Question

Two masses, Ma= 35.0kg and Mb = 40.0 kg, are connected by a rope that hangs over a pulley (as in the figure ). The pulley is a uniform cylinder of radius 0.381m and mass 3.4kg . Initially Ma is on the ground and Mb rests 2.3m above the ground.
If the system is released, use conservation of energy to determine the speed of just before it strikes the ground. Assume the pulley bearing is frictionless.

Answer 1

Concepts and reason

The concepts required to solve this problem are moment of inertia, potential energy, kinetic energy, and conservation of energy.

Initially, calculate the rotational kinetic energy of the pulley by using the concept of moment of inertia, angular velocity and the expression for the rotational kinetic energy.

Finally, calculate the speed of the mass B by using the concept of the law of conservation of energy.

Fundamentals

Moment of inertia is the product of the mass of the rotating body and its radius of gyration.

The expression for the moment of inertia about its rotational axis is as follows:

$I = \frac{1}{2}m{k^2}$

Here, $m$ is the mass of the body and $k$ is the radius of the gyration about its axis.

The energy stored within the body by its position is known as potential energy.

The expression for the potential energy of the body is as follows:

$U = mgh$

Here, $m$ is the mass of the body, $h$ is the height of the body and $g$ is the acceleration due to gravity.

The energy of experienced by the body due its motion is known as kinetic energy.

The expression for the kinetic energy of the body is as follows:

$K = \frac{1}{2}m{v^2}$

Here, $v$ is the velocity of the body.

The expression for the rotational kinetic energy of the body is as follows:

${K_R} = \frac{1}{2}I{\omega ^2}$

Here, $I$ is the moment of inertia and $\omega$ is the angular velocity.

According to the law of conservation of energy, the energy can neither be created nor be destroyed but it can change from one form to other form. That is, the initial total energy is equal to final total energy.

${E_i} = {E_f}$

Here, ${E_i}$ is the total initial energy and ${E_f}$ is the final energy.

The expression for the rotational kinetic energy of the pulley is as follows:

${K_R} = \frac{1}{2}I{\omega ^2}$

Here, I is the moment of inertia of the cylinder about its central axis and the $\omega$ is the angular velocity of the cylinder.

Substitute $\frac{v}{{{R_0}}}$ for $\omega$ and $\frac{1}{2}{m_P}{R_0}^2$ for $I$ in the equation ${K_R} = \frac{1}{2}I{\omega ^2}$ .

$\begin{array}{c}\\{K_R} = \frac{1}{2}\left( {\frac{1}{2}{m_P}R_0^2} \right){\left( {\frac{v}{{{R_0}}}} \right)^2}\\\\{K_R} = \frac{1}{4}{m_P}{v^2}\\\end{array}$

According to conservation of energy, the initial energy of the system is equal to final energy of the system.

That is, initial energy of the system is equal to the final energy of the system.

$\begin{array}{c}\\{E_i} = {E_f}\\\\\frac{1}{2}{m_A}{v^2} + \frac{1}{2}{m_A}{v^2} + \frac{1}{4}{m_p}{v^2} + {m_A}gh = {m_B}gh\\\\\left( {{m_B} - {m_A}} \right)gh = \frac{1}{2}\left( {{m_A} + {m_B} + \frac{{{m_P}}}{2}} \right){v^2}\\\\{v^2} = \frac{{2\left( {{m_B} - {m_A}} \right)gh}}{{\left( {{m_A} + {m_B} + \frac{{{m_P}}}{2}} \right)}}\\\end{array}$

Substitute $40.0\,{\rm{kg}}$ for ${m_B}$ , $35.0\,{\rm{kg}}$ for ${m_A}$ , $9.81\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ , $2.3\,{\rm{m}}$ for $h$ , and $3.4\;{\rm{kg}}$ for ${m_P}$ in the equation ${v^2} = \frac{{2\left( {{m_B} - {m_A}} \right)gh}}{{\left( {{m_A} + {m_B} + \frac{{{m_P}}}{2}} \right)}}$ and solve for v.

$\begin{array}{c}\\{v^2} = \frac{{2\left( {40.0\,{\rm{kg}} - 35.0\;{\rm{kg}}} \right)\left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right)\left( {2.3\,{\rm{m}}} \right)}}{{\left( {40.0\;{\rm{kg}} + 35.0\;{\rm{kg}} + \frac{{3.4\;{\rm{kg}}}}{2}} \right)}}\\\\{v^2} = 2.94\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}\\\\v = \sqrt {2.94\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\\\ = 1.72\,{\rm{m/s}}\\\end{array}$

Ans:

The speed of mass $B$ just before striking the ground is $1.72\,{\rm{m/s}}$