How Does a Lift's Acceleration Affect the Pulley System Dynamics?

[Victoria_235]

Homework Statement

Two blocks are attached by strings of negligible mass to a pulley with two radii R1=0.06 m and R2=0.08 m. The strings are wrapped around their respective radios so that the masses can move either up or down. The pulley has a moment of inertia I=0.0023 kg

m2 , and is supported by a bearing with negligible friction.
1)If block 1 has a mass of 0.6 kg and block 2 has a mass of 0.5 kg, what is the magnitude of the angular acceleration of the system?
2) Idem, but if the system is moving up with an acceleration a_0.
(My problem is part two.)

Homework Equations

I have solved the part 1) without problems.

The equations would be:

\begin{equation}
T_1 - m_1g = m_1a_1 \\
m_1g - T_2 = m_2a_2\\

Where:\\
a_1 = \alpha R_1 \\
a_2 = \alpha R_2\\
\end{equation}
For the rotation motion of the pulley:
\begin{equation}
\sum M = I \alpha\\
\sum M = F \times r\\

T_2 R_2 -T_1R_1 = I \alpha\\
\end{equation}
Solving the system equation, I have obtained: \\
\begin{equation}
T_1 = 6.06 N\\
T_2 =4.69N\\
\alpha = 5.117 rad/s²\\
\end{equation}
Where $\alpha$ is the angular acceleration.

My problem comes in the part 2. Ths system inside a lift with an acceleration up.
My attempt solution...

The Attempt at a Solution

The only thing I think I can do is add to $\vec{g}$ the acceleration $\vec{a_0}$.\\
The equations in the lift would be:

\begin{equation}
T_1 - m_1(\vec{g}+\vec{a_0})= m_1(\vec{a_1}+\vec{a_0}) \\

m_1(\vec{g}+\vec{a_0}) - T_2 = m_2(\vec{a_2}-\vec{a_0})\\
\end{equation}

(In here I am not sure if I have to add a_0 and subtract it, depending if the mass is going down or up. Also if I have to add a_0 to g in the term:$$m_1(\vec{g}+\vec{a_0})$$ and $$m_2(\vec{g}+\vec{a_0})$$.For the rotation motion of the pulley:\\
\begin{equation}

T_2 R_2 -T_1R_1 = I \alpha\\

\end{equation}

Should I add the $\vec{a_0}$ in the rotation equations?
PS: I am sorry for my poor LaTeX, I will try to improve it.

 

[Orodruin]

You can always go to the accelerated frame. In this frame the effective gravitational field will indeed be $\vec g -\vec a_0$ (note that $\vec g$ and $\vec a_0$ are directed in opposite directions, the first down and the second up) and the problem will be exactly equivalent to the first, but with a different gravitational field.

 

[Victoria_235]

So, I understand that the equations (4), about dinamic motion, would be:

\begin{equation}
T_1 -m_1 (\vec{g}- \vec{a_0}) =m_1a_1\\
m_2 (\vec{g}- \vec{a_0}) -T_2 =m_2a_2\\
\end{equation}
When the lift is going up. Because someone inside the lift would fell the gravity harder if the lift is going up, for this reason I don't understand why should be $$(\vec{g}+\vec{a_0})$$, as you say...

 

[PeroK]

So, I understand that the equations (4), about dinamic motion, would be:

\begin{equation}
T_1 -m_1 (\vec{g}- \vec{a_0}) =m_1a_1\\
m_2 (\vec{g}- \vec{a_0}) -T_2 =m_2a_2\\
\end{equation}
When the lift is going up. Because someone inside the lift would fell the gravity harder if the lift is going up, for this reason I don't understand why should be $$(\vec{g}+\vec{a_0})$$, as you say...

$\vec{g}$ and $\vec{a_0}$ are vectors. In general, the effective gravity will be $\vec{g_{eff}} = \vec{g} - \vec{a_0}$.

If we put in some numbers. Let $\vec{g} = -10 m/s^2$ and $\vec{a_0} = + 2m/s^2$, then $\vec{g_{eff}} = \vec{g} - \vec{a_0} = -12m/s^2$.

But, in this case, in terms of magnitudes: $g_{eff} = g+a_0 = 12m/s^2$