Rotational and Circular Motion

Therefore, natural frequency of vibration of the system is, $$\int_n = \frac{\omega_n}{2 \pi} = \frac{1}{2 \pi} \sqrt{\frac{2k}{3m}}$$

$\text{Now we have to find cutting force $(F_c)$ and frictional force $(f_t)$from merchant's theory,}\\ \hspace{1cm}2\phi+\beta-\alpha=90^0 \\ \hspace{3cm}\beta=90^0+\alpha-2\phi \\ \hspace{3cm}=90^0+7-2\times28=41^0$

From the given data the component of force at joint A along $AO_2$ is necessary to find the joint reaction at $O_2$. So, it is not possible to find the magnitude of the joint reaction at $O_2$

The pitch circle is defined as an imaginary circle that, by pure rolling action, would give the same motion as the actual gear. It is a fundamental concept in gear geometry.

It slips when.
$\mu mg = m{\omega ^2}r$
$ \Rightarrow \mu g = {\omega ^2}r$
$ \Rightarrow \mu g = {\omega ^2}40 - \left( 1 \right)$
$Now,\,\omega  = 2\omega $
$ \Rightarrow \mu g = {\left( {2\omega } \right)^2}r \Rightarrow 4{\omega ^2}r - \left( 2 \right)$
$\therefore {\omega ^2} \times 40 = 4{\omega ^2}r$
$ \Rightarrow r = 10cm$

For a solid cylinder to roll without slipping on an incline, the friction force f = (1/3)mg*sin(theta). The condition f <= mu*N = mu*mg*cos(theta) leads to mu >= (1/3)tan(theta).

An elastic metal rod will change its length when it is pulled along its length by a force acting at one end or rotates about an axis at one end. Since, deforming force exceeds the elastic limit.

Uniform circular motion is a periodic motion but not simple harmonic

The correct option is (a)