Honors Physics I Chapter 11 Rolling, Torque and Angular Momentum

Rolling as Translation and Rotation Combined

$v_{\text{com}}=\omega R$

For any point : $v_i=v_{\text{com}}+(r_i\omega)\hat\theta_i$

Differential the first equation, we get:

$a_{\text{com}}=\alpha R$

NOTE : Only for the smooth rolling motion.

$\vec f_s=+f\cdot\hat x\\[8px] -mg\sin\theta+f_s=ma_{\text{com}}\\[7px] \tau_{F_s}=Rf_s=I_{\text{com}}\alpha\\[7px] a_{\text{com}}=-\alpha R$

And final, we get:

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

Conclusion : Equation for Rolling

$-Mg\sin\theta+f_s=-mR\alpha\\ f_sR=I_{\text{com}}\alpha$

The ramp down of angle is $90\degree$ , and the $\sin\theta$ becomes $1$ .

$\vec l =\vec r\times \vec p=m(\vec r\times \vec v)$

We rewrite Newton’s second law as:

$\vec\tau_{\text{net}}=\frac {d\vec l}{dt}$

For the total angular momentum:

$L=\sum \vec l_i$

Therefore:

$\frac {d\vec L}{dt}=\sum_{i=1}^n\vec\tau_{\text{net},i}$

Define rigorously:

$L_Z=\omega\left(\sum_{i=1}^n\Delta m_ir_{\perp i}^2\right)=I\omega$

If $\tau_{\text{net}}=0$ , then $\vec L$ is a constant, so:

$L_i=L_f\;\;(\text{for a isolated system})$

This is the law of conservation of angular momentum.

The self Rotation give torque to the spinning Rotation. And the self rotation is come from its falling down(Like a yo-yo).

The precession rate is given by:

$\Omega=\frac{Mgr}{I\omega}$