Q. 28. Derive an expression for the kinetic energy of a body rotating
with constant angular velocity. State how it depends on the
moment of inertia, and frequency and period of rotation.
Answers
Rotational inertia takes the place of mass in the rotational version of Newton's 2ⁿᵈ law.
Consider a mass mmm attached to one end of a massless rod. The other end of the rod is hinged so that the system can rotate about the central hinge point as shown in Figure 2.
We now start rotating the system by applying a tangential force F_TFTF, start subscript, T, end subscript to the mass. From Newton’s 2ⁿᵈ law,
We now start rotating the system by applying a tangential force F_TFTF, start subscript, T, end subscript to the mass. From Newton’s 2ⁿᵈ law,F_T = m a_TFT=maTF, start subscript, T, end subscript, equals, m, a, start subscript, T, end subscript.
We now start rotating the system by applying a tangential force F_TFTF, start subscript, T, end subscript to the mass. From Newton’s 2ⁿᵈ law,F_T = m a_TFT=maTF, start subscript, T, end subscript, equals, m, a, start subscript, T, end subscript.this can also be written as
We now start rotating the system by applying a tangential force F_TFTF, start subscript, T, end subscript to the mass. From Newton’s 2ⁿᵈ law,F_T = m a_TFT=maTF, start subscript, T, end subscript, equals, m, a, start subscript, T, end subscript.this can also be written asF_T = m (r \alpha)FT=m(rα)F, start subscript, T, end subscript, equals, m, left parenthesis, r, alpha, right parenthesis.
We now start rotating the system by applying a tangential force F_TFTF, start subscript, T, end subscript to the mass. From Newton’s 2ⁿᵈ law,F_T = m a_TFT=maTF, start subscript, T, end subscript, equals, m, a, start subscript, T, end subscript.this can also be written asF_T = m (r \alpha)FT=m(rα)F, start subscript, T, end subscript, equals, m, left parenthesis, r, alpha, right parenthesis.Newton's 2ⁿᵈ law relates force to acceleration. In rotational mechanics torque \tauτtau takes the place of force. Multiplying both sides by the radius gives the expression we want.
We now start rotating the system by applying a tangential force F_TFTF, start subscript, T, end subscript to the mass. From Newton’s 2ⁿᵈ law,F_T = m a_TFT=maTF, start subscript, T, end subscript, equals, m, a, start subscript, T, end subscript.this can also be written asF_T = m (r \alpha)FT=m(rα)F, start subscript, T, end subscript, equals, m, left parenthesis, r, alpha, right parenthesis.Newton's 2ⁿᵈ law relates force to acceleration. In rotational mechanics torque \tauτtau takes the place of force. Multiplying both sides by the radius gives the expression we want.\begin{aligned} F_T r &= m (r \alpha) r\\ \tau &= m r^2 \alpha \\ \tau &= I \alpha\end{aligned}FTrττ=m(rα)r=mr2α=Iα
We now start rotating the system by applying a tangential force F_TFTF, start subscript, T, end subscript to the mass. From Newton’s 2ⁿᵈ law,F_T = m a_TFT=maTF, start subscript, T, end subscript, equals, m, a, start subscript, T, end subscript.this can also be written asF_T = m (r \alpha)FT=m(rα)F, start subscript, T, end subscript, equals, m, left parenthesis, r, alpha, right parenthesis.Newton's 2ⁿᵈ law relates force to acceleration. In rotational mechanics torque \tauτtau takes the place of force. Multiplying both sides by the radius gives the expression we want.\begin{aligned} F_T r &= m (r \alpha) r\\ \tau &= m r^2 \alpha \\ \tau &= I \alpha\end{aligned}FTrττ=m(rα)r=mr2α=IαThis expression can now be used to find the behavior of a mass in response to a known torque.
