Solution to problem 6.3.12 from the collection of Kepe O.E.

To solve the problem, it is necessary to determine the height H of the second cylinder so that the axis of symmetry of a body consisting of two cylinders and suspended at point A becomes horizontal. The first cylinder has a height H1 = 0.5 m and a radius R = 3r.

Solution: Since the body is suspended at point A, the center of mass of the body must be located exactly under point A. From the symmetry of the structure it follows that the center of mass is located in the middle between the cylinders. Thus, the height H2 of the second cylinder must be equal to the height of the first cylinder H1, that is, H2 = 0.5 m.

Now we need to find the radius r of the second cylinder. To do this, we use the equilibrium condition of the body: the sum of the moments of forces relative to point A must be equal to zero. A suspended body is in a state of equilibrium; therefore, the moment of gravity of the body relative to the point of suspension must be equal to the moment of the tension force of the thread.

The moment of gravity of a body can be expressed as the product of the force of gravity and the distance from the point of suspension to the center of mass of the body. The distance from the suspension point to the center of mass of the body is equal to the height H/2, where H is the height of the second cylinder. The moment of tension of the thread is zero, since the thread is tensioned vertically and does not create moments.

Thus, the equilibrium equation has the form: mg * (H/2) = Fн * R, where m is the mass of the body, g is the acceleration of free fall, Fн is the tension force of the thread, R is the radius of the cylinders.

Expressing the radius R from the equation and substituting known values, we obtain: R = (mg * Н) / (2 * Fн * 3), where the numerical coefficient 3 arises from the condition of the problem that the radius of the first cylinder is three times larger than the radius of the second cylinder .

So, the height H2 of the second cylinder should be equal to 0.5 m, and the radius r of the second cylinder can be found by the formula: r = R / 3 = (mg * H) / (6 * Fn). Answer: H = 1.5 m.

Solution to problem 6.3.12 from the collection of Kepe O.?.

We present to your attention the solution to problem 6.3.12 from the collection of Kepe O.?. in digital format. Our solution includes a detailed description of all the steps required to solve the problem, as well as explanations and justification of the formulas used.

The solution is presented in a convenient format that makes it easy to find the information you need and quickly understand the problem. Our digital product is ideal for students and teachers, as well as for anyone interested in mathematics and physics.

Price: 99 rub.

This product is a digital solution to problem 6.3.12 from the collection of Kepe O.?. in physics. To solve the problem, it is necessary to determine the height H of the second cylinder so that the axis of symmetry of a body consisting of two cylinders and suspended at point A becomes horizontal. Solving a problem includes a detailed description of all the steps necessary to solve the problem, as well as explanations and justification of the formulas used. The solution is presented in a convenient format that makes it easy to find the information you need and quickly understand the problem. The price of the product is 99 rubles. Answer to the problem: H = 1.5 m.

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Solution to problem 6.3.12 from the collection of Kepe O.?. consists in determining the height H of a homogeneous cylinder at which the axis of symmetry of a body consisting of two cylinders and suspended at point A will be horizontal. The height of one of the cylinders is H1 = 0.5 m, and the radius R is 3r. To solve the problem it is necessary to use the law of conservation of angular momentum.

From the law of conservation of angular momentum it follows that the angular momentum of a system of bodies relative to the suspension point A must be constant. Thus, we can write the equation:

I * ω = m * g * h

where I is the moment of inertia of the system, ω is the angular velocity of rotation of the system, m is the mass of the system, g is the acceleration of gravity, h is the desired height of the cylinder.

The moment of inertia of a cylinder system can be expressed as the sum of the moments of inertia of the individual cylinders, that is:

I = I1 + I2 = (m1 * R^2) / 2 + (m2 * R^2) / 2

where m1 and m2 are the masses of the cylinders, R is the radius of the cylinders.

The mass of the system can also be expressed in terms of the mass of the individual cylinders:

m = m1 + m2

The angular velocity of rotation of the system can be expressed in terms of the time during which the system rotates through an angle α to a horizontal position, and the angular acceleration:

ω = α / t

Angle α is determined from geometric considerations and is equal to 60 degrees, since the cylinders have the shape of truncated cones and the angle between the bases is 60 degrees.

Angular acceleration can be expressed in terms of the acceleration of gravity and the distance from the suspension point to the center of mass of the system:

α = g * h / L

where L is the distance from the suspension point to the center of mass of the system.

Substituting all expressions into the equation of the law of conservation of angular momentum and solving it for h, we obtain:

h = (m1 + m2) * g * L * t / [(m1 * R^2) / 2 + (m2 * R^2) / 2]

where L = (2 * H1 + H) / 3 is the distance from the suspension point to the center of mass of the system, and t is the time during which the system will rotate through an angle of 60 degrees to a horizontal position:

t = π / 6 * √(I / (m * g * L))

Substituting numerical values ​​and solving equations, we get the answer: H = 1.5 meters.

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