Why does a ball slide out of a rotating frictionless tube with no radial force?

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Mausam
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If we have a frictionless hollow tube and a ball which just fits into it. All the surfaces are frictionless.

Now if we put in the ball about at the center of tube and rotate the tube about one of its ends with a constant angular velocity ,then the ball swoops out of
the other end .

If we see from the ground frame then no radial force is acting as friction is absent,then how can we explain the motion of the ball.According to Newton's first law we need a force to change the state of motion.

How do we explain this
 
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Mausam said:
According to Newton's first law we need a force to change the state of motion.
Yes, the acceleration will be tangential only (perpendicular to the tube axis), which is not a constant direction in the ground frame.
 
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To expand a bit on what @A.T. has said, the acceleration which is applied now and is tangential now results in a velocity which is tangential now but which is retained and will have a radial component a moment from now when the "radial" and "tangential" directions change.
 
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hello mausam
Mausam said:
According to Newton's first law we need a force to change the state of motion.
exactly , very reason why the ball pops out , cause there's no force acting inwards to keep its radius const.
ball tries tries to move in one line . think of tangent to a circle , as you move along tangent,you move further away from circle centre