Solution to problem 19.2.4 from the collection of Kepe O.E.

19.2.4 A variable force F = 9t is applied to rack 2 with mass m = 2.5 kg². It is necessary to find the angular acceleration of gear 1 at time t = 1 s, if the radius of the gear is r = 0.4 m, and the moment of inertia relative to the axis of rotation is I₁ = 2 kg • m². To solve the problem, you can use the formula for calculating the moment of force: M = F * r, where M is the moment of force, F is force, r is radius. The angular acceleration of the gear can be found using the formula: α = M / I₁, where α is angular acceleration, I₁ – moment of inertia relative to the axis of rotation. Substituting the known values, we get: M = 9 * 1² * 0.4 = 3.6 N * m, α = 3.6 / 2 = 1.8 rad / s². Thus, the angular acceleration of gear 1 at time t = 1 s is 1.8 rad/s².

Solution to problem 19.2.4 from the collection of Kepe O..

We present to your attention a unique digital product - the solution to problem 19.2.4 from the collection of Kepe O.. This product will help you understand the topic of mechanics and learn how to solve similar problems.

In this solution, we used the formula to calculate the moment of force and the formula to find the angular acceleration of the gear. We did all the calculations and got the answer, which is 1.8 rad/s².

This solution will be useful for both students and teachers, as well as for anyone interested in mechanics and physics. It is available in digital format, making it convenient to use on your computer or mobile device.

Don't miss the opportunity to purchase this unique solution and improve your mechanical knowledge!

We present to your attention a unique digital product - the solution to problem 19.2.4 from the collection of Kepe O.?. This product will help you understand the topic of mechanics and learn how to solve similar problems.

In this problem, it is necessary to find the angular acceleration of gear 1 at time t = 1 s when a variable force F = 9t2 is applied to a rack of mass m = 2.5 kg. The radius of the gear is r = 0.4 m, and the moment of inertia about the axis of rotation I1 = 2 kg • m2.

To solve the problem, you can use the formula for calculating the moment of force: M = F * r, where M is the moment of force, F is force, r is radius. The angular acceleration of the gear can be found using the formula: α = M / I1, where α is the angular acceleration, I1 is the moment of inertia about the axis of rotation.

Substituting the known values, we get: M = 9 * 1^2 * 0.4 = 3.6 N * m, α = 3.6 / 2 = 1.8 rad / s2. Thus, the angular acceleration of gear 1 at time t = 1 s is 1.8 rad / s2.

This solution will be useful for both students and teachers, as well as for anyone interested in mechanics and physics. It is available in digital format, making it convenient to use on your computer or mobile device. Don't miss the opportunity to purchase this unique solution and improve your mechanical knowledge!

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Problem 19.2.4 from the collection of Kepe O.?. is formulated as follows:

Given a gear rack 2 with a mass m = 2.5 kg, to which a variable force F = 9t2 is applied. It is required to determine the angular acceleration of gear 1 at time t = 1 s, if the radius of the gear is r = 0.4 m, and the moment of inertia relative to the axis of rotation I1 = 2 kg • m2.

To solve this problem it is necessary to use the equation of dynamics of rotational motion:

I1 * α = M,

where I1 is the moment of inertia of gear 1 relative to the axis of rotation, α is the angular acceleration of gear 1, M is the moment of force acting on gear 1.

The moment of force acting on gear 1 can be determined by the law of conservation of energy:

ΔE = ΔK + ΔП = 0,

where ΔE is the change in the total mechanical energy of the system, ΔK is the change in the kinetic energy of gear 1, ΔП is the change in the potential energy of rack 2 and gear 1.

The change in kinetic energy of gear 1 can be expressed as follows:

ΔK = K2 - K1 = (I1 * ω1^2) / 2 - 0,

where K1 is the initial kinetic energy of gear 1 (it is zero, since the gear is at rest), K2 is the final kinetic energy of gear 1, ω1 is the angular velocity of gear 1.

The change in potential energy of rack 2 and gear 1 can be expressed as follows:

ΔП = П2 - П1 = (m * g * h2) - (m * g * h1) = - m * g * (h1 - h2),

where P1 is the initial potential energy of rack 2 and gear 1 (it is zero, since the rack and gear are at rest), P2 is the final potential energy of rack 2 and gear 1, g is the acceleration of gravity, h1 is the height of the initial position of rack 2 and gear 1 (it is equal to zero), h2 is the height of the final position of rack 2 and gear 1.

The height of the final position of rack 2 and gear 1 can be found, knowing that the movement of the center of mass of rack 2 and gear 1 during time t is equal to:

Δh = (F * r * t^2) / (m * g).

Then the final potential energy of the system will be equal to:

П2 = m * g * h2 = m * g * (h1 + Δh) = m * g * (F * r * t^2) / (m * g) = F * r * t^2.

Thus, the change in the total mechanical energy of the system will be equal to:

ΔE = ΔK + ΔП = (I1 * ω1^2) / 2 - F * r * t^2.

Since ΔE is zero, we can write:

(I1 * ω1^2) / 2 = F * r* t^2.

Expressing the angular acceleration α from this equation, we obtain:

α = M / I1 = (F * r) / I1 = 9t^2 * r / I1.

Substituting specific values ​​from the problem conditions (t = 1 s, r = 0.4 m, I1 = 2 kg • m2), we obtain:

α = 9 * 1^2 * 0.4 / 2 = 1.8 rad/s2.

Thus, the angular acceleration of gear 1 at time t = 1 s is equal to 1.8 rad/s2, which does not correspond to the answer in the problem condition (1.5). It is possible that the problem statement contains an incorrect answer or an error was made in the calculation.

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