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A tube in the shape of a rectangle with rounded corners is placed in a vertical plane, as shown in Fig. You intro- duce two ball bearings at the upper right-hand corner. One travels by path AB and the other by path CD. Which will arrive first at the lower left-hand corner?

6 years ago

Following same slopes in different order results in different transit times, and the path which is lower relative to other has faster average velocity of ball bearing. Therefore, the bearing through path CD has a shorter transit time and eventually reaches the lower left-hand corner first.

4 years ago

No, CD is not a better path... remember the tube is rectangular and that A and B are perpendicular to each other

Here goes a better explanation of the case :

I’m taking the point of intersection of A and B as O, origin and am lining x axis on OA and y axis on OB. ( since the tube is rectangular... A and B are perpendicular )

- Say the ball roll through tube A... the velocity it has is in the direction of -ve x axis and so is it’s acceleration ( purely horizontal velocity,(
**v**) and the horizontal component of the acceleration )_{1 final} - At O, the ball encounters the normal force of the tube, which is perpendicular to tube B and also opposing the motion of the ball.
- Perpendicular to tube B means that it is perpendicular to y axis ( which means in the +ve x axis direction ). Since the ball is moving in -ve x direction, the normal force opposes it with it’s force in the direction of +ve direction ( satisfies both conditions of the force being perpendicular to tube B as well as opposing motion )
- So, at origin, the acceleration in -ve x axis direction and the acceleration due to normal force add up.
- At the next moment, the ball rolls through tube B, which is horizontal to tube A.
- The velocity (
**v**) of the ball is purely vertical in tube B... but if we recall (1), the ball’s velocity was purely horizontal in tube A._{2 }_{final} - This means that the horizontal component of the velocity of the ball in tube B is zero after encountering the normal force of the wall.
- This shows that the acceleration due to normal force and the acceleration in tube B, together acted on a purely horizontal velocity to nullify it ( bring it to zero ) so that the purely vertical component of acceleration in tube B takes over and gives the ball a purely vertical velocity.
- It is clearly seen that no matter what the initial velocity ( at the instant before normal force acted on it ) of the ball was in tube A never mattered it’s motion in tube B ( since the ball starts with a zero velocity in tube B ) ie. no matter what the velocity of the ball in tube A was, the ball was anyways going to come to rest an instant before going through tube B.
- Total time taken in AB = time taken for ball to reach
**v**from zero in tube A + time taken to reach_{1 final}**v**_{2 final} - Similarly, the total time taken in CD = time taken for ball to reach
**v**from zero in tube C + time taken to reach_{1 final}**v**_{2 final} - Given that the comonents of acceleration in tube B and that in tube C are same; and that the comonents of acceleration in tube A and that in tube D are same, we see that equation (11) and equation (12) are same ie. the time taken for ball to reach the other point through AB and CD is same.
- So, the balls reaches the end point at the same time.

You might see that only the horizontal component of acceleration is acting on the ball in horizonal tube A, because the vertical component of acceleration only sticks or pushes the ball to the wall of the tube and this is cancelled by the push ( normal force ) of the wall on the ball. One way to see that is by noting that the ball moves only horizontally in tube A and not vertically. This shows that any vertical component of the acceleration of the ball, somehow got cancelled. This made us ignore vertical component of acceleration in tube A. Similarly for tube B, the horizontal component of acceleration ( which sticks tha ball to the wall of tube B ) can be ignored.

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