# Sum of forces divided by total mass

I was watching a video in YouTube and the guy was using a very strange way to calculate the acceleration of a system 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.

• 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? – Steeven Nov 16 '18 at 15:17
• 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 – Naifqar Nov 16 '18 at 15:22

$$\let\a=\alpha \def\Wp{W_{\mathrm{1p}}} \def\half{{\textstyle{1 \over 2}}}$$ 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.

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