Question

На цилиндр радиусом 4 см который может вращаться вокруг неподвижной горизонтальной оси, совпадающей с осью цилиндра намотана нить. К концу нити привязали круг и предоставили ему опускаться двигаясь равноускорено груз за 3 секунды опустился на 1.5 м. Определить угловое ускорение цилиндра и число оборотов сделанных им за это время.

Answer 1

ДАНОt =3h =1,5 мr =4 см.-----------------------w'  - ?
РЕШЕНИЕдвижение равноускоренное  v0=0h=v0*t+at^2/2= at^2/2a=2h/t^2угловое ускорение w'=a/r=2h/(t^2*r)=2*1.5/(3^2*0.04)=8.(3) = 8.3 рад/с2
Ответ 8.3 рад/с2

Answer 2

Problem Analysis

We are given a cylinder with a radius of 4 cm that can rotate around a fixed horizontal axis coinciding with the axis of the cylinder. A thread is wound around the cylinder, and a circle is attached to the end of the thread. The circle is allowed to descend with uniform acceleration, and in 3 seconds, it descends 1.5 m. We need to determine the angular acceleration of the cylinder and the number of revolutions it makes during this time.

Solution

To solve this problem, we can use the kinematic equations of motion for rotational motion. Let's denote the angular acceleration of the cylinder as α and the number of revolutions made by the cylinder as n.

Using the equation for the displacement of an object undergoing uniform acceleration, we have:

s = ut + (1/2)at^2

In this case, the displacement of the circle attached to the thread is 1.5 m, the initial velocity is 0 m/s, and the time is 3 seconds. Therefore, we can write:

1.5 = (1/2)α(3^2)

Simplifying the equation, we get:

α = (2 * 1.5) / (3^2)

Now, we can calculate the angular acceleration of the cylinder.

To find the number of revolutions made by the cylinder, we can use the equation:

θ = ωt + (1/2)αt^2

where θ is the angle rotated, ω is the initial angular velocity (which is 0 in this case), t is the time, and α is the angular acceleration.

Since the initial angular velocity is 0, the equation simplifies to:

θ = (1/2)αt^2

Substituting the values, we have:

θ = (1/2) * ((2 * 1.5) / (3^2)) * (3^2)

Simplifying the equation, we get:

θ = 1.5

Therefore, the number of revolutions made by the cylinder is 1.5.

Answer

The angular acceleration of the cylinder is (2 * 1.5) / (3^2) and the number of revolutions made by the cylinder is 1.5.

Please note that the above solution assumes that the thread is unwinding smoothly from the cylinder without slipping or any other external factors affecting the motion.