- #1
Victoria_235
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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
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.