Step 1

We have given the following values:

The mass of the spool is $m = 100\;{\rm{kg}}$.

The coefficient of kinetic friction is ${\mu _k} = 0.1$.

The radius of gyration about the center of mass is ${k_G} = 0.25\;{\rm{m}}$.

We are asked to determine the angular velocity of the spool when $t = 4\;{\rm{s}}$ after it released from rest.

 
Step 2

The weight W of the spool is given by:

\[W = mg\]

Substitute 100 kg for m and $9.81\;{\rm{m}}/{{\rm{s}}^2}$ for g in the above equation:

\[\begin{array}{c} W = \left( {100\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m}}/{{\rm{s}}^2}} \right) \times \frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}}}\\ = 981\;{\rm{N}} \end{array}\]
 
Step 3

Draw a kinetic diagram of the spool.

Here, IC is the instantaneous center, $\omega $ is the angular velocity of the spool, ${v_G}$ is the velocity of the center of mass G, ${r_{G/IC}}$is the distance of center G from instantaneous center IC.

 
Step 4

Consider figure (a), the velocity of center G is given by:

\[{v_G} = \omega {r_{G/IC}}\]

Substitute 0.2 m for ${r_{G/IC}}$ in the above equation:

\[{v_G} = \omega \left( {0.2\;{\rm{m}}} \right)\]
 
Step 5

The mass moment of inertia of the spool about its mass center is given by:

\[{I_G} = mk_G^2\]

Substitute the value of m and ${k_G}$ in the above equation:

\[\begin{array}{c} {I_G} = \left( {100\;{\rm{kg}}} \right){\left( {0.25\;{\rm{m}}} \right)^2}\\ = 6.25\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]
 
Step 6

Draw a free-body diagram of the spool:

Here, T is the tension in the tension in the cable, f is the kinetic friction, and N is the normal reaction force.

 
Step 7

Consider figure (b) and apply principle of impulse and momentum along the y-axis:

\[\begin{array}{c} m{\left[ {{{\left( {{v_G}} \right)}_y}} \right]_1} + \sum {\int\limits_{{t_1}}^{{t_2}} {{F_y}dt} } = m{\left[ {{{\left( {{v_G}} \right)}_y}} \right]_2}\\ 0 + N\left( {4\;{\rm{s}}} \right) - W\cos 30^\circ \left( {4\;{\rm{s}}} \right) = 0\\ N = W\cos 30^\circ \end{array}\]

Substitute the value of W in the above equation:

\[\begin{array}{l} N = \left( {981\;{\rm{N}}} \right)\cos 30^\circ \\ N = 849.57\;{\rm{N}} \end{array}\]
 
Step 8

Consider figure (b) and apply principle of impulse and momentum along the x-axis:

\[\begin{array}{c} m{\left[ {{{\left( {{v_G}} \right)}_x}} \right]_1} + \sum {\int\limits_{{t_1}}^{{t_2}} {{F_x}dt} } = m{\left[ {{{\left( {{v_G}} \right)}_x}} \right]_2}\\ 0 + T\left( {4\;{\rm{s}}} \right) + f\left( {4\;{\rm{s}}} \right) - W\sin 30^\circ \left( {4\;{\rm{s}}} \right) = m{v_G}\\ T\left( {4\;{\rm{s}}} \right) + {\mu _k}N\left( {4\;{\rm{s}}} \right) - W\sin 30^\circ \left( {4\;{\rm{s}}} \right) = m{v_G} \end{array}\]

Substitute the value of parameters in the above equation:

\[\begin{array}{c} \left\{ \begin{array}{l} T\left( {4\;{\rm{s}}} \right) + \left( {0.1} \right)\left( {849.57\;{\rm{N}}} \right)\left( {4\;{\rm{s}}} \right)\\ - \left( {981\;{\rm{N}}} \right)\sin 30^\circ \left( {4\;{\rm{s}}} \right) \end{array} \right\} = 100\;{\rm{kg}}\left[ { - \omega \left( {0.2\;{\rm{m}}} \right)} \right]\\ T + 5\omega \;{\rm{m/s}} = 405.54\;{\rm{N}}\\ T = 405.54\;{\rm{N}} - 5\omega \;{\rm{kg}} \cdot {\rm{m/s}} \end{array}.\].....(1)
 
Step 9

Consider figure (b) and apply principle of impulse and momentum about center G:

\[\begin{array}{c} {I_G}{\omega _1} + \sum {\int\limits_{{t_1}}^{{t_2}} {{M_G}dt} } = {I_G}{\omega _2}\\ 0 + {\mu _k}N\left( {0.4\;{\rm{m}}} \right)\left( {4\;{\rm{s}}} \right) - T\left( {0.2\;{\rm{m}}} \right)\left( {4\;{\rm{s}}} \right) = - 6.25\omega \;{\rm{kg}} \cdot {{\rm{m}}^2}\\ 0.8T - 6.25\omega \;{\rm{kg}} \cdot {\rm{m}}/{\rm{s}} = 1.6{\mu _k}N \end{array}\]

Substitute the value of parameters in the above equation:

\[\begin{array}{l} 0.8T - 6.25\omega \;{\rm{kg}} \cdot {\rm{m}}/{\rm{s}} = 1.6\left( {0.1} \right)\left( {849.57\;{\rm{N}}} \right)\\ 0.8T - 6.25\omega \;{\rm{kg}} \cdot {\rm{m}}/{\rm{s}} = 135.93\;{\rm{N}} \end{array}.\].....(2)
 
Step 10

Substitute the value of T from equation (1) in the equation (2):

\[\begin{array}{c} 0.8\left( {405.54\;{\rm{N}} - 5\omega \;{\rm{kg}} \cdot {\rm{m}}/{\rm{s}}} \right) - 6.25\omega \;{\rm{kg}} \cdot {\rm{m}}/{\rm{s}} = 135.93\;{\rm{N}}\left( {\frac{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1{\rm{ N}}}}} \right)\\ \omega = 18.39\;{\rm{rad}}/{\rm{s}} \end{array}\]
 
Step 11

Substitute the value of $\omega $ in equation (1):

\[\begin{array}{c} T = \left( {405.54 - 5\left( {18.39} \right)} \right)\;{\rm{N}}\\ T = 313.59\;{\rm{N}} \end{array}\]