Conservation of angular momentum - linear velocity The angular momentum of a single particle is $mr^2 \omega$, and its linear velocity is $\omega r$. Suppose that $m=1$, distance from the rotation axis $r=1$ and $\omega=1$.
What happens when the distance from the axis of rotation is reduced to $r=0.5$? This is analogous to a rotating ice skater that pulls their arms inwards, there is no external force that changes the momentum of particles, so the angular momentum remains constant. According to the formula it means it has to be equal to $1$, and therefore the angular velocity will be 4 times the original angular velocity $\omega$. Moreover the linear speed of the particle will double.
How is that possible? Doesn't it contradict the law that states if there's no external force acing on the particle, its linear momentum should remain constant?
 A: Angular momentum $\mathbf L$ is given by:
$$\mathbf L = \mathbf r \times \mathbf P$$
Now the torque $\mathbf {\tau} $ acting on an object is given by: 
$$\boldsymbol {\tau }  = \frac {d \mathbf L}{dt}$$
$$\Rightarrow \boldsymbol {\tau } = \frac {d (\mathbf r \times \mathbf P) }{dt}$$
$$\Rightarrow \boldsymbol {\tau } = \mathbf r \times \mathbf F + \mathbf v \times  \mathbf P$$
Since $\mathbf v $ and $\mathbf P$ are in the same direction therefore 
$$\mathbf v \times  \mathbf P = 0$$
i.e.,
$$\Rightarrow \boldsymbol {\tau } = \mathbf r \times \mathbf F $$
Now as you correctly say that $\mathbf L$ is constant therefore 
$$\boldsymbol {\tau } = \mathbf r \times \mathbf F =0$$
Clearly since $\mathbf r \neq 0$ then this means that 
$\mathbf F$ at all instants points towards the centre (i.e., to say that line of action of the force passes through the centre). 
Note that force cannot be zero as a centripetal force is always required for maintaining circular motion.

Now as you can see from the figure during transition from outer orbit to inner orbit the force and velocity vector have acute angle between them therefore the force acts to accelerate the object and hence increase it's speed.
Also it should be noted that for angular momentum to be conserved the absence of force isn't a necessary criteria (example elliptical orbits of planets).

Regarding the comment:

So if I calculated the speed of particle after the radius has been reduced down to $0.5$ it would turn out to be $2v$, where $v$ is the original speed? Is there a formula that I can use? 

Yes as you know that 
$$\mathbf L = \mathbf r \times \mathbf P$$
$$\frac {\mathbf L}{m} =  \mathbf r \times \mathbf v \tag 1$$
Clearly after the radius reduces to $\mathbf {r'}$ then the velocity becomes $\mathbf {v'}$
$$\Rightarrow \frac {\mathbf L}{m} =  \mathbf {r'} \times \mathbf {v'} \tag 2$$
Therefore from $(1)$ and $(2)$ (also noting that after the particle comes into orbit the vectors have $90°$ angle between them) 
$$\Rightarrow v'r'=vr$$
Then 
$$v'= v \frac {r}{r'}$$

Note that this has been explained in the following Vsauce video (explanation starts from 10:05 to 13:15):


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*Laws and Causes
A: This all depends on what you define to be an external force or, equivalently, what is and is not part of your system. If your system is just the particle and nothing else, then there definitely is an external force. In fact, there's an external force even before the radius is reduced, namely the centripetal force maintaining the particle's circular motion. If your system is more than just the particle, then there must be something else in the system providing both the centripetal force as well as any force required to decrease the radius of rotation. In this case, the total momentum (angular and linear) of the whole system is conserved, and whatever momentum the particle appears to have gained or lost will be matched by an equal and opposite gain or loss experienced by the rest of the system.
