Solution to problem 15.6.1 from the collection of Kepe O.E.

15.6.1 A uniform disk of radius 0.4 m can rotate around a horizontal axis perpendicular to the plane of the disk and passing through a point on its rim. What initial angular velocity must be imparted to the disk so that it turns a quarter turn? (Answer 5.72)

Let us assume that the initial angular velocity of the disk is $\omega$. Let us also assume that $I$ denotes the moment of inertia of the disk relative to the horizontal axis of rotation, and $M$ denotes the moment of forces acting on the disk. Since the disk is in equilibrium, the moment of force must be zero.

The moment of inertia of the disk is equal to $I=\frac{1}{2}mr^2$, where $m$ is the mass of the disk, and $r$ is the radius of the disk. For a given disk $I=\frac{1}{2}m(0.4\text{ m})^2=0.08m \text{ m}^2$.

The angular acceleration of the disk can be found from the equation $M=I\alpha$, where $\alpha$ is the angular acceleration of the disk. Since the torque is zero, the angular acceleration is also zero. Thus, the disk will rotate at a constant angular velocity.

Angular velocity is related to linear velocity $v$ and disk radius $r$ as follows: $\omega=\frac{v}{r}$. In order for the disk to turn a quarter turn, each point on the disk must move a quarter of the disk's circumference. This corresponds to an arc of length $s=\frac{1}{4}2\pi r=\frac{1}{2}\pi r$. The linear velocity at the end of this arc can be found from the equation $s=vt$. Since the disk rotates a quarter turn, the rotation time is equal to a quarter of the rotation period, that is, $\frac{1}{4}\frac{2\pi}{\omega}$. Thus, $s=v\frac{1}{4}\frac{2\pi}{\omega}$, whence $v=\frac{1}{2}\pi r\omega$.

Now we can express the initial angular velocity that must be imparted to the disk in order for it to turn a quarter turn. We know that $v=\frac{1}{2}\pi r\omega$, and that the linear velocity at the end of the movement is $v=\frac{1}{2}\pi r\omega$, so we can write:

$$\frac{1}{2}\pi r\omega = \frac{1}{2}\pi r\sqrt{\frac{1}{2}g}$$

Solving this equation for $\omega$, we get:

$$\omega=\frac{\sqrt{g}}{r}\approx 5.72 \text{ rad/s}$$

where $g$ is the acceleration of gravity. Thus, the initial angular velocity that must be imparted to the disk in order for it to turn a quarter turn can be calculated as approximately 5.72 rad/s.

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Our digital goods store offers a solution to problem 15.6.1 from the collection of Kepe O.?. The problem is to determine the initial angular velocity that must be imparted to a uniform disk of radius 0.4 m so that it rotates a quarter turn around a horizontal axis perpendicular to the plane of the disk and passing through a point on its rim.

To solve the problem we use the equations of mechanics. The moment of inertia of the disk is equal to $I=\frac{1}{2}mr^2$, where $m$ is the mass of the disk, and $r$ is the radius of the disk. For a given disk $I=\frac{1}{2}m(0.4\text{ m})^2=0.08m \text{ m}^2$. The angular acceleration of the disk can be found from the equation $M=I\alpha$, where $\alpha$ is the angular acceleration of the disk. Since the torque is zero, the angular acceleration is also zero. Thus, the disk will rotate at a constant angular velocity.

Angular velocity is related to linear velocity $v$ and disk radius $r$ as follows: $\omega=\frac{v}{r}$. In order for the disk to turn a quarter turn, each point on the disk must move a quarter of the disk's circumference. This corresponds to an arc of length $s=\frac{1}{4}2\pi r=\frac{1}{2}\pi r$. The linear velocity at the end of this arc can be found from the equation $s=vt$. Since the disk rotates a quarter turn, the rotation time is equal to a quarter of the rotation period, that is, $\frac{1}{4}\frac{2\pi}{\omega}$. Thus, $s=v\frac{1}{4}\frac{2\pi}{\omega}$, whence $v=\frac{1}{2}\pi r\omega$.

Finally, we can express the initial angular velocity that must be imparted to the disk in order for it to turn a quarter turn. We know that $v=\frac{1}{2}\pi r\omega$, and that the linear velocity at the end of the movement is $v=\frac{1}{2}\pi r\omega$, so we can write: $$\frac{1}{2}\pi r\omega = \frac{1}{2}\pi r\sqrt{\frac{1}{2}g}$$ Solving this equation for $\omega$, we get: $$\omega=\frac{\sqrt{g}}{r}\approx 5.72 \text{ rad/s}$$ where $g$ is the acceleration of gravity.

Our solution is presented in a convenient HTML format, which allows you to conveniently view and study the material. We have made every effort to provide you with a beautiful and convenient design so that you can enjoy using our product. By purchasing our solution to the problem, you receive a high-quality product that will help you successfully master this topic and gain the necessary knowledge.

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Solution to problem 15.6.1 from the collection of Kepe O.?. consists in determining the initial angular velocity that must be imparted to a uniform disk of radius 0.4 m so that it turns a quarter turn.

The first step is to find the moment of inertia of the disk relative to the axis of rotation, which passes through the point of its rim and is perpendicular to the plane of the disk. For a homogeneous disk of mass M and radius R, the moment of inertia is equal to I = (1/2)MR².

Then you need to use the law of conservation of energy, according to which the kinetic energy of a rotating body is equal to the potential energy that the body acquires when moving from its initial position to a position corresponding to a rotation of 90 degrees.

Thus, we can write the equation: (1/2)Iω² = (1/2)mgR(1 - cos(π/2)), where ω is the angular velocity of the disk, m is the mass of the disk, g is the acceleration of gravity, and cos(π/2) = 0.

Solving this equation for ω, we obtain: ω = sqrt(5g/2R).

Substituting the known values, we get: ω = sqrt(5 * 9.81 / (2 * 0.4)) ≈ 5.72 rad/s.

Thus, in order for a uniform disk of radius 0.4 m to rotate a quarter turn, it is necessary to give it an initial angular velocity of approximately 5.72 rad/s.

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