19.5 Role of friction in rolling  (Page 4/5)

Problem : Two forces " $F_1$ " and" $F_2$ " are applied on a spool of mass “M”, moment of inertia “I” and radius “R” as shown in the figure. If the spool is rolling on the surface, find the ratio of forces, $\raisebox{1ex}{$F_1$}\!\left/ \!\raisebox{-1ex}{$F_2$}\right.$ , such that friction between spool and the surface is zero.

Solution : A spool consists of two disk joined by a cylinder and is used to store flexible cables, ropes etc. Notably, this question involves more than one external force. However, one simplifying aspect here (as given in the question) is that friction is not required for the overall analysis.

For the overall analysis, friction of rolling is zero for the combined effect of two forces.

(i) For translation :

Applying Newton’s second law for translation,

(ii) For rotation :

Applying Newton’s second law for rotation,

Note that force“ $F_1$ “ constitutes a clockwise torque (negative in sign), whereas force“ $F_2$ “ constitutes an anticlockwise torque (positive in sign).

(iii) For rolling :

Putting this value of “α” in equation – 5,

$$\begin{array}{l}-F_{1}R+F_{2}x\frac{R}{2}=Ix\frac{a_C}{R}\end{array}$$

Putting the value of linear acceleration “ $a_C$ “ from equation – 4, we have :

$$\begin{array}{l}-F_{1}R+F_{2}x\frac{R}{2}=Ix\frac{F_1+F_2}{MR}\end{array}$$

Rearranging,

$$\begin{array}{l}\Rightarrow F_1(R-\frac{I}{MR})=F_2(\frac{R}{2}+\frac{I}{MR})\end{array}$$

Dividing both sides by“ $F_2$ “ ,

$$\begin{array}{l}\Rightarrow \frac{F_1}{F_2}=\frac{(\frac{R}{2}+\frac{I}{MR})}{(R-\frac{I}{MR})}\end{array}$$

$$\begin{array}{l}\Rightarrow \frac{F_1}{F_2}=\frac{(M{R}^2+2I)}{2(M{R}^2-I)}\end{array}$$