On an immovable circle of radius a, another circle, which lies in the same plane as the first one, of the same radius rolls without slipping. A point A' is fixed on the moving circle, which at the initial moment of time coincides with the point A of the non-moving circle. Find the acceleration of a material point as a function of its distance from point A, if the moving circle rotates around the immovable one with the constant angular velocity w.

Ответ:

Объяснение:

Let's consider a point B on the rolling circle which is in contact with the stationary circle at point A. Let O be the center of the stationary circle, and O' be the center of the rolling circle.

At any instant, the velocity of point B can be resolved into two components: one is the velocity of the rolling circle at point B, which is perpendicular to OB and has magnitude w times the distance OB, and the other is the velocity due to the rolling motion, which is tangential to the rolling circle and has magnitude OB times w.

Let v be the velocity of point A' on the rolling circle at any instant, and let x be the distance between A' and A. Then, the velocity of point B can be expressed as:

vB = v + w*(OB) + w*(OB)x

Since the rolling motion is without slipping, the velocity of point B must be perpendicular to AB. Therefore, we can resolve the velocity vB into two components: one along AB, which is the acceleration of the material point, and the other perpendicular to AB, which is equal to w*(OB + x*OB) and does not contribute to the acceleration.

Let a(x) be the acceleration of the material point at a distance x from A. Then, we have:

a(x) = dvB/dt = dv/dt + w*OB

where dv/dt is the tangential acceleration due to the rolling motion. Since the rolling motion is without slipping, the tangential acceleration is equal to v*omega, where omega is the angular velocity of the rolling circle.

Therefore, we have:

a(x) = vomega + wOB

Now, we need to express v and OB in terms of x. Since the rolling motion is without slipping, the distance traveled by the rolling circle is equal to the distance traveled by the material point on the circle. Therefore, we have:

x = OB*theta

where theta is the angle rotated by the rolling circle. Differentiating both sides with respect to time, we get:

v = OBd(theta)/dt = OBomega

Substituting this into the expression for a(x), we get:

a(x) = OBomega^2 + wOB

Simplifying, we get:

a(x) = (omega^2 + w)*OB

Finally, substituting OB = a*sin(theta), we get:

a(x) = a*(omega^2 + w)*sin(theta)

where a is the radius of the stationary circle, and theta is related to x as x = a*theta.