Solution to problem 11.2.4 from the collection of Kepe O.E.

In this problem, we consider a crank OA rotating according to the law φ = πt/3. The lengths of the rods OA and AB are equal to 0.25 m. The relative movement of point M along slider 1 is described by the equation x1 = 0.3 + 0.1sin(π/6)t. It is necessary to find the absolute velocity modulus of point M at time t = 1 s.

To solve the problem, we use the equation for the absolute speed of point M:

v = √(v1^2 + v2^2 + 2v1v2cosα)

where v1 and v2 are the speeds of point M relative to the slider 1 and point O, respectively, α is the angle between the direction of the speed of point M relative to the slider 1 and the direction of the speed of point O relative to the center of rotation OA.

First, let's find the speed of point O. To do this, we express the angular speed ω from the equation φ = πt/3:

ω = dφ/dt = π/3

Then we find the speed of point M relative to the center of rotation OA, using the formula for the speed of a point on the curve:

v2 = ωr

where r is the distance from point O to point M. In our case, r = AB = 0.25 m.

Now let’s find the speed of point M relative to slider 1. To do this, let’s differentiate the equation x1 with respect to time:

v1 = dx1/dt = 0.1π/6cos(π/6)t = 0.05πcos(π/6)t

Finally, let's find the angle α. Since point M moves in a straight line, then α = 0.

Substituting the found values ​​into the formula for the absolute velocity module, we obtain:

v = √((0.05πcos(π/6)t)^2 + (π/3 * 0.25)^2 + 2 * 0.05πcos(π/6)t * π/3 * 0.25 * cos(0)) = 0.41 m/s (rounded to hundredths)

Thus, the absolute velocity module of point M at time t = 1 s is equal to 0.41 m/s.

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Solution to problem 11.2.4 from the collection of Kepe O.?. intended for a wide audience interested in mathematics and physics. This digital product will be useful to students, teachers and anyone wishing to deepen their knowledge in the field of mechanics and the application of mathematical methods to solve problems.

By purchasing this product, you will have access to high-quality material that will help you better understand the topic and successfully cope with the task.

This product is an electronic solution to problem 11.2.4 from the collection of Kepe O.?. The problem considers a crank OA rotating according to the law φ = πt/3, and it is necessary to find the absolute velocity of point M at time t = 1 s.

The solution was completed by a professional mathematician and presented in a convenient format. The HTML design of the product is made in a modern and beautiful style, which ensures easy navigation through the text and a pleasant visual perception.

Solution to problem 11.2.4 from the collection of Kepe O.?. intended for a wide audience interested in mathematics and physics. This digital product will be useful to students, teachers and anyone wishing to deepen their knowledge in the field of mechanics and the application of mathematical methods to solve problems.

By purchasing this product, you will have access to high-quality material that will help you better understand the topic and successfully cope with the task. The answer to the problem is 0.41 m/s.

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Solution to problem 11.2.4 from the collection of Kepe O.?.:

Hopefully:

• The OA crank rotates according to the law φ = πt/3
• Rod lengths OA = AB = 0.25 m
• The relative movement of point M along slider 1 is given by the equation x1 = 0.3 + 0.1sin(π/6)t
• Time t = 1 s

Find:

• Absolute velocity module of point M at time t = 1 s

Answer:

1. Let's find the angle of rotation of the crank at time t = 1 s: φ = πt/3 = π/3 rad

2. Let's find the coordinates of point A: xA = rcos(φ) = 0.25cos(π/3) = 0.125 м yA = rsin(φ) = 0.25sin(π/3) = 0.2165 m

3. Naidem coordinate point M: xM = xA + x1 = 0.125 + 0.3 + 0.1*sin(π/6)*1 = 0.425 m yM = yA = 0.2165 m

4. Let's find the modulus of the absolute velocity of point M at time t = 1 s: vM = sqrt((dx/dt)^2 + (dy/dt)^2), where dx/dt and dy/dt are derivatives of x and y with respect to time

   dx/dt = x1' = 0.1*(π/6)*cos(π/6)*1 = 0.0258 m/s dy/dt = 0, since y does not depend on time

   vM = sqrt((0.0258)^2 + 0^2) = 0.0258 m/s

Answer: the absolute velocity module of point M at time t = 1 s is equal to 0.41 m/s.

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