Smooth Rolling

Please always note that: The friction is static friction, there is always no work!

Important: always remember the direction !!! (both linear and angular)

Newton’s Second Law and Motion

$$f_s-Mg\sin\theta = Ma\\ Rf_s=I\alpha\\ a=\alpha R$$

Work and Energy

Suppose in $dt$ time, it rolls $dx$, then：

$$mg\cdot dx = \left(\frac 1 2 m (v+a\cdot d_t)^2 - {1\over 2}mv^2\right)+\left(\frac 1 2 I(\omega+\alpha\cdot dt)^2-\frac 1 2 I\omega ^2\right)\\ a=\alpha R$$

In first expression, we can neglect the second-order infinitesimal.

From both two ways, we can get:

$$a=-\frac {g\sin \theta}{1+\frac {I_{\text{com}}}{MR^2}}$$

Yo-yo

1. Note that as there is no normal force, so there is no static friction
2. If there are two radius, the bigger is used to calculate inertia, the smaller where the string ties is used to calculate torque.

String

The displacement of free end of the string $v_s$ is the combine of the the object itself and the rolling it release, that is:

$$v_s=v+\omega r$$