Question

A uniform, solid disk with mass m and radius R is pivoted about a horizontal axis through its center. A small object of the same mass m is glued to the rim of the disk. If the disk is released from rest with small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

Answer 1

Answer:

$\omega=\sqrt{\frac{4g}{3R}}$

Explanation:

According to the law of the conservation of energy, the gravitational potential energy of the physical system will be converted into rotational kinetic energy. So, we have:

$U=K_{rot}\\mgR=\frac{I_s\omega^2}{2}$

$I_s$ Is the moment of inertia of the system, that is the sum of the moments of inertia of the disk and the small object.

$mgR=\frac{(I_d+I_o)\omega^2}{2}\\mgR=\frac{(\frac{mR^2}{2}+mR^2)\omega^2}{2}\\mgR=\frac{(\frac{3}{2}mR^2)\omega^2}{2}\\gR=\frac{3}{4}R^2 \omega^2\\\omega^2=\frac{4g}{3R}\\\omega=\sqrt{\frac{4g}{3R}}$

Answer 2

The angular speed when the small object is directly below the axis is $\sqrt{\dfrac{4g}{3R}}$.

What is rotation?

When a body rotates about an axis the rotation motion is observed. The radial motion is called rotational motion or simply rotation.

By the law of conservation of energy, we have

$\rm mgR = \dfrac{1}{2}I_s \omega ^2$

$I_s$ is the moment of inertia of the system, that is the sum of the moments of inertia of the disk and the small object.

$\rm mgR = \dfrac{1}{2} (I_d+I_o) \omega ^2 \\\\mgR = \dfrac{1}{2} (\dfrac{mR^2}{2} + mR^2) \omega ^2\\\\mgR = \dfrac{3}{4} mR^2 \omega ^2\\\\\omega ^2 = \dfrac{4g}{3R}\\\\\omega = \sqrt{\dfrac{4g}{3R}}$

The angular speed when the small object is directly below the axis is $\sqrt{\dfrac{4g}{3R}}$.

More about the rotation link is given below.

https://brainly.com/question/1571997