A spool with a thread wound on it is placed on an inclined smooth plane set at an angle $\alpha = 30

Question

A spool with a thread wound on it is placed on an inclined smooth plane set at an angle $\alpha = 30^\circ$ to the horizontal. The free end of the thread is attached to the wall. The mass of the spool is $m = 200 	ext{ g}$, its moment of inertia relative to its own axis $I = 0.45 	ext{ g}\cdot	ext{m}^2$, and the radius of the wound thread layer $r = 3.0 	ext{ cm}$. Find the acceleration of the spool axis in $	ext{m/s}^2$. (Take $g = 9.8 	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

Let tension be $T$. Force equation along incline: $mg \sin\alpha - T = mw$. Torque equation about CM: $Tr = I\alpha_{ang}$. Constraint for no slipping of the thread: $w = \alpha_{ang} r \Rightarrow \alpha_{ang} = w/r$. Substituting $T = Iw/r^2$ into the force equation gives $mg \sin\alpha - Iw/r^2 = mw \Rightarrow w = \frac{mg \sin\alpha}{m + I/r^2}$. \Using the formula derived: $w = \frac{mg \sin\alpha}{m + I/r^2}$. Convert units: $m = 0.2 	ext{ kg}$, $I = 0.45 	imes 10^{-3} 	ext{ kg}\cdot	ext{m}^2$, $r = 0.03 	ext{ m}$. Thus, $I/r^2 = \frac{0.45 	imes 10^{-3}}{0.0009} = 0.5 	ext{ kg}$. Plugging in the values: $w = \frac{0.2 	imes 9.8 	imes 0.5}{0.2 + 0.5} = \frac{0.98}{0.7} = 1.4 	ext{ m/s}^2$.