Answer to Question #201835 in Mechanics | Relativity for fexypotato

Question

Consider a “round” rigid body with moment of inertia I = BMR², where M is the body’s mass, R is the body’s radius, and B 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.

A. Determine the angular frequency and the period for small oscillations of the round rigid body. Express your answer in terms of B.

B. Among the four “round” rigid bodies shown at the table, for the same masses and radii, which among them will have the most number of cycles per second.

Answer 1

A.

$Ma=M\ddot{x}=-kx-F_{fr}$

$I\varepsilon=F_{fr}R$

For rolling without slipping:

$a=\varepsilon R$

Then:

$F_{fr}=Ia/R^2$

$Ma=-kx-Ia/R^2$

$a(M+I/R^2)+kx=0$

$a+\frac{kR^2}{I+MR^2}x=0$

So, we get differential equation:

$\ddot{x}+\frac{kR^2}{I+MR^2}x=0$

Solve it:

$m^2+\frac{kR^2}{I+MR^2}=0$

Let:

$\frac{kR^2}{I+MR^2}=\omega^2$

Then:

$m=\pm i\omega$

$x(t)=c_1cos\omega t+c_2sin\omega t$

The angular frequency:

$\omega=R\sqrt{\frac{k}{I+MR^2}}=R\sqrt{\frac{k}{\beta MR^2+MR^2}}=\sqrt{\frac{k}{M(\beta +1)}}$

The period for small oscillations:

$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{M(\beta+1)}{k}}$

B.

The most number of cycles per second is for maximal angular frequency. So, have to get minimal

value of $\beta$ from the table:

$\beta=2/5$ is for uniform rigid sphere.