Question

A spool of thread consists of a cylinder of radius${\mathit{R}}_{\mathbf{1}}$with end caps of radius${\mathit{R}}_{\mathbf{2}}$as depicted in the end view shown in Figure P10.91. The mass of the spool, including the thread, is$\mathit{m}$, and its moment of inertia about an axis through its center is l. The spool is placed on a rough, horizontal surface so that it rolls without slipping when a force$\overrightarrow{\mathbf{T}}$acting to the right is applied to the free end of the thread. (a) Show that the magnitude of the friction force exerted by the surface on the spool is given by

$\mathit{f}\mathbf{=}\mathbf{\left(}\frac{{\displaystyle I+m{R}_1{R}_2}}{{\displaystyle I+m{R}_2^2}}\mathbf{\right)}\mathit{T}$

(b) Determine the direction of the force of friction.

Short answer

The solution is,

a)$\alpha =\frac{a}{R_2}$

b)$f=\left(\frac{{\displaystyle I+m{R}_1{R}_2}}{{\displaystyle I+m{R_2}^2}}\right)T$

The frictional force is left, hence it is directed to the left.

Step-by-step solution

About the problem

VISUALIZE: From the figure on the rough horizontal surface, visualize the spool's motion, and calculate the friction force exerted by the surface on the spool.

CATEGORIZE: The second law of motion applies to this problem.

Given information

We are given that,

Radius of spool is $R_1$.

Radius of spool along with end caps is$R_2$.

The mass of the spool including the thread is m.

The moment of inertia of the spool about an axis through its centre is l.

Force applied is T.

The magnitude of the friction force exerted by the surface on the spool

(a)

As stated in Newton's second law of motion.

$\begin{array}{l}\Sigma F_x=m{a}_x\\ -f+T=ma\mathrm{........}\left(1\right)\end{array}$

We can utilize if we want to calculate torques around the mass center.

$\begin{array}{l}\Sigma \tau =I\alpha \\ +f{R}_2-T{R}_1=I\alpha \end{array}$

We get when we can roll without slipping

$\alpha =\frac{a}{R_2}$

Substituting the values, we get,

$f{R}_2-T{R}_1=\frac{Ia}{R_2}$

$\begin{array}{c}=\frac{I}{R_{2}m}(T-f)\left[\text{From\hspace{0.17em}equation\hspace{0.17em}1}\right]\\ f{R}_2^{2}m-T{R}_1{R}_{2}m=IT-If\\ f(I+m{R}_2^2)=T(I+m{R}_1{R}_2)\end{array}$

As a result, frictional forces are,

$f=\left(\frac{{\displaystyle I+m{R}_1{R}_2}}{{\displaystyle I+m{R}_2^2}}\right)T$

The direction of the force of friction

(b)

The frictional force is left, hence it is directed to the left, according to the answer.

We can utilize the impulse-momentum theorem to calculate the spool's translational motion while disregarding the fact that it is rotating.