The total kinetic energy of a uniform rod rotating about an axis is calculated using both rotational and translational kinetic energy formulas. The correct expression is derived from the equations deltaKrot = 1/2Iw^2 and deltaKtrans = 1/2mv^2, where I is the moment of inertia and w is the angular velocity. The rod's kinetic energy includes contributions from both its rotation about the center of mass and the translational motion of the center of mass. The final answer aligns with option B, confirming the necessity to express all variables in terms of angular velocity (ω).
PREREQUISITESPhysics students, educators, and anyone studying mechanics, particularly those focusing on rotational motion and energy calculations.
You can view the rod as having both rotational KE about the center of mass and translational KE of the center of mass. Or you can view the rod as in pure rotation about the given axis. Your choice. Either way, the answer will be the same.ccarit3007 said:
Two points: (1) How did you get that expression for the rotational KE? (2) Realize that ω and v are related. Express everything in terms of ω.ccarit3007 said:
