Solution to problem 16.1.10 from the collection of Kepe O.E.

16.1.10 A homogeneous rod, the mass of which is m = 2 kg and the length AB = 1 m, rotates around the Oz axis under the action of a pair of forces with a moment M1 and a moment of resistance forces M2 = 12 N m according to the law? = 3t2. It is necessary to determine the modulus of moment M1 of the applied pair of forces at time t = 1 s.

Solution: To solve this problem, we use the equation of dynamics of rotational motion: ΣM = Iα, where ΣM is the sum of the moments of forces acting on the body, I is the moment of inertia of the body, α is the angular acceleration of the body.

Considering that a uniform rod rotates around its central axis, the moment of inertia can be expressed as I = (1/12) * m * l2, where l is the length of the rod, m is its mass.

Also, taking into account that the moment of the resistance force is directed against the direction of rotation of the rod, it will have a negative value: M2 = -12 N m.

Then the equation for the dynamics of rotational motion will take the form: M1 - 12 = (1/12) * 2 * 12 * α, where α = dω/dt is the angular acceleration, and ω = dφ/dt is the angular velocity (φ is the angle of rotation of the rod) .

Differentiating the given law of motion, we find the angular velocity at time t = 1 s: ω = dφ/dt = 2t = 2 rad/s.

Substituting the obtained values ​​into the equation of the dynamics of rotational motion, we find the moment modulus M1: M1 = (1/12) * 2 * 12 * α + 12 = 16 N m.

Answer: 16.

Solution to problem 16.1.10 from the collection of Kepe O.?.

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This solution is based on the equation of rotational motion dynamics and contains a detailed description of the formulas and calculation methods used. The solution is presented in a convenient HTML page format, designed using a beautiful design.

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This product is a solution to problem 16.1.10 from the collection of problems in physics by Kepe O.?. The problem is to determine the modulus of the moment M1 of the applied pair of forces at the moment of time t = 1 s for a homogeneous rod, the mass of which is m = 2 kg and the length AB = 1 m, rotating around the Oz axis under the action of a pair of forces with a moment M1 and a moment of resistance forces M2 = 12 N m according to the law? = 3t2.

The solution to the problem is based on the equation of dynamics of rotational motion and contains a detailed description of the formulas and calculation methods used. It is presented in a convenient HTML page format, designed using a beautiful design.

This product may be useful to students and teachers studying physics and preparing for exams and testing. By purchasing this product, you get access to its contents anytime and anywhere.

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Solution to problem 16.1.10 from the collection of Kepe O.?.:

Given: Rod mass m = 2 kg Rod length AB = 1 m Moment of resistance M2 = 12 N m Law of change in angular velocity? = 3t2 at t = 1 s

Find: Modulus of moment M1 of the applied pair of forces at time t = 1 s

Answer:

1. Let's find the angular acceleration of the rod: ? = d?/dt = 6t, at t = 1 s: ? = 6 rad/s2

2. Let us find the moment of inertia of the rod relative to the axis of rotation Oz: I = ml2/12 = 1/12 * 2 * 12 = 1 kgm2

3. Let's find the moment of resistance force: M2 = I*?' Where ?' - angular acceleration, derivative of ? by time. ? = 3t2, ?' = 6t, at t = 1 s: ?' = 6 rad/s2 M2 = 1 * 6 = 6 N m

4. Let's find the modulus of moment M1: M1 = I*? - M2 M1 = 1 * 6 - 12 = -6 N m

Answer: the modulus of moment M1 of the applied pair of forces at time t = 1 s is equal to 6 N m.

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