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. Find the force of friction $f$, in terms of tension $T$.

• A. $\frac{T(I-mRr)}{I+m{R}^2}$
• B. $\frac{T(I+mRr)}{I+m{R}^2}$
• C. $\frac{T(I+2mRr)}{2I+m{R}^2}$
• D. $\frac{T(2I+mRr)}{I-m{R}^2}$

Answer 1

The correct option is B $\frac{T(I+mRr)}{I+m{R}^2}$

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 as 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
$a=-\frac{TR(R-r)}{I+m{R}^2}$
Substituting this value into the equation for $f_s$ we get
$f_s=\frac{T(I+mRr)}{I+m{R}^2}$