91影视

Given in the question,

The length of the rod is\(L = 80.0\,\,{\rm{cm}}\)

The mass of the rod is\({m_l} = \,0.120\,\,{\rm{kg}}\)

The mass of the first sphere,\({m_1} = \,0.0200\,\,{\rm{kg}}\)

The mass of the second sphere, \({m_2} = \,0.0500\,\,{\rm{kg}}\)

Law of conservation of energy

According to the conservation of energy, the energy of a systemcan neither be created nor destroyed it can only be converted from one form of energy to another, so the total energy of the system is always constant.

\({E_F} = {E_I}\)

From the conservation of energy,

The total initial energy of the system is equal to the final total energy

Therefore, we can write

Initial potential energy +initial kinetic energy= final potential energy + final kinetic energy.

Change in potential energy = change is kinetic energy

\(\Delta U = \Delta K\)

Since the masses are attached to the rod, therefore, the change in potential energy

\(\Delta U = {m_1}g{h_1} + {m_2}g{h_2}\)

Considering the center of the rod as the origin

\({h_1} = - 0.4\,{\rm{m}}\) and \({h_1} = 0.4\,{\rm{m}}\)

Change in potential energy

\(\begin{array}{c}\Delta U = {m_1}g{h_1} + {m_2}g{h_2}\\ = \left( {\,0.0200\,\,{\rm{kg}}} \right)\left( {9.8\,\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( { - 0.4\,{\rm{m}}} \right) + \left( {0.0500\,\,{\rm{kg}}} \right)\left( {9.8\,\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {0.4\,{\rm{m}}} \right)\\ = 0.118\,\,{\rm{J}}\end{array}\)

Since the system is initially at the rest so the change in kinetic energy is equal to the rotational kinetic energy of the system

\(\begin{array}{c}\Delta K = {K_{rotational}}\\ = \frac{1}{2}{I_{total}}\omega \end{array}\)

Where I is the moment of inertia and \(\omega \) is the angular velocity

Total moment of inertia = Inertia of rod + Inertial of both the spheres

\({I_{total}} = {I_l} + {I_{s1 + s2}}\)

Moment of inertia of the sphere

\(\begin{array}{c}{I_{s1 + s2}} = {m_1}{r^2} + {m_2}{r^3}\\ = \left( {\,0.0200\,\,{\rm{kg}}} \right){\left( {0.4\,{\rm{m}}} \right)^2} + \left( {0.0500\,\,{\rm{kg}}} \right){\left( {0.4\,{\rm{m}}} \right)^2}\\ = 0.0112\,\,{\rm{kg}}.{{\rm{m}}^2}\end{array}\)

The moment of inertia of a rod can be given as

\(I = \frac{1}{{12}}m{L^2}\)

Where m is the mass of the rod and L is the length of the rod

Substituting the values

\(\begin{array}{c}{I_l} = \frac{1}{{12}}\left( {\,0.120\,\,{\rm{kg}}} \right){\left( {0.80\,{\rm{m}}} \right)^2}\\ = 0.0064\,\,{\rm{kg}}.\,{{\rm{m}}^2}\end{array}\)

So, the total moment of inertia

\(\begin{array}{c}{I_{total}} = {I_l} + {I_{s1 + s2}}\\ = 0.0064\,\,{\rm{kg}}.\,{{\rm{m}}^2} + 0.0112\,\,{\rm{kg}}.{{\rm{m}}^2}\\ = 0.0176\,\,{\rm{kg}}.{{\rm{m}}^2}\end{array}\)

Now from the conservation of energy

\(\begin{array}{l}\Delta U = \Delta K\\\Delta U = \frac{1}{2}{I_{total}}\omega \end{array}\)

Now substituting the values into the equation

\(\begin{array}{l}0.118\,\,{\rm{J}} = \frac{1}{2}\left( {0.0176\,\,{\rm{kg}}.{{\rm{m}}^2}} \right){\omega ^2}\\\omega = 3.66\,{{\,{\rm{rad}}} \mathord{\left/{\vphantom {{\,{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{array}\)

The linear speed of the sphere can be given as

\(1.46{{\,{\rm{m}}} \mathord{\left/ {\vphantom {{\,{\rm{m}}} {\rm{s}}}} \right.} {\rm{s}}}\)