A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in a vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end, the speed of the free end of the rod when it passes thro…

Question

A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in a vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end, the speed of the free end of the rod when it passes through its lowest position is m s⁻¹. (Take g=10 m/s².)

Accepted Answer

6. The rod swings about its clamped lower end, from straight up to straight down. Its centre of mass therefore drops from +L/2 to -L/2, a total fall of L=0.6 m. Moment of inertia about the clamped end: I=(ML²)/3=((2)(0.36))/3=0.24 kg m². Energy: MgL=1/2Iω². (2)(10)(0.6)=1/2(0.24)ω², so 12=0.12 ω² and ω²=100. ω=10 rad s⁻¹. The free end is a distance L from the axis, so v=ω L=(10)(0.6)=6 m s⁻¹.

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