Question

Two spheres A and B of the same mass and the same radius are placed on a rough horizontal surface. A is a uniform hollow sphere and B is a uniform solid sphere. Both A and B can roll without sliding on the floor. They are also tied centrally to a light spring of spring constant $ k $ . A and B are released when the extension in the spring is $ A $ . Find the time period of oscillations and the individual amplitudes of A and B.

Answer 1

Hint: The extension in the string is equal to the amplitude of both the spheres. Friction acting at the surface sphere interface will create a net torque and allow the spheres to roll. The angular momentum of any such system is conserved.

Formula used: In this solution we will be using the following formulae;
 $ Fr = I\alpha $ where $ F $ is the force acting on an object, $ r $ is distance from a reference point, $ I $ is moment of inertia of a body and $ \alpha $ is angular acceleration.
The quantity $ Fr $ is called torque $ F = ke $ where $ F $ is the force exerted by a spring, $ k $ is the spring constant, and $ e $ is the extension of the spring.
 $ \dfrac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = 0 $ where $ \dfrac{{{d^2}x}}{{d{t^2}}} $ denote the differential form of acceleration, $ \omega $ is the angular frequency.
 $ \omega = \dfrac{{2\pi }}{T} $ where $ T $ is the time period of oscillation.
 $ I = \dfrac{5}{3}m{R^2} $ where $ I $ is the moment of inertia of a hollow spherical object about an axis on its surface, $ m $ and $ R $ are the mass and radius respectively.
 $ I = \dfrac{7}{5}m{R^2} $ where $ I $ is the moment of inertia of a solid spherical object about an axis on its surface.

Complete Step-by-Step solution
The spheres are on a horizontal surface connected to each other through their centres. Both are allowed to oscillate, and we are to find the time period of that oscillation. To do so, we note that the force exerted by the spring on each on both the spheres will be due to the displacement of both the spheres, hence, the force exerted is
 $ F = ke = k\left( {{x_1} + {x_2}} \right) $ where $ k $ is the spring constant, and $ e $ is the extension of the spring, $ {x_1} $ is the instantaneous extension of hollow sphere and $ {x_2} $ is the instantaneous extension of solid sphere.
Now, since the spheres are not sliding but rolling, we use the torque
Hence, from $ Fr = I\alpha $ where $ F $ is the force acting on an object, $ r $ is distance from a reference point, $ I $ is moment of inertia of a body and $ \alpha $ is angular acceleration, we have
 $ - k\left( {{x_1} + {x_2}} \right)R = I\alpha $ this is negative because the direction of force and that of the extension are always in opposite directions.
For hollow sphere
Hence, torque on hollow sphere is
 $ {I_1} = \dfrac{5}{3}m{R^2} $ where $ {I_1} $ is the moment of inertia of the hollow sphere about a point on its surface, $ m $ is the mass of the sphere $ R $ is the radius of the substance. Hence,
 $ \Rightarrow - k\left( {{x_1} + {x_2}} \right) = \dfrac{5}{3}mR{\alpha _1} $
But $ a = R\alpha $ where $ a $ is the linear acceleration
Hence,
 $ - k\left( {{x_1} + {x_2}} \right) = \dfrac{5}{3}m{a_1} $
Now from conservation of angular momentum (given by $ L = I\dfrac{v}{R} $ where $ v $ means linear velocity) we have,
 $ {I_1}\dfrac{{{v_1}}}{R} = {I_2}\dfrac{{{v_2}}}{R} $ where $ {I_2} $ is the moment of inertia of inertia for the solid sphere about a point at its surface)
 $ {I_2} = \dfrac{7}{5}m{R^2} $
Hence, by inserting both moment of inertia and simplifying we have
 $ \dfrac{5}{3}{v_1} = \dfrac{7}{5}{v_2} $
 $ \Rightarrow \dfrac{5}{3}{x_1} = \dfrac{7}{5}{x_2} $ (since $ v = \dfrac{x}{t} $ )
Making $ {x_2} $ subject,
 $ {x_2} = \dfrac{{25}}{{21}}{x_1} $ from $ k\left( {{x_1} + {x_2}} \right) = \dfrac{5}{3}m{a_1} $ we have
 $ - k\left( {{x_1} + \dfrac{{25}}{{21}}{x_1}} \right) = \dfrac{5}{3}m{a_1} $
 $ \Rightarrow - \dfrac{{46}}{{21}}k{x_1} = \dfrac{5}{3}m{a_1} $
Hence, making $ a $ subject, we have
 $ {a_1} = - \dfrac{{46}}{{35}}\dfrac{k}{m}{x_1} $ ,
In differential form, we have
 $ \dfrac{{{d^2}x}}{{d{t^2}}} + \dfrac{{46}}{{35}}\dfrac{k}{m}{x_1} = 0 $
Comparing with the equation, of a SHM $ \dfrac{{{d^2}{x_1}}}{{d{t^2}}} + {\omega ^2}{x_1} = 0 $
We see that $ {\omega ^2} = \dfrac{{46}}{{35}}\dfrac{k}{m} $
Then
 $ \omega = \sqrt {\dfrac{{46}}{{35}}\dfrac{k}{m}} $
But $ T = \dfrac{{2\pi }}{\omega } $ , hence
 $ {T_1} = 2\pi \sqrt {\dfrac{{35m}}{{46k}}} $
To find the individual amplitudes, we note that the total amplitude is the sum of the individual amplitudes, i.e.
 $ {A_1} + {A_2} = A $ ,
Now, just like $ {x_1} = \dfrac{{21}}{{25}}{x_2} $ , so is $ {A_1} = \dfrac{{21}}{{25}}{A_2} $
Hence,
 $ \dfrac{{21}}{{25}}{A_2} + {A_2} = A $
By adding and making $ {A_2} $ subject, we have
 $ {A_2} = \dfrac{{25}}{{46}}A $
Then similarly for $ {A_1} $ , we get
 $ {A_1} = A - \dfrac{{25}}{{46}}A $
 $ \Rightarrow {A_1} = \dfrac{{25}}{{46}}A $ .

Note
For clarity, we do not calculate the time periods for the second sphere because the time periods will be identical. This is because they are both derived from the linear acceleration and this is the same for both spheres since the same force pulls on them and they are both the same mass. But angular acceleration is different due to difference in moment of inertia (which affects only the rotational behaviour of objects).