mi at a distance ri from the rotation axis: vi = ri. So, each mass mi has kinetic energy Ki = ()mi(vi)2. So, The Rotational Kinetic Energy is: KR = [()mi(vi)2] = ()mi(ri)22 = ()mi(ri)22 2 goes outside the sum, since its the same everywhere in the bodyDefine the moment of inertia of the object, I mi(ri)2 KR = ()I2 (Analogous to ()mv2)

Translation-Rotation Analogues & Connections Translation RotationDisplacementx VelocityvAccelerationaForce (Torque)F Mass(moment of inertia)mIKinetic Energy (K) ()mv2 ()I2 CONNECTIONSs = r, v = r, at= r, ac = (v2/r) = 2r

Example 10.3: Four Rotating Objects4 tiny spheres on ends of 2 massless rods. Arranged in the x-y plane as shown. Sphere radii are very small. Masses are treated as point masses. Calculate moment of inertia I when

(A) The system is rotated about the y axis, as in Fig. a.

(B) The system is rotated in the x-y plane as in Fig. b.

Bottom line for rotational kinetic energy Analogy between the kinetic energies associated with linear motion: K = ()mv2, & the kinetic energy associated with rotational motion: KR = ()I2.

NOTE: Rotational kinetic energy is not a new type of energy. But the form of this kinetic energy is different because it applies to a rotating object.

Of course, the units of rotational kinetic energy are Joules (J)

Sect. 10.5: Moments of InertiaDefinition of moment of inertia: I mi(ri)2Dimensions of I are ML2 & its SI units are kg.m2Use calculus to calculate the moment of inertia of an object: Assume it is divided into many small volume elements, each of mass mi. Rewrite expression for I in terms of mi. & take the limit as mi 0. So Make a small volume segment assumption & the sum changes to an integral: density of the object. If is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known. With this assumption, I can be calculated for simple geometries.

Notes on Densities Volume Mass Density Solid mass m in volume V: r = mass per unit volume: r = (m/V)Surface Mass Density Mass m on a thin surface of thickness t: = mass per unit thickness: = rtLinear Mass Density Mass m in a rod of length L & cross sectional area A: = mass per unit length: = (m/L) = rA

Moment of Inertia of a Uniform Thin HoopThis is a thin hoop, so all mass elements are the same distance from the center

Moment of Inertia of a Uniform Thin RodThe hatched area has mass dm = dx = (M/L)dxLinear Mass Density: Mass per unit length of a rod of uniform cross-sectional area = M/L = ASo, the moment of inertia is

Moment of Inertia of a Uniform Solid CylinderDivide cylinder into concentric shells of radius r, thickness dr, length L. Volume Mass Density: = (m/V) Then, I becomes:

Moments of Inertia of Some Rigid Objects

The Parallel Axis Theorem NOTE! I depends on the rotation axis! Suppose we need I for rotation about the off-center axis in the figure:If the axis of interest is a distance d from aparallel axis through the center of mass, the moment of inertia has the form: d

Moment of Inertia of a Rod Rotating Around an EndWeve seen: Moment of inertia ICM of a rod of length L about the center of mass is:

What is the moment of inertia I about one end of the rod? Shift rotation axis by D = ()L& use Parallel Axis Theorem:

Figure 10.8: (Example 10.3) Four spheres form an unusual baton. (a) The baton is rotated about the y axis. (b) The baton is rotated about the z axis.