Two blocks are connected by a light string passing over a pulley of radius 0.15 m and the moment of inertia I. The blocks move (towards the right) with an acceleration of \({\bf{1}}{\bf{.00}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) along their frictionless inclines (see Fig. 8–51). (a) Draw free-body diagrams for each of the two blocks and the pulley. (b) Determine the tensions in the two parts of the string. (c) Find the net torque acting on the pulley, and determine its moment of inertia, I.

FIGURE 8-51

Problem 46

Torque is the product of the moment of inertia and the square of the distance. For this problem, find the forces on the string using Newton’s second law and then find the torque on the pulley.

The mass of block A is \({m_{\rm{A}}} = 8.0\;{\rm{kg}}\).

The mass of block B is \({m_{\rm{B}}} = 10.0\;{\rm{kg}}\).

The acceleration of the blocks is \(a = 1.00\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

The radius of the pulley is \(r = 0.15\;{\rm{m}}\).

The angle of the left incline is \(\theta = {32^ \circ }\).

The angle of the right incline is \(\phi = {61^ \circ }\).

The moment of inertia of the pulley is I.

Part (a)

The free-body diagrams of the blocks are shown as

Part (b)

For the forces on block A along the x-axis,

\(\begin{align}{F_{{\rm{TA}}}} - {m_{\rm{A}}}g\sin \theta &= {m_{\rm{A}}}a\\{F_{{\rm{TA}}}} &= {m_{\rm{A}}}a + {m_{\rm{A}}}g\sin \theta \\{F_{{\rm{TA}}}} = {m_{\rm{A}}}\left( {a + g\sin \theta } \right)\end{align}\).

Now, substituting the values in the above equation,

\(\begin{align}{F_{{\rm{TA}}}} &= \left( {8.0\;{\rm{kg}}} \right) \times \left( {1.00\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} + \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\sin {{32}^ \circ }} \right)\\ &= 49.55\;{\rm{N}}\end{align}\).

For the forces on the block B along the x-axis,

\(\begin{align}{m_{\rm{B}}}g\sin \phi - {F_{{\rm{TB}}}} &= {m_{\rm{B}}}a\\{F_{{\rm{TB}}}} &= {m_{\rm{B}}}g\sin \phi - {m_{\rm{B}}}a\\{F_{{\rm{TB}}}} &= {m_{\rm{B}}}\left( {g\sin \phi - a} \right)\end{align}\).

Now, substituting the values in the above equation,

\(\begin{align}{F_{{\rm{TB}}}} &= \left( {10.0\;{\rm{kg}}} \right) \times \left\{ {\left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\sin {{61}^ \circ } - \left( {1.00\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)} \right\}\\ &= 75.71\;{\rm{N}}\end{align}\).

Hence, the values of \({F_{{\rm{TA}}}}\) and \({F_{{\rm{TB}}}}\) are \(49.55\;{\rm{N}}\) and \(75.71\;{\rm{N}}\), respectively.

Part (c)

The angular acceleration of the pulley is \(\alpha = \frac{a}{r}\).

Then, the net torque acting on the pulley is

\(\begin{align}\tau &= \left( {{F_{{\rm{TB}}}} - {F_{{\rm{TA}}}}} \right)r\\I\alpha &= \left( {{F_{{\rm{TB}}}} - {F_{{\rm{TA}}}}} \right)r\\I\frac{a}{r} &= \left( {{F_{{\rm{TB}}}} - {F_{{\rm{TA}}}}} \right)r\\I &= \left( {{F_{{\rm{TB}}}} - {F_{{\rm{TA}}}}} \right)\frac{{{r^2}}}{a}\end{align}\).

Now, substituting the values in the above equation,

\(\begin{align}I &= \left( {75.71\;{\rm{N}} - 49.55\;{\rm{N}}} \right) \times \frac{{{{\left( {0.15\;{\rm{m}}} \right)}^2}}}{{1.00\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}\\ &= 0.59\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{align}\).

Hence, the moment inertia of the pulley is \(0.59\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).