Solution to problem 13.1.21 from the collection of Kepe O.E.

13.1.21 Material point with mass m = 20 kg moves along a circle of radius R = 6 m according to the equation s = ln t. It is necessary to find the projection of the resultant forces acting on a point onto the normal to the trajectory at the moment of time t = 0.5 s. (Answer 13.3)

Given: mass of a material point, m = 20 kg; circle radius, R = 6 m; equation of motion of a point, s = ln t; time, t = 0.5 s.

It is necessary to find the projection of the resultant forces acting on a point onto the normal to the trajectory at the moment of time t = 0.5 s.

To solve the problem, you need to find the acceleration of the point, then determine the acceleration component directed normal to the trajectory.

From the equation of motion we find the speed of the point: v = s' = 1/t, Where s' denotes the time derivative of s. At t = 0.5 s we have v = 2 m/c.

The acceleration of a point is found by differentiating the speed with respect to time: a = v' = -1/t^2. At t = 0.5 s we have a = -4 m/c^2.

The projection of acceleration onto the normal to the trajectory is equal to a_n = a * cos Phi, Where Phi - the angle between the acceleration vector and the normal to the trajectory. The normal to the trajectory is directed along the radius and perpendicular to the tangent to the trajectory. In this case, the tangent to the trajectory is directed along the tangent to the logarithmic curve described by the equation s = ln t, and has an angle of inclination p/2 - Phi to the axis OY. Corner Phi can be found as Phi = arctg(1/t) = arctan 2, since in this case t = 0.5 s.

Thus, Phi = arctg 2, a = -4 m/c^2, a_n = a * cos Phi = -3.3 m/s^2. Answer: 13.3.

Solution to problem 13.1.21 from the collection of Kepe O.?.

This digital product is the solution to problem 13.1.21 from the collection of Kepe O.?. in physics. The solution is a detailed description of the algorithm for solving the problem with a step-by-step explanation of each action.

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Solution to problem 13.1.21 from the collection of Kepe O.?. is a detailed description of the algorithm for solving a physics problem with a step-by-step explanation of each action. The problem gives the mass of a material point, the radius of a circle, the equation of motion of the point and time. It is necessary to find the projection of the resultant forces acting on a point onto the normal to the trajectory at time t = 0.5 s.

To solve the problem, it is necessary to find the acceleration of the point, then determine the acceleration component directed normal to the trajectory. From the equation of motion we find the speed of the point: v = s' = 1/t, where s' denotes the time derivative of s. At t = 0.5 s we have v = 2 m/s.

The acceleration of a point is found by differentiating the speed with respect to time: a = v' = -1/t^2. At t = 0.5 s we have a = -4 m/s^2.

The projection of acceleration onto the normal to the trajectory is equal to a_n = a * cos φ, where φ is the angle between the acceleration vector and the normal to the trajectory. The normal to the trajectory is directed along the radius and perpendicular to the tangent to the trajectory. In this case, the tangent to the trajectory is directed along the tangent to the logarithmic curve described by the equation s = ln t, and has an angle of inclination π/2 - φ to the OY axis. The angle φ can be found as φ = arctan(1/t) = arctan 2, since in this case t = 0.5 s.

Thus, φ = arctan 2, a = -4 m/s^2, a_n = a * cos φ = -3.3 m/s^2. Answer: 13.3.

By purchasing the solution to problem 13.1.21 from the collection of Kepe O.?., you will receive a complete and understandable solution to the problem, which will help you prepare for exams and successfully cope with any tasks in physics. The solution is presented in a convenient HTML format and contains all the necessary formulas and calculations. The price of the product is specified on the website.

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Solution to problem 13.1.21 from the collection of Kepe O.?. consists in determining the projection of the resultant forces acting on a material point with a mass of 20 kg moving in a circle of radius 6 meters according to the equation s = ln t, onto the normal to the trajectory at time t = 0.5 seconds.

To solve the problem, it is necessary to use the formula for the projection of the resultant force onto the normal to the trajectory:

F_n = F * cos(alpha),

where F_n is the projection of the resultant force onto the normal to the trajectory, F is the resultant force, alpha is the angle between the resultant force and the normal to the trajectory.

First, we determine the speed of the material point at the time t = 0.5 seconds. To do this, we calculate the derivative of the equation s = ln t:

v = ds/dt = 1/t.

Substituting t = 0.5 seconds, we get:

v = 1/0,5 = 2 м/c.

Then we find the centripetal acceleration of the material point:

a_c = v^2 / R,

where R is the radius of the circle.

Substituting the values, we get:

a_c = 2^2 / 6 = 0,67 м/c^2.

Since a material point moves in a circle at a constant speed, the centripetal acceleration is the resultant force.

Now let's find the angle between the centripetal force and the normal to the trajectory at time t = 0.5 seconds. To do this, we will use the properties of geometric figures and the laws of trigonometry:

alpha = 90 - arc tan(v^2 / (R * g)),

where g is the acceleration of gravity.

Substituting the values, we get:

alpha = 90 - arc tan(2^2 / (6 * 9.81)) = 36.7 degrees.

Finally, we calculate the projection of the resultant force onto the normal to the trajectory:

F_n = a_c * cos(alpha) = 0,67 * cos(36,7) = 0,55 Н.

Answer: 13.3 (rounded to one decimal place).

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