Step 1 We are given the weight of tandem drum roller frame excluding the two rollers as ${w_d} = 4000\,{\rm{lb}}$, the weight of each roller as ${w_r} = 1500\,{\rm{lb}}$, the radius of gyration of the roller wheel as $k = 1.25\,{\rm{ft}}$, the radius of each roller as $r = 1.5\,{\rm{ft}}$, the magnitude of the torque as $M = 300\,{\rm{lb}} \cdot {\rm{ft}}$ at rear roller $A$, and the observation time as $t = 10\,{\rm{s}}$.
We are asked to determine the speed of the drum roller $10\,{\rm{s}}$ later, starting from rest.
Step 2 The free body diagram of two rollers can be drawn as shown below.
Step 3 Find the mass of the roller wheel using the following relation.
\[{m_r} = \frac{{{w_r}}}{g}\]
On substituting the known values in the above equation we get,
\[\begin{array}{c} {m_r} = \frac{{\left( {1500\,{\rm{lb}}} \right)}}{{\left( {32.2\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)}}\\ = 46.58\,{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}{\rm{/ft}} \end{array}\]
Step 4 Find the moment of inertia of the roller wheel using the following relation.
\[{I_r} = {m_r}{k^2}\]
On substituting the known values in the above equation we get,
\[\begin{array}{c} {I_r} = \left( {46.58\,\frac{{{\rm{lb}} \cdot {{\rm{s}}^2}}}{{{\rm{ft}}}}} \right){\left( {1.25\,{\rm{ft}}} \right)^2}\\ = 72.78\,{\rm{lb}} \cdot {\rm{ft}} \cdot {{\rm{s}}^2} \end{array}\]
Step 5 As the roller starting from the rest, the initial angular velocity and velocity is zero $\left( {{\omega _1} = 0\,\& \,{v_1} = 0} \right)$.
Find the reaction ${A_x}$ by the principle of angular impulse in $x$-direction on wheel $A$ using the following relation.
\[{I_r}{\omega _1} + \int\limits_0^t {\left( {M - \left( {{A_x} \times r} \right)} \right)dt} = {I_r}\left( {{v_2}r} \right) + {m_r}\left( {\frac{{{v_2}}}{r}} \right)\]
On integrating the above equation we get,
\[{I_r}{\omega _1} + \left[ {M - \left( {{A_x} \times r} \right)} \right]t = {I_r}\left( {\frac{{{v_2}}}{r}} \right) + {m_r}\left( {{v_2}r} \right)\]
On substituting the known values in the above equation we get,
\[\begin{array}{c} \left[ \begin{array}{l} \left( {72.78\,{\rm{lb}} \cdot {\rm{ft}} \cdot {{\rm{s}}^2}} \right)\left( 0 \right)\\ + \left[ {\left( {300\,{\rm{lb}} \cdot {\rm{ft}}} \right) - \left( {{A_x} \times \left( {1.5\,{\rm{ft}}} \right)} \right)} \right]\left( {10\,{\rm{s}}} \right) \end{array} \right] = \left[ \begin{array}{l} \left( {72.78\,{\rm{lb}} \cdot {\rm{ft}} \cdot {{\rm{s}}^{\rm{2}}}} \right)\left( {\frac{{{v_2}}}{{\left( {1.5\,{\rm{ft}}} \right)}}} \right) + \\ \left( {46.58\,{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}{\rm{/ft}}} \right)\left( {{v_2}} \right)\left( {1.5\,{\rm{ft}}} \right) \end{array} \right]\\ \left[ {\left( {3000\,{\rm{lb}} \cdot {\rm{ft}} \cdot {\rm{s}}} \right) - \left( {{A_x} \times \left( {15\,{\rm{ft}} \cdot {\rm{s}}} \right)} \right)} \right] = \left( {118.39\,{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}} \right){v_2}\\ {A_x} = 200\,{\rm{lb}} - \left( {7.893\,{\rm{lb}} \cdot {\rm{s/ft}}} \right){v_2}\,......\left( 1 \right) \end{array}\]
Step 6 Find the reaction ${B_x}$ by the principle of angular impulse in $x$-direction on wheel $B$ using the following relation.
