Question

A yo-yo has mass $m$, inner radius $r$, and outer radius $R$. Its moment of inertia is $I$ about its center. The yo-yo rolls without slipping on a horizontal table, and is pulled along by a horizontal string wound around its inner radius. The pulling by the string with tension $T$ gives rise to an acceleration $a$. The yo-yo rolls without slipping.

What is the magnitude of acceleration a ?

• A. $\frac{TR(R-r)}{I+m{R}^2}$
• B. $\frac{TR(R+r)}{I-m{R}^2}$
• C. $\frac{TR(2r+R)}{I+m{R}^2}$
• D. None of these

Answer 1

The correct option is A $\frac{TR(R-r)}{I+m{R}^2}$

Using force equilibrium we have
$\sum F_x=m{a}_x$
$T-f_s=Ma$
$\sum F_y=m{a}_y$
$N-mg=0$
$\sum {\tau }_{cm}=I_{cm}{a}_{cm}$
$rT-R{f}_s=I\alpha$
Also as it rolls without slipping we get $a=R\alpha$
The torque equation is written in terms of friction $f_s$. Thus
$f_s=(\frac{r}{R}T)-(\frac{I}{R^2})a$
We substitute this in the equation of $\sum F_x$, we get
$T-\left[\right(\frac{r}{R}T)-(\frac{I}{R^2}\left)a\right]=Ma$
or
$T\frac{(R-r)}{R}=(M+\frac{I}{R^2})a$
or
$a=\frac{TR(R-r)}{I+M{R}^2}$