Question

An Atwood machineconsists of two masses,\({m_A} = {\bf{65 kg}}\) and\({m_B} = {\bf{75 kg}}\) connected by a massless inelastic cord that passes over a pulley free to rotate, Fig. 8 52. The pulley is a solid cylinder of radius\(R = {\bf{0}}{\bf{.45 m}}\) and mass 6.0 kg. (a) Determine the acceleration of each mass. (b) What % error would be made if the moment of inertia of the pulley is ignored? (Hint: The tensions\({F_{TA}}\) and\({F_{TB}}\)are not equal. We discussed the Atwood machine in Example 4–13, assuming I = 0 for the pulley.)

FIGURE 8-52 Problem 47.Atwood machine.

Short answer

1. The acceleration of mass A is \(0.686{\rm{ m/}}{{\rm{s}}^2}\) in the upward direction. The acceleration of mass B is \(0.686{\rm{ m/}}{{\rm{s}}^2}\) in the downward direction.
2. The percentage error if the moment of inertia of the pulley is ignored is \(2.14\% \).

Step-by-step solution

Identification of given data 

The given data can be listed below as:

• The mass of block A is\({m_A} = 65{\rm{ kg}}\).
• The mass of block B is\({m_B} = 75{\rm{ kg}}\).
• The radius of the pulley is\(R = 0.45{\rm{ m}}\).
• The mass of the pulley is\(M = 6.0{\rm{ kg}}\).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Understanding the motion of the Atwood machine

The Atwood machine has two masses attached to the pulley. Mass B moves downward as its mass is more than mass A.

In this problem, first, draw the free-body diagram of each mass. At equilibrium, resolve the forces acting on each mass along the horizontal and vertical axes. Then, apply Newton’s second law to evaluate the accelerations.

Analysis of forces on block B

The free-body diagram of block B can be shown as:

Here,\({F_{TB}}\)is the tension in the cord and g is the acceleration due to gravity. Mass B is moving downward with acceleration (a).

At the equilibrium condition, the forces along the vertical direction of block B can be expressed as:

\(\begin{align}\sum {F_y} &= 0\\{m_B}g - {F_{TB}} &= {m_B}a\end{align}\)

\({F_{TB}} = {m_B}g - {m_B}a\) … (i)

Here, a is the acceleration of mass B acting in the downward direction.

Analysis of forces on block A

The free-body diagram of block A can be shown as:

Here, mass A is moving upward with acceleration (a).

At the equilibrium condition, the forces along the vertical direction of block A can be expressed as:

\(\begin{align}\sum {F_y} &= 0\\{F_{TA}} - {m_A}g &= {m_A}a\end{align}\)

\({F_{TA}} = {m_A}g + {m_A}a\) … (ii)

Here, a is the acceleration of mass A acting in the upward direction.

(a) Determination of the acceleration of each mass

At the equilibrium condition, the summation of torques in the Atwood machine pulley can be expressed as:

\(\begin{align}\sum T &= I\alpha \\{F_{TB}}R - {F_{TA}}R &= I \times \frac{a}{R}\end{align}\)

\({F_{TB}} - {F_{TA}} = \frac{{Ia}}{{{R^2}}}\) … (iii)

Here, Iis the moment of inertia of the solid cylinder-shaped pulley, R is the radius of the solid cylindrical-shaped pulley, and a is the acceleration of the pulley.

Substitute the values from equations (i) and (ii) in equation (iii).