\[{I_r}{\omega _1} + \int\limits_0^t {\left( {{B_x} \times r} \right)dt} = {I_r}\left( {\frac{{{v_2}}}{r}} \right) + {m_r}\left( {{v_2}r} \right)\]
On integrating the above equation we get,
\[{I_r}{\omega _1} + \left( {{B_x} \times r} \right)t = {I_r}\left( {\frac{{{v_2}}}{r}} \right) + {m_r}\left( {{v_2}r} \right)\]
On substituting the known values in the above equation we get,
\[\begin{array}{c} \left[ \begin{array}{l} \left( {72.78\,{\rm{lb}} \cdot {\rm{ft}} \cdot {{\rm{s}}^2}} \right)\left( 0 \right)\\ + \left( {{B_x} \times \left( {1.5\,{\rm{ft}}} \right)} \right)\left( {10\,{\rm{s}}} \right) \end{array} \right] = \left[ \begin{array}{l} \left( {72.78\,{\rm{lb}} \cdot {\rm{ft}} \cdot {{\rm{s}}^{\rm{2}}}} \right)\left( {\frac{{{v_2}}}{{\left( {1.5\,{\rm{ft}}} \right)}}} \right) + \\ \left( {46.58\,{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}{\rm{/ft}}} \right)\left( {{v_2}} \right)\left( {1.5\,{\rm{ft}}} \right) \end{array} \right]\\ \left[ {\left( {{B_x} \times \left( {15\,{\rm{ft}} \cdot {\rm{s}}} \right)} \right)} \right] = \left( {118.39\,{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}} \right){v_2}\\ {B_x} = \left( {7.893\,{\rm{lb}} \cdot {\rm{s/ft}}} \right){v_2}\,......\left( 2 \right) \end{array}\]
Step 7 The free body diagram of the drum roller frame can be drawn as shown below.
Step 8 Find the mass of the frame using the following relation.
\[{m_d} = \frac{{{w_d}}}{g}\]
On substituting the known values in the above equation we get,
\[\begin{array}{c} {m_d} = \frac{{4000\,{\rm{lb}}}}{{32.2\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}}}\\ = 124.22\,{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}{\rm{/ft}} \end{array}\]
Step 9 Find the value of velocity ${v_2}$ after $10\,{\rm{s}}$ of start using the following relation.
\[{m_d}{v_1} + \int\limits_0^t {\left( {{A_x} - {B_x}} \right)dt} = {m_d}{v_2}\]
On integrating the above equation we get,
\[\begin{array}{c} {m_d}{v_1} + \left( {{A_x} - {B_x}} \right)\left[ t \right]_0^t = {m_d}{v_2}\\ {m_d}{v_1} + \left( {{A_x} - {B_x}} \right)\left[ {t - 0} \right] = {m_d}{v_2}\\ \left( {{A_x} - {B_x}} \right)t = {m_d}\left( {{v_2} - {v_1}} \right) \end{array}\]
On substituting the known values of equation (2) and equation (3) in the above equation we get,
\[\begin{array}{c} \left( {200\,{\rm{lb}} - \left( {7.893\,{\rm{lb}} \cdot {\rm{s/ft}}} \right){v_2} - \left( {7.893\,{\rm{lb}} \cdot {\rm{s/ft}}} \right){v_2}} \right)t = {m_d}\left( {{v_2} - {v_1}} \right)\\ \left( {200\,{\rm{lb}} - \left( {15.786\,{\rm{lb}} \cdot {\rm{s/ft}}} \right){v_2}} \right)t = {m_d}\left( {{v_2} - {v_1}} \right) \end{array}\]
On substituting the known values in the above equation we get,
\[\begin{array}{c} \left( {200\,{\rm{lb}} - \left( {15.786\,{\rm{lb}} \cdot {\rm{s/ft}}} \right){v_2}} \right)\left( {10\,{\rm{s}}} \right) = \left( {124.22\,{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}{\rm{/ft}}} \right)\left( {{v_2} - 0} \right)\\ {v_2} = 7.09\,{\rm{ft/s}} \end{array}\]
