A spool consists of a cylinder of radius R_1 | vedclass

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(B) Let $a$ be the linear acceleration of the center of mass and $\alpha$ be the angular acceleration of the spool.$1$. Force equation for translation

Vedclass · Accepted Answer

$f = \left( \frac{I + mR_1R_2}{I + mR_2^2} \right)T$,towards left

Vedclass · Answer

(B) Let $a$ be the linear acceleration of the center of mass and $\alpha$ be the angular acceleration of the spool.$1$. Force equation for translational motion:$T - f = ma$  --- $(i)$$2$. Torque equation about the center of mass:$f R_2 - T R_1 = I \alpha$  --- $(ii)$$3$. Condition for rolling without slipping:$a = \alpha R_2 \implies \alpha = \frac{a}{R_2}$  --- $(iii)$Substituting $(iii)$ into $(ii)$:$f R_2 - T R_1 = I \left( \frac{a}{R_2} \right)$$f R_2^2 - T R_1 R_2 = Ia$$a = \frac{f R_2^2 - T R_1 R_2}{I}$  --- $(iv)$Substitute $(iv)$ into $(i)$:$T - f = m \left( \frac{f R_2^2 - T R_1 R_2}{I} \right)$$I(T - f) = m f R_2^2 - m T R_1 R_2$$IT - If = m f R_2^2 - m T R_1 R_2$$IT + m T R_1 R_2 = If + m f R_2^2$$T(I + m R_1 R_2) = f(I + m R_2^2)$$f = \left( \frac{I + m R_1 R_2}{I + m R_2^2} \right) T$Since the result is positive,the assumed direction of friction (towards the left) is correct.

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