Irodov Problem 1.53

(a) The velocities at the three points A,B and C are illustrated in the figure beside. At each of these points the net velocity vector is the sum of two vectors i) the linear motion of the ball and ii) the rotational velocity along the rim. The magnitude of both these vectors will be v, since the ball is not slipping. The direction of the velocity vector will always be at a tangent to the rotating circle, whereas the linear velocity vector will be directed along the direction of linear motion.





(b)
Every point on the ball experiences three different kinds of acceleration components, i) centripetal acceleration directed towards the center of the ball, ii) tangential rotational acceleration w directed along the tangent to the surface of the ball and iii) the linear acceleration w due to the linear motion of the ball.
The net acceleration at any point is thus the sum of all these three vectors. Now we can find the accelerations at the asked points on the balls as,

Irodov Problem 1.52

(a) This part does not need a lot of explanation I feel. We know that the wheel is moving at a constant velocity in the linear direction, thus there is no acceleration due to this motion. The point A however is moving in a circle, thus it will be undergoing the centripetal acceleration given by .
Nevertheless a completely mathematical treatment is possible and I shall provide it. It will also better explain the origin of centripetal acceleration. The position vector of a any point on the rim can be expressed as,














(b) The distance traversed by the wheel rim is given by, . As seen in the figure, when the rim point is at an angle , the magnitude of the net velocity vector (AB in the figure), using basic trigonometry, is given by,



Thus, the distance traveled by the rim of the wheel between two points of contact is given by,