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I was watching a video in YouTube and the guy was using a very strange way to calculate the acceleration of a system enter image description here Is this always valid? If so, can you please provide the proof?

Note that he multiplies the mass of the disk by the constant in its inertia.

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  • $\begingroup$ It is just Newton's 2nd law. Would you mind clarifying what you mean by "he multiply the mass of the disk by the constant in its inertia"? I guess you are referring to the $cM$ term in the denominator. What is the definition of these parameters in the video? $\endgroup$ – Steeven Nov 16 '18 at 15:17
  • $\begingroup$ No I'm asking specifically about the change that he added because of the rotational motion which is cM where c is the constant of the disk inertia and M is the mass of the disk .. sorry if my question isn't clear $\endgroup$ – Naifqar Nov 16 '18 at 15:22
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$\let\a=\alpha \def\Wp{W_{\mathrm{1p}}} \def\half{{\textstyle{1 \over 2}}}$ Unfortunately you don't inform us as to your degree of confidence with Newtonian mechanics. So it's difficult to calibrate an answer. Surely the proposed solution is right but far from a simple application of 2nd law. You have to decompose the problem into three sub-parts:

  1. mass on the incline with attached rope

  2. the pulley with the ropes

  3. mass vertically falling with vertical rope.

1) Mass $m_1$ is subjected to two active forces (parallel to plane): component $\Wp$ of its weight and tension $T_1$ of the rope. Its acceleration upwards is given by $$m_1\,a = T_1 - \Wp.\tag1$$

2) Mass $m_2$ has the same acceleration, but downwards. If $T_2$ is rope's tension there $$m_2\,a = m_2\,g - T_2.\tag2$$

3) As to disk we must apply equation for angular momentum (I don't know its usual name in english): $$I\,\a = r\,T_2 - r\,T_1 \tag3$$ where $\a$ is angular acceleration = $a/r$, ($r$ = disk radius), $I=c\,M\,r^2$, $c=1/2$ for a homogeneous disk.

Adding eqs. (1) and (2), and substituting for $T_1-T_2$ its value taken from (3) $$(m_1 + m_2)\,a = m_2\,g - \Wp - {I \a \over r} = m_2\,g - \half\,m_1\,g - \half\,M\a\,r = m_2\,g - \half\,m_1\,g - \half\,M\,a$$ $$a = {m_2\,g - \half\,m_1\,g \over m_1 + m_2 + \half\,M}.$$ This example should help you to solve similar problems.

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  • $\begingroup$ Very very good. I have upvoted a very nice answer. $\endgroup$ – Sebastiano Nov 16 '18 at 16:53

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