A pendulum consists of a light rigid rod of length \(250 \mathrm{~mm}\), with two identical uniform solid spheres of radius \(50 \mathrm{~mm}\) attached one on either side of its lower end. Find the period of small oscillations (a) perpendicular to the line of centres, and (b) along it.

First, we need to find the moments of inertia for both situations. We will denote the length of the pendulum rod as L, the radius of the spheres as R, and the mass of each sphere as m. For situation (a), oscillation perpendicular to the line of centers: We can consider the moment of inertia of the rod to be negligible because its mass is much less than the mass of each sphere. So, we will only calculate the moment of inertia of the two spheres. In this situation, each sphere rotates about an axis parallel to the rod passing through its center. The moment of inertia for one sphere (I_sphere) can be calculated as: \(I_{sphere} = \frac{2}{5}mR^2\) Since both spheres are identical, the total moment of inertia (I_perp) is the sum of the individual moments: \(I_{perp} = 2 \times \frac{2}{5}mR^2 = \frac{4}{5}mR^2\) For situation (b), oscillation along the line of centers: In this case, the axis of rotation is the same as the line of centers, passing through the end of the rod and the center of one sphere. The moment of inertia (I_along) can be calculated using the parallel axis theorem: \(I_{along} = I_{cm} + md^2\) In this situation, d = L + R for one sphere, and d = L - R for the other sphere. Let´s calculate the moment of inertia for each sphere separately, and then add them together: One sphere has moment of inertia: \(I_1 = \frac{2}{5}mR^2 + m(L + R)^2\) The other sphere has moment of inertia: \(I_2 = \frac{2}{5}mR^2 + m(L - R)^2\) \(I_{along} = I_1 + I_2\)

Now, we will use the formula for the period of oscillation for a physical pendulum: \(T = 2\pi \sqrt{\frac{I}{mgr}}\) where T is the period, I is the moment of inertia, m is the mass of the pendulum, g is the acceleration due to gravity, and r is the distance from the pivot point to the center of mass. For situation (a), the distance r is equal to the length of the rod (L), and for situation (b), the distance r is L + R for one sphere, and L - R for the other sphere.

Now we can plug the values of I and r into the formula for each situation: (a) For oscillation perpendicular to the line of centers, the period T_perp is: \(T_{perp} = 2\pi \sqrt{\frac{I_{perp}}{mgr}} = 2\pi \sqrt{\frac{\frac{4}{5}mR^2}{mgL}}\) Simplify: \(T_{perp} = 2\pi \sqrt{\frac{4R^2}{5L}}\) Using the given values for L and R: \(T_{perp} = 2\pi \sqrt{\frac{4(50 \mathrm{~mm})^2}{5(250 \mathrm{~mm})}}\) \(T_{perp} \approx 1.14\ s\) (b) For oscillation along the line of centers, we calculate the period T_along: \(T_{along} = 2\pi \sqrt{\frac{I_{along}}{mgr}} = 2\pi \sqrt{\frac{I_1 + I_2}{mg(L+R) + mg(L-R)}}\) Simplify and plug in the values for L and R: \(T_{along} = 2\pi \sqrt{\frac{(\frac{2}{5}mR^2 + m(L + R)^2) + (\frac{2}{5}mR^2 + m(L - R)^2)}{mg(2L)}}\) \(T_{along} = 2\pi \sqrt{\frac{(\frac{2}{5}(50 \mathrm{~mm})^2 + (250 \mathrm{~mm} + 50 \mathrm{~mm})^2) + (\frac{2}{5}(50 \mathrm{~mm})^2 + (250 \mathrm{~mm} - 50 \mathrm{~mm})^2)}{(2)(250 \mathrm{~mm})}}\) \(T_{along} \approx 2.16\ s\) Thus, the period of small oscillations (a) perpendicular to the line of centers is approximately 1.14 seconds, and (b) along the line of centers is approximately 2.16 seconds.