Question

A roll of toilet paper is held by the first piece and allowed to unfurl. The role has an outer radius $R = 6\,cm$ , an inner radius $r = 1.8\,cm$ , a mass $m = 200\,g$ , and falls a distance of $s = 3\,m$ . Assuming the outer diameter of the rod does not change significantly during the fall, then angular acceleration of roll is approximately $k$ , then the value $\dfrac{k}{2}$ is:

Answer 1

We can use the law of conservation of mechanical energy to solve this problem. According to this law the total initial mechanical energy will be equal to the total final mechanical energy. By using this we can find the value of angular velocity. Then the angular acceleration can be found by using the equation of motion connecting initial and final angular velocity, angular acceleration and the angular displacement.

Complete step by step answer:
Here the potential energy that the roll has at the top will be concentrated into the kinetic energy due to the translation and kinetic energy due to rotation at the bottom.
This is because we know that according to conservation of energy, total mechanical energy should be conserved.
Initially the roll has only potential energy since it is at rest and finally it has only kinetic energy. Thus, we can write,
$\Rightarrow U = {K_T} + {K_R}$ (1)
Where, $U$ is the initial potential energy, ${K _T}$ is the translated kinetic energy and ${K_R}$ is the kinetic energy due to rotation.
Kinetic energy due to translation is given as
${K_T} = \dfrac{1}{2}m{v^2}$
Where, $m$ is the mass $v$ is the velocity.
We know that the relation between linear velocity and angular velocity is given as
$v = \omega R$
Where, $v$ is the linear velocity, $\omega$ is the angular velocity and $R$ is the perpendicular distance.
$\Rightarrow {K_T} = \dfrac{1}{2}m{\omega ^2}{R^2}$
Kinetic energy due to rotation is given as
${K_R} = \dfrac{1}{2}I{\omega ^2}$
Where, $I$ is the moment of inertia.
We know that the moment of inertia of hollow cylinder is given as
$I = \dfrac{1}{2}m\left( {{R^2} + {r^2}} \right)$
Where, $R$ is the outer radius and $r$ is the inner radius
$\therefore {K_R} = \dfrac{1}{2}\left( {\dfrac{1}{2}m\left( {{R^2} + {r^2}} \right)} \right)$
Potential energy is given as
$U = mgh$
Where, m is the mass, $h$ is the height and $g$ is the acceleration due to gravity.
Here, the height is equal to the distance through which it falls
$\Rightarrow h = s$
On substituting all these values in equation 1, we get
$\Rightarrow mgs = \dfrac{1}{2}mR{\omega ^2} + \dfrac{1}{4}m{R^2}{\omega ^2} + \dfrac{1}{4}m{r^2}{\omega ^2}$
$\Rightarrow 4gs = \left( {3{R^2} + {r^2}} \right){\omega ^2}$
$\Rightarrow \omega = {\left( {\dfrac{{4gs}}{{3{R^2} + {r^2}}}} \right)^{\dfrac{1}{2}}}$ (2)
Given,
The outer radius, $r = 0 \cdot 018\,m$
mass, $m = 0 \cdot 2\,kg$
distance, $s = 3\,m$
we know acceleration due to gravity is
$g = 9 \cdot 8m/{s^2}$
On substituting all these values we get,
$\Rightarrow \omega = {\left( {\dfrac{{4 \times 9 \cdot 8 \times 3}}{{{{\left( {3 \times 0 \cdot 06} \right)}^2} + {{\left( {0 \cdot 018} \right)}^2}}}} \right)^{\dfrac{1}{2}}}$
$\therefore \omega = 103\,rad/s$
We know that equation of motion connecting angular acceleration and angular velocity given as
${\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta$
Where, ${\omega _0}$ is the initial angular velocity, $\alpha$ is the angular acceleration and $\theta$ is the angular displacement.
Since initial angular velocity
${\omega _0} = 0$
We get
$\Rightarrow {\omega ^2} = 2\alpha \Delta \theta$
Where, $\Delta \theta = \dfrac{s}{R}$
Since angular displacement is the ratio of distance to the radius.
$\therefore {\omega ^2} = 2\alpha \dfrac{s}{R}$
$\Rightarrow \alpha = \dfrac{{R{\omega ^2}}}{{2s}}$
On substituting the values we get,
$\Rightarrow \alpha = \dfrac{{0 \cdot 06 \times 103}}{{2 \times 3}}$
$\therefore \alpha = 106\,\,rad/{s^2}$
It is given that angular acceleration $\alpha = k$
$\therefore k = 106\,rad/{s^2}$
So, the value of $\dfrac{k}{2}$ will be $\dfrac{{106}}{2}\,rad/{s^2}$
$\therefore \dfrac{k}{2} = 53\,rad/{s^2}$

Note: The equation of motion can be applied only when the angular acceleration is a constant. When the angular acceleration varies continuously the situation will be more complex that we cannot solve it using simple equations of motion.
Remember that the conservation law of total mechanical energy is applicable only if all the forces acting on it are conservative. When friction is also taken into consideration, we cannot apply the conservation of energy because friction is a non-conservative force.