\(\begin{align}{m_B}g - {m_B}a - \left( {{m_A}g + {m_A}a} \right) &= \frac{{Ia}}{{{R^2}}}\\{m_B}g - {m_B}a - {m_A}g - {m_A}a = \frac{{Ia}}{{{R^2}}}\\g\left( {{m_B} - {m_A}} \right) &= \frac{{Ia}}{{{R^2}}} + a\left( {{m_A} + {m_B}} \right)\end{align}\)

\(g\left( {{m_B} - {m_A}} \right) = \left\{ {\frac{I}{{{R^2}}} + \left( {{m_B} + {m_A}} \right)} \right\}a\) … (iv)

This can be further solved as:

\(\begin{align}a &= \frac{{g\left( {{m_B} - {m_A}} \right)}}{{\left\{ {\frac{I}{{{R^2}}} + \left( {{m_B} + {m_A}} \right)} \right\}}}\\ &= \frac{{g\left( {{m_B} - {m_A}} \right)}}{{\left\{ {\frac{{\left( {\frac{{M{R^2}}}{2}} \right)}}{{{R^2}}} + \left( {{m_B} + {m_A}} \right)} \right\}}}\\ &= \frac{{g\left( {{m_B} - {m_A}} \right)}}{{\left\{ {\frac{M}{2} + \left( {{m_B} + {m_A}} \right)} \right\}}}\end{align}\)

Substitute the values in the above equation.

\(\begin{align}a &= \frac{{9.81{\rm{ m/}}{{\rm{s}}^2} \times \left( {75{\rm{ kg}} - 65{\rm{ kg}}} \right)}}{{\left\{ {\frac{{6.0{\rm{ kg}}}}{2} + \left( {75{\rm{ kg}} + 65{\rm{ kg}}} \right)} \right\}}}\\ &= \frac{{98.1}}{{143}}{\rm{ m/}}{{\rm{s}}^2}\\ &= 0.686{\rm{ m/}}{{\rm{s}}^2}\end{align}\)

Thus, the acceleration of mass A is \(0.686{\rm{ m/}}{{\rm{s}}^2}\) in the upward direction. The acceleration of mass B is \(0.686{\rm{ m/}}{{\rm{s}}^2}\) in the downward direction.

Determination of the acceleration of each mass if the moment of inertia of the pulley is ignored

The moment of inertia \(\left( I \right)\) of the pulley becomes zero. Therefore, from equation (iv), the equation can be written as:

\(\begin{align}g\left( {{m_B} - {m_A}} \right) &= \left\{ {\frac{0}{{{R^2}}} + \left( {{m_B} + {m_A}} \right)} \right\}{a_1}\\g\left( {{m_B} - {m_A}} \right) &= \left( {{m_B} + {m_A}} \right){a_1}\\{a_1} &= \frac{{g\left( {{m_B} - {m_A}} \right)}}{{\left( {{m_B} + {m_A}} \right)}}\end{align}\)

Here, the acceleration of the masses becomes\(\left( {{a_1}} \right)\)when the moment of inertia of the pulley is ignored.

Substitute the values in the above equation.

\(\begin{align}a &= \frac{{9.81{\rm{ m/}}{{\rm{s}}^2} \times \left( {75{\rm{ kg}} - 65{\rm{ kg}}} \right)}}{{\left( {75{\rm{ kg}} + 65{\rm{ kg}}} \right)}}\\ &= \frac{{98.1}}{{140}}{\rm{ m/}}{{\rm{s}}^2}\\ &= 0.701{\rm{ m/}}{{\rm{s}}^2}\end{align}\)

Thus, the acceleration of mass A when the pulley’s inertia is ignored is \(0.701{\rm{ m/}}{{\rm{s}}^2}\) in the upward direction. The acceleration of mass B is \(0.701{\rm{ m/}}{{\rm{s}}^2}\) when the pulley’s inertia is ignored in the downward direction.

(b) Determination of the percentage error in the acceleration for both cases if the moment of inertia of the pulley is ignored

The percentage error in both accelerations can be expressed as:

\(\% E = \frac{{{a_1} - a}}{{{a_1}}} \times 100\)

Substitute the values in the above equation.

\(\begin{align}\% E &= \frac{{0.701{\rm{ m/}}{{\rm{s}}^2} - 0.686{\rm{ m/}}{{\rm{s}}^2}}}{{0.701{\rm{ m/}}{{\rm{s}}^2}}} \times 100\% \\ &= 0.0214 \times 100\% \\ &= 2.14\% \end{align}\)

Thus, the percentage error, if the moment of inertia of the pulley ignored, is \(2.14\% \).