Question

A large 16.0 kg roll of paper with radius R = 18.0 cm rests against the wall and is held in place by a bracket attached to a rod through the center of the roll (Fig. P10.61). The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is

0.260 kg ${\mathit{m}}^{\mathbf{2}}$. The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of 30.0° with the wall. The weight of the bracket is negligible. The coefficient of kinetic

friction between the paper and the wall is ${\mu }_{k=}0.25$. A constant vertical force F = 60.0 N is applied to the paper, and the paper unrolls. What is the magnitude of (a) the force that the rod exerts on the paper as it unrolls; (b) the angular acceleration of the roll?

Short answer

(a) The force that the rod exert is $\boxed{293\hspace{0.33em}N}$.

(b) The angular acceleration is equal to $\boxed{16.2rad/s^2}$.

Step-by-step solution

Solution

At the point of contact with the wall, the wall is going to exert a friction force f directed in the downward direction and a normal force n to the right,

Thus,

There is a net force of zero network on the roll.

The figure is given below

Here, F is the force ,x is the horizontal axis, y is the vertical exercise, n is the normal reaction, is the net downward force, and f is the force of friction

So the balancing vertical and horizontal forces would be correct, that is,

$$\begin{gathered} F_{rod}cos\theta =f+w+f \\ And, \end{gathered}$$

$F_{rod}sin\theta =n$

The expression for the force of friction is as follow:

$f={\mu }_{k}n$

Substitute ${\mu }_{k}n$for in the equation

$$\begin{gathered} F_{rod}cos\theta =f+w+F \\ {F}_{rod}cos\theta ={\mu }_{k}n+w+F \\ Substitute{F}_{rod}fornintheequation{F}_{rod}cos\theta ={\mu }_{k}n+w+faandsolvefor{F}_{rod} \\ {F}_{rod}cos\theta ={\mu }_k\left(F_{rod}\sin \theta \right)+w+f \\ {F}_{rod}cos\theta -{\mu }_k\left(F_{rod}sin\theta \right)=w+f \\ {F}_{rod}=(w+f)/(cos\theta -{\mu }_{k}sin\theta ) \end{gathered}$$

(a) Find the force that the rod exerts

Calculate the magnitude of the force that rod exert on the paper as follow:

$$\begin{gathered} F_{rod}=\frac{w+F}{cos\theta -{\mu }_{k}sin\theta } \\ \mathrm{Substituting}\mathrm{mg}\mathrm{for}\mathrm{w}\mathrm{in}\mathrm{the}\mathrm{equation} \\ {F}_{rod}=\frac{w+F}{cos\theta -{\mu }_{k}sin\theta } \\ {F}_{rod}=\frac{mg+F}{cos\theta -{\mu }_{k}sin\theta } \end{gathered}$$

$$\begin{gathered} F_{rod}=(mg+F)/(cos\theta -{\mu }_{k}sin\theta ) \\ {F}_{r}od=\left(\right(16kg\left)\right(9.8\hspace{0.33em}m/s^2)+60\hspace{0.33em}N)/(cos{30}^{\circ }-(0.25\left)sin{30}^{\circ }\right) \\ =\left(\right(16kg\left)\right(9.8\hspace{0.33em}m/s^2)+60\hspace{0.33em}N)/\left(\right(cos{30}^{\circ }-(0.25)sin{30}^{\circ }\left)\right) \\ =292.6\hspace{0.33em}N \\ =293\hspace{0.33em}N \end{gathered}$$

Hence, the force that the rod exert is localid="1667809565437" $\boxed{293N}$.

(b) Find the angular acceleration of the roll

With respect to the center of the roll, both the rod and the normal force exert zero torque.

The magnitude of net torque is,

$$\begin{gathered} \tau =(F-f)R \\ Alsotheexpressionfor\tau isasfollow: \\ \tau =I\alpha \\ Equatingtheequation\tau =(F-f)Rand\tau =I\alpha . \\ (F-f)R=I\alpha \end{gathered}$$

$$\begin{gathered} Thefrictionforceis,f={\mu }_{k}n \\ ={\mu }_k\left(F_{rod}sin\theta \right) \\ \mathrm{Substitute}0.25\mathrm{for}{\mathrm{\mu }}_{\mathrm{k}},293\mathrm{N}\mathrm{for}{\mathrm{F}}_{\mathrm{rod}},\mathrm{and}{30}^{\circ }\mathrm{for}\mathrm{\theta }\mathrm{in}\mathrm{the}\mathrm{equation} \\ \mathrm{f}={\mathrm{\mu }}_{\mathrm{k}}\left({\mathrm{F}}_{rod}sin\theta \right) \\ f=(0.25)\left(\right(293\hspace{0.33em}N\left)sin{30}^{\circ }\right) \\ =36.63\hspace{0.33em}N \\ \mathrm{Rearrange}\mathrm{the}\mathrm{equation}(F-f)R=I\alpha for\alpha \\ \alpha =\frac{(F-f)R}{I} \end{gathered}$$

$$\begin{gathered} \mathrm{Substituting}60NforFand36.63Nforf,18cmforR,and0.260\hspace{0.33em}kg.m^{2}forIin \\ \mathrm{the}\mathrm{equation}\alpha =\frac{(F-f)R}{I} \\ \mathrm{\alpha }=\frac{(60\hspace{0.33em}\mathrm{N}-36.63\hspace{0.33em}\mathrm{N})\left(18\hspace{0.33em}\mathrm{cm}\left({\displaystyle \frac{{10}^2\mathrm{m}}{1\hspace{0.33em}\mathrm{cm}}}\right)\right)}{0.260\hspace{0.33em}\mathrm{kg}.{\mathrm{m}}^2} \\ =16.179\mathrm{rad}/{\mathrm{s}}^2 \\ =16.2\hspace{0.33em}\mathrm{rad}/{\mathrm{s}}^2 \\ \mathrm{Hence},\mathrm{the}\mathrm{angular}\mathrm{acceleration}\mathrm{is}\mathrm{equal}\mathrm{to}\boxed{16.2\hspace{0.33em}\mathrm{rad}/{\mathrm{s}}^2}. \end{gathered}$$