uniform circular motion

part a：
因為個圖是錯的。
應該 OA = r， AB = L 而不是 OB = L
所以 A 的轉動半徑是 r，而 B 的轉動半徑是 r + Lpart b
若兩物沒繩相連，B 是較易滑行，當 AB 用繩連連著，B 便可以借著繩的張力拉著，使它要較大的轉速才會滑動，因此兩物滑動前，繩是呈緊張狀態而產生張力拉著 A 和 B 兩物。依張力的性質是指離開物體。 對物體 A，繩是在外方，所以張力是向外，指離圓心，它的方向和摩擦力相反。 對物體 B，繩是向內，所以張力是向內，指向圓心，它的方向和摩擦力相同。向心力主要是由摩擦力產生，所以摩擦力是指向圓心。

Q: part a: 點解block b既radius唔係l,而係L+r

A: Block B is at a distance of L from block A. Hence, the distance of block B from the centre of rotation is (L+r).
Just read carefully the statement in the question: "... A is of distance r from the rotation axis, and of distance L from B..."
The figure is wrongly labelled.

Q: 點解block A 個friction同個tension係相反方向?點解block b 個friction同個tension又同方向?

A: Just imagine the rope behaves like an elastic string (like a rubber band), once it is stretched, its two ends would try to shrink in opposite directions. During the shrinking, it would pull block A outward and block B inward.

Q: 其個個friction係邊一個方向?點樣睇?

A: In this example, friction always acts inward towards the centre.

When the plane rotates, the centrifugal force attempts to pull both blocks radially outward. Hence, there is a tendency for the block to start an outward motion. Because friction always opposes motion, the frictional forces thus acts inward to balance the outward motion of the blocks arising from centrifugal force. The two blocks could then remain on the rotating plane without any slipping.