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The moment of inertia depends on the mass distribution. Rather than consider our rigid body as a series of point masses, we can describe it with a mass or density distribution.
21.2.1 Angular momentum and the CM. vCM{z} |wrt CM{z }J of CM translationWhat we have learned so farNewton’s laws relate to ro. ting systems in the same way that the laws relate to transitional motion.For any system of particles, the rate of change of internal angular momentum about an o. gin is equal to the total torque of the external ...
Purdue University – ME365 – Rotational Mechanical Systems • EOM of a simple Mass-Spring-Damper System We want to look at the energy distribution of the system.
Deriving the inertia formulas (1) Prove the angular momentum formula h + (2) Prove the kinetic energy formula + —O.IGCÙ (3) Prove the parallel axis theorem 10 (4) Prove the rotation formula 10 = RIOR (5) Derive the time derivative of the inertia matrix low
13.1.1 Examples of rigid bodies. Our first example of a rigid body is of a wheel rolling with constant angular velocity ̇φ = ω, and without slipping, This is shown in Fig. 13.1. The no-slip condition is dx = R dφ, so ̇x. CM = V = Rω. The velocity of a point within the wheel is. v = VCM + ω × r , (13.3) 1.
Example 10.2 A cable unwinding from a winch. Find: (a) Magnitude of the angular acceleration (b) Final angular velocity (c) Final speed of cable Solution: The torque by the tension force = Fl = (9.0 N)×(0.06 m) = 0.54 N•m. The moment of inertia about the given axis 2I 2= 1 2 𝑅2=1 2 (50 kg)(0.06 m) = 0.090 kg•m
The moment of inertia of a body rotating around an arbitrary axis is equal to the moment of inertia of a body rotating around a parallel axis through the center of mass
