Solution to problem 16.2.15 from the collection of Kepe O.E.

A wheel weighing 20 kg is under the action of a horizontal force Q = 120 N with a friction coefficient Ftr = 40 N. The radius of the wheel is R = 0.3 m, and the moment of inertia ICz = 0.9 kg • m2. It is necessary to find the module of angular acceleration e of the wheel.

To solve the problem, we use the formula connecting the moment of force and angular acceleration:

M = ICz • е,

where M is the moment of force, ICz is the moment of inertia of the wheel, and e is the angular acceleration.

First, let's find the moment of force acting on the wheel. To do this we use the formula:

Ftr = μ • N,

where μ is the coefficient of friction and N is the normal force acting on the wheel. Since the wheel is in equilibrium, then N = m • g, where m is the mass of the wheel, and g is the acceleration of gravity.

Then Ftr = μ • m • g = 40 N.

Force Q creates a moment of force M = Q • R, then:

M = 120 N • 0.3 m = 36 N • m.

Now we can find the angular acceleration e using the first equation:

f = M / ICz = 36 N • m / 0.9 kg • m2 = 40 rad / s2.

Thus, the modulus of angular acceleration of the e wheel is 40 rad / s2, which is approximately equal to 13.3.

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Our digital product is a detailed solution to problem 16.2.15 from the collection of Kepe O.?. in physics. The problem considers a wheel with a mass of 20 kg, which is under the action of a horizontal force Q = 120 N and a friction coefficient Ftr = 40 N. The radius of the wheel is R = 0.3 m, and the moment of inertia ICz = 0.9 kg • m2. It is necessary to find the module of angular acceleration e of the wheel.

To solve the problem, we use the formula connecting the moment of force and angular acceleration: M = ICz • e, where M is the moment of force, ICz is the moment of inertia of the wheel, and e is the angular acceleration. First, we find the moment of force acting on the wheel using the formula Ftr = μ • N, where μ is the friction coefficient, and N is the normal force acting on the wheel. Since the wheel is in equilibrium, then N = m • g, where m is the mass of the wheel, and g is the acceleration of gravity. Then Ftr = μ • m • g = 40 N.

Force Q creates a moment of force M = Q • R, then M = 120 N • 0.3 m = 36 N • m. Now we can find the angular acceleration e using the first equation: e = M / ICz = 36 N • m / 0 .9 kg • m2 = 40 rad / s2. Thus, the modulus of angular acceleration of the e wheel is 40 rad / s2, which is approximately equal to 13.3.

Our digital product includes a detailed description of each step in solving the problem, all the necessary formulas and intermediate calculations, as well as drawings and diagrams for a better understanding of the solution process. We are confident that our solution to the problem will become an indispensable assistant for everyone who studies physics and will help significantly speed up their preparation for exams.

Our digital product is the solution to problem 16.2.15 from the collection of Kepe O.?. in physics. In this problem, it is necessary to find the modulus of angular acceleration e of the wheel if it is acted upon by a horizontal force Q = 120 N with a friction coefficient Ftr = 40 N, and the wheel radius R = 0.3 m and the moment of inertia ICz = 0.9 kg are given. m2.

Our solution to the problem includes a detailed description of each step of the solution, all the necessary formulas and intermediate calculations. We have also attached all the necessary pictures and diagrams so that you can better understand the solution process.

First, we find the moment of force acting on the wheel using the formula Ftr = μ • N, where μ is the coefficient of friction and N is the normal force acting on the wheel. Since the wheel is in equilibrium, then N = m • g, where m is the mass of the wheel, and g is the acceleration of gravity. Then we find the angular acceleration e using the formula relating the moment of force and angular acceleration: M = ICz • e, where M is the moment of force, ICz is the moment of inertia of the wheel.

In total, the module of angular acceleration of the e wheel is 40 rad/s2, which is approximately equal to 13.3. Our digital product will help you quickly and easily solve this problem and prepare for physics exams.

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Solution to problem 16.2.15 from the collection of Kepe O.?. consists in determining the modulus of angular acceleration of the wheel when a horizontal force is applied and taking into account the friction force. To do this, you need to use the formula of Newton’s second law for rotational motion:

Q - Ftr = ICz * is,

where Q is the applied horizontal force, Ftr is the friction force, ICz is the moment of inertia of the wheel relative to its axis of rotation, R is the radius of the wheel, e is the angular acceleration.

Expressing e from this formula, we get:

is = (Q - Ftr) / ICz,

substituting the values ​​of Q, Ftr, ICz and R, we obtain:

e = (120 N - 40 N) / 0.9 kg m² = 80 N m / 0.9 kg m² = 88.89 rad/s².

The modulus of angular acceleration of the wheel is equal to the absolute value of e, that is, e = |88.89 rad/s²| = 88.89 rad/s².

Answer: the modulus of angular acceleration of the wheel is 88.89 rad/s², which is rounded to 13.3 (to one decimal place).

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