The radius of the roll of paper shown in Fig. 8–67 is 7.6 cm and its moment of inertia is \(I = 3.3 \times {10^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^2}\). A force of 3.5 N is exerted on the end of the roll for 1.3 s, but the paper does not tear so it begins to unroll. A constant friction torque of \(I = 0.11\;{\rm{m}} \cdot {\rm{N}}\) is exerted on the roll which gradually brings it to a stop. Assuming that the paper’s thickness is negligible, calculate (a) the length of paper that unrolls during the time that the force is applied (1.3 s) and (b) the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving

The radius is\(R = 7.6\;{\rm{cm}}\).

The moment of inertia is\(I = 3.3 \times {10^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^2}\).

The force is\(F = 3.5\;{\rm{N}}\).

The time is\(t = 1.3\;{\rm{s}}\).

The frictional torque is \({\tau _{{\rm{fr}}}} = 0.11\;{\rm{N}} \cdot {\rm{m}}\).

In rotational motion, according to Newton’s second law, the angular acceleration of a body is proportional to the total torque applied to it.

Consider clockwise as the positive direction and anticlockwise as the negative direction. In this problem, apply Newton’s second law for rotational motion and the definition of radial angle.

The free-body diagram of the roll of paper is as follows:

The kinematic equation can be written as:

\(\begin{aligned}{l}\Delta \theta = \omega t + \frac{1}{2}\alpha {t^2}\\\Delta \theta = \left( 0 \right) + \frac{1}{2}\alpha {t^2}\\\Delta \theta = \frac{1}{2}\alpha {t^2}.\end{aligned}\)

Here,\(\omega \)is the initial angular velocity, whose value is zero, and\(\alpha \)is the angular acceleration.

The relation for the rate of change in angular momentum can be written as shown below:

\(\begin{aligned}{c}\Sigma \tau = I\alpha \\FR - {\tau _{{\rm{fr}}}} = I\alpha \\\alpha = \left( {\frac{{FR - {\tau _{{\rm{fr}}}}}}{I}} \right)\end{aligned}\)

The relation for distance can be written as shown below:

\(\begin{aligned}{l}{s_1} = R\Delta \theta \\{s_1} = R\left( {\frac{1}{2}\alpha {t^2}} \right)\\{s_1} = R\left( {\frac{1}{2}{t^2}} \right)\left( {\frac{{FR - {\tau _{{\rm{fr}}}}}}{I}} \right)\end{aligned}\)

Put the values in the above relation.

\(\begin{aligned}{l}{s_1} = \left( {7.6\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\left( {\frac{1}{2}{{\left( {1.3\;{\rm{s}}} \right)}^2}} \right)\left( {\frac{{\left( {3.5\;{\rm{N}}} \right)\left( {7.6\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right) - \left( {0.11\;{\rm{N}} \cdot {\rm{m}}} \right)}}{{\left( {3.3 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^2}} \right)}}} \right)\\{s_1} = 3.03\;{\rm{m}}\end{aligned}\)

Thus, \({s_1} = 3.03\;{\rm{m}}\) is the required length of the paper.

The kinematic equation is written below:

\(\begin{aligned}{l}{\omega _0} = \omega + \alpha t\\{\omega _0} = 0 + \left( {\frac{{FR - {\tau _{{\rm{fr}}}}}}{I}} \right)t\\{\omega _0} = \left( {\frac{{FR - {\tau _{{\rm{fr}}}}}}{I}} \right)t\end{aligned}\)

Put the values in the above relation.

\(\begin{aligned}{l}{\omega _0} = \left( {\frac{{\left( {3.5\;{\rm{N}}} \right)\left( {7.6\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right) - \left( {0.11\;{\rm{N}} \cdot {\rm{m}}} \right)}}{{\left( {3.3 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^2}} \right)}}} \right)\left( {1.3\;{\rm{s}}} \right)\\{\omega _0} = 61.4\;{\rm{rad/s}}\end{aligned}\)

The relation for frictional torque is

\(\begin{aligned}{c} - {\tau _{{\rm{fr}}}} = I\alpha '\\\alpha ' = \left( {\frac{{ - {\tau _{{\rm{fr}}}}}}{I}} \right).\end{aligned}\)

The relation for angular displacement is \(\Delta \theta ' = \frac{{{\omega ^2} - {\omega _0}^2}}{{2\alpha '}}\).

The relation for the length of the paper can be written as shown below:

\(\begin{aligned}{l}{s_2} = R\Delta \theta '\\{s_2} = R\left( {\frac{{{\omega ^2} - {\omega _0}^2}}{{2\alpha '}}} \right)\\{s_2} = R\left( {\frac{{\left( {{\omega ^2} - {\omega _0}^2} \right)I}}{{ - 2{\tau _{{\rm{fr}}}}}}} \right)\end{aligned}\)

Put the values in the above relation.

\(\begin{aligned}{l}{s_2} = \left( {7.6\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\left( {\frac{{{{\left( 0 \right)}^2} - {{\left( {61.4\;{\rm{rad/s}}} \right)}^2}}}{{ - 2\left( {0.11\;{\rm{N}} \cdot {\rm{m}}} \right)}}} \right)\left( {3.3 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\\{s_2} = 4.3\;{\rm{m}}\end{aligned}\)

Thus, \({s_2} = 4.3\;{\rm{m}}\) is the required length of the paper.