Confusion with Conservation of Momentum and 2nd law I apologize if something like this has been asked before but I have been unable to find an answer through searching.
If a weight is carefully added to a rolling cart then the cart should slow down due to conservation of momentum. How is it that the cart has accelerated without a horizontal force? 
This issue originally came up for me while discussing rotational motion. If a figure skater is spinning and brings his hands in he will speed up. How can he accelerate without any torque?
There must be something very basic that I am missing here, so thank you to anybody who can help with an answer.
 A: If the mass is added to the cart at a non-zero velocity relative to the cart, then friction will impart an impulse to the mass. The mass will impart an equal and opposite impulse to the cart. The net momentum change of the cart-mass system will be zero, but the cart itself must accelerate because of the impulse (its momentum changed) and the mass accelerates opposite until they have identical velocities.  There is a horizontal force on the cart.
If the mass is added to the cart, moving with the same velocity as the cart (zero relative velocity), the cart doesn't change velocity because the mass has enough momentum to join the cart without an impulse.
In the case of the spinning skater, the angular momentum is conserved so there is no torque. The moment of inertia is changing.
A: As for your second question on the spinning skater. We have "Newton's second Law for Rotations," which says
$\tau = \frac{d L}{dt} = I \alpha$.
(Recall $L = I \omega$, and the derivative of velocity is acceleration). Notably, the second equality in the line above ONLY holds if $I$ is NOT a function of $t$. In the case of the figure skater, $I$ IS a function of $t$, which means $\tau$, and $\alpha$ are not directly proportional to each other. 
To do this formally, 
$\tau = \frac{dL}{dt} = \frac{d}{dt} (I(t) \omega(t)) = I'(t) \omega(t) + I(t)\alpha(t)$.
Solving for $\alpha$, we get
$\alpha(t) = \frac{\tau - I'(t) \omega(t)}{I(t)}$.
Now assume a frictionless rink with no other external torques:
$\alpha(t) = -\frac{I'(t) \omega(t)}{I(t)}$.
Taa-daa! You can have an angular acceleration in the absence of torque so long as (1)The Moment of Inertia is Changing and (2) the rate of rotation is not zero.
