Combined Translation And Rotation

Question

A solid cylinder of mass M and radius R unwinds from a light string that is wrapped around it and fixed at the top, falling vertically under gravity. Taking g as 9.8 m/s^2, what is the downward acceleration of its centre?

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As the cylinder falls, the string tension both opposes gravity and exerts a torque that spins the cylinder, with the no-slip condition linking the two motions through $a = R\alpha$. Newton's second law along the vertical direction is $Mg - T = Ma$, while the torque equation about the centre is $TR = I\alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R}$, giving $T = \frac{1}{2}Ma$. Combining the two relations, $Mg - \frac{1}{2}Ma = Ma$, which leads to $g = \frac{3}{2}a$, hence $a = \frac{2g}{3} = \frac{2 	imes 9.8}{3} \approx 6.53	ext{ m/s}^2$. The value 9.8 m/s$^2$ ignores the tension and treats the cylinder as in free fall. The value 4.9 m/s$^2$ halves $g$ without justification. The value 3.27 m/s$^2$ corresponds to using the wrong inertia factor. This is the classic NCERT unwinding-cylinder problem. As a plausibility check, the acceleration must lie below $g$ because the tension retards the fall, and $6.53 < 9.8$ confirms it. Energetically, the descending cylinder converts gravitational potential energy into both translational and rotational kinetic energy, and since one-third of that energy is diverted into spin, only two-thirds drives the downward motion, fully consistent with the reduced acceleration of $\frac{2g}{3}$ obtained from the force analysis.

Combined Translation And Rotation

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