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The moment of inertia is a key concept in understanding how objects rotate. It can be thought of as the rotational equivalent of mass in linear motion. Just like mass determines how easy or hard it is to change the speed of an object moving in a straight line, the moment of inertia determines how easy or hard it is to change the rotation of an object. For different shapes, the moment of inertia depends on both the mass of the object and how that mass is distributed with respect to the axis of rotation. In the case of a solid sphere, like the one in our exercise, the moment of inertia is given by the formula: \[ I = \frac{2}{5} m r^2 \]Here, \(I\) stands for the moment of inertia, \(m\) is the mass, and \(r\) is the radius of the sphere. This formula shows that the larger the radius or the mass, the bigger the moment of inertia, which means more energy is required to spin the sphere at a certain speed.

Rotational energy, similar to kinetic energy in linear motion, refers to the energy possessed by an object due to its rotation. It can be calculated with the formula:\[ E = \frac{1}{2} I \omega^2 \] where \(E\) is the rotational energy, \(I\) is the moment of inertia calculated earlier, and \(\omega\) is the angular velocity. Angular velocity \(\omega\) is the speed of rotation and is usually measured in radians per second; in our exercise, it is given by \(\omega = 2\pi n\), where \(n\) represents rotations per second. This tells us that rotational energy increases with both a higher moment of inertia and a higher angular velocity. For our metal sphere, this rotational energy initially helps us determine how much energy can be potentially converted to heat when the sphere is stopped.

Specific heat is a property of a material that indicates how much heat energy is required to change the temperature of a given mass of that material by one degree. It’s usually given in units of Joules per kilogram per degree Celsius (J/kg°C).For the metal sphere in our exercise, the specific heat, represented by \(S\), tells us how much energy is needed to increase the sphere's temperature. When some energy of rotation is converted to heat, the specific heat allows us to link the energy change to the resulting temperature change:\[ E_{heat} = m S \Delta T \]Here, \(E_{heat}\) is the energy converted to heat, \(m\) is the mass, \(S\) is the specific heat, and \(\Delta T\) is the change in temperature. By rearranging this equation, we can find the temperature rise, \(\Delta T\), given the energy input and specific heat.

When a rotating object like our sphere stops, part of its energy gets converted into heat. Specifically, in our exercise, 50% of the rotational energy translates into thermal energy, causing a temperature rise.To find out how much the temperature increases, we use the relationship between heat energy and temperature via specific heat. Starting from the formula:\[ \Delta T = \frac{E_{heat}}{mS} \]we substitute the expression for the heat energy from step 5 in the solution:\[ E_{heat} = \frac{1}{5} m r^2 \pi^2 n^2 \]With this substitution, we can derive the expression for the temperature rise:\[ \Delta T = \frac{\pi^2 n^2 r^2}{5 S} \]Thus, which aligns with the choice (a). This tells us how efficiently the rotational energy is being transformed into heat, raising the temperature of the sphere. The links between rotational motion and thermal dynamics become apparent here, illustrating intricate physics concepts in a tangible problem.