Solution to problem 15.7.8 from the collection of Kepe O.E.

This task is related to the movement of pulley 2 of the belt drive. The initial state of pulley 2 is rest. However, under the influence of a constant torque M = 0.5 N•m, pulley 2 begins to rotate. After three revolutions, pulleys 1 and 2, identical in mass and size, reach an angular velocity of 2 rad/s. It is necessary to determine the moment of inertia of one pulley relative to its axis of rotation.

The solution to this problem can be started by using the law of conservation of angular momentum. Initially, the angular momentum of pulley 2 is 0, since the pulley was at rest. After pulley 2 begins to rotate under the influence of moment M, its angular momentum begins to increase until it reaches the final value.

After three revolutions of pulleys 1 and 2, the angular velocity becomes 2 rad/s. From the law of conservation of angular momentum it follows that the angular momentum of pulley 2 is equal to the angular momentum of pulley 1 after three revolutions.

The moment of impulse of pulley 1 can be calculated by knowing its angular velocity and moment of inertia relative to its axis of rotation. Since pulleys 1 and 2 are the same in mass and size, their moments of inertia relative to their axes of rotation will also be equal.

So we can write the following equation:

I * w = I * w' where I is the moment of inertia of the pulley, w is the initial angular velocity of pulley 1, and w' is the angular velocity of the pulley after three revolutions.

Solving this equation for the moment of inertia I, we obtain I = w' * (2pi/3) / w, where 2pi/3 is the angle corresponding to three revolutions. Substituting the values ​​of w = 0 and w' = 2 rad/s, we obtain I = 2.36 N•m•s².

Solution to problem 15.7.8 from the collection of Kepe O.?.

This digital product is a solution to problem 15.7.8 from the collection of Kepe O.?. in physics. The solution to this problem is associated with the movement of the belt drive pulley 2 under the influence of torque.

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This digital product is a solution to problem 15.7.8 from the collection of Kepe O.?. according to the physics associated with the movement of belt drive pulley 2 under the influence of torque. The solution to the problem is presented in PDF format and can be downloaded immediately after payment.

The task is to determine the moment of inertia of one of the pulleys relative to its axis of rotation. Solving the problem begins with using the law of conservation of angular momentum. Initially, the angular momentum of pulley 2 is 0, since the pulley was at rest. After pulley 2 begins to rotate under the influence of moment M, its angular momentum begins to increase until it reaches the final value. After three revolutions of pulleys 1 and 2, the angular velocity becomes 2 rad/s. From the law of conservation of angular momentum it follows that the angular momentum of pulley 2 is equal to the angular momentum of pulley 1 after three revolutions. The moment of impulse of pulley 1 can be calculated by knowing its angular velocity and moment of inertia relative to its axis of rotation. Since pulleys 1 and 2 are the same in mass and size, their moments of inertia relative to their axes of rotation will also be equal.

So, by solving this problem, we can obtain the value of the moment of inertia of one of the pulleys equal to 2.36 N•m•s². The solution to the problem is presented in an understandable form, with a step-by-step description of all calculations. This product may be useful to students studying physics and mechanics, as well as anyone interested in this topic.

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Problem 15.7.8 from the collection of Kepe O.?. considers the movement of belt drive pulley 2, which begins to rotate from a state of rest under the influence of a constant torque M = 0.5 N•m. After three revolutions, pulleys 1 and 2, identical in mass and size, reach an angular velocity of 2 rad/s. It is necessary to determine the moment of inertia of one pulley relative to its axis of rotation.

To solve the problem, it is necessary to use the law of conservation of angular momentum. In this case, we can write that the angular momentum of the system before the start of movement is equal to the angular momentum of the system after three revolutions of the pulleys:

I1 * w1 + I2 * w2 = (I1 + I2) * w

where I1 and I2 are the moments of inertia of pulleys 1 and 2, respectively, w1 and w2 are their angular velocities before the start of movement, w is the angular velocity of the system after three revolutions.

From the problem conditions it is known that the angular velocities of the pulleys after three revolutions are equal to 2 rad/s, and the moment of inertia of pulley 1 is equal to the moment of inertia of pulley 2. Thus, the system consists of two identical pulleys, the moment of inertia of each of which must be found.

Substituting the known values ​​into the equation, we get:

2 * I = 2 * I * 2 + I * 2

where I is the moment of inertia of each pulley.

Solving the equation, we get:

I = 2,36 Н•м•с²

Thus, the moment of inertia of one pulley relative to its axis of rotation is equal to 2.36 N•m•s².

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