Answer to Question #203012 in Mechanics | Relativity for toot

Question

Consider a “round” rigid body with moment of inertia 𝑰 = 𝜷𝑴𝑹 𝟐 , where 𝑀 is the body’s mass, 𝑅 is the body’s radius, and 𝛽 is a constant depending on the type of the body. The center of the “round” rigid body is attached to a spring of force constant 𝑘, and then the body is made to roll without slipping on a rough horizontal surface. Due to the spring, it is expected the body will oscillate by rolling back and forth from its resting position. Determine the angular frequency and the period for small oscillations of the round rigid body. Express your answer in terms of 𝛽.  

Answer 1

Let $x$ be the displacement of the body at any time $t$ .

The equation of motion of the body is

$Ma=-kx-F_{fr}\ ........(1)$ ,

where $M$ = mass of the body,

$a=\ddot x$ is the linear acceleration of the body,

$k$ = force constant of the spring, and

$F_{fr}$ = force of friction.

If $a$ be the linear acceleration, the angular acceleration is $\alpha = a/R$ , where $R$ is the radius of the body.

The torque acting about the center of the body, $\tau = I\alpha$ , where $I$ is the moment of inertia of the body.

Again, the torque about the center due to the frictional force is $F_{fr}R$ .

$\therefore I\alpha = F_{fr}R\\ \Rightarrow F_{fr}=I\alpha/R$

From Eq.(1),

$Ma=-kx-I\alpha/R\\ \Rightarrow Ma=-kx-Ia/R^2$ (Since $\alpha=a/R$ )

$\Rightarrow Ma+Ia/R^2+kx=0\\ \Rightarrow (M+I/R^2)a+kx=0\\ \Rightarrow (M+I/R^2)\ddot x+kx=0\\ \Rightarrow \ddot x+\frac{k}{M+I/R^2} x=0\\ \Rightarrow \ddot x+\omega^2x=0\ ..........(2)$

where $\omega^2=\frac{k}{M+I/R^2}$

The solution of Eq.(2) is $x=c_1 \sin\omega t+c_2\cos\omega t$ , where $c_1, c_2$ are constants and can be determined from the boundary conditions.

The angular frequency of oscillation is $\omega = \sqrt{\frac{k}{M+I/R^2}}$

Substitute $I=\beta MR^2$

$\therefore \omega = \sqrt{\frac{k}{M+\beta MR^2/R^2}}\\ \Rightarrow \boxed {\omega = \sqrt{\frac{k}{M(1+\beta)}}}$

The period of oscillation is $\boxed{ T=\frac{2\pi}{\omega} = 2\pi \sqrt{ \frac{M(1+\beta)}{k} } }$