Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length.
(a) Find the initial angular acceleration of the rod.
(b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg.
Solution
Total moment of inertia of the system about the axis of rotation,
\[I_{net} = \left( m_1 r_1^2 + m_2 r_2^2 + \frac{m l^2}{12} \right)\]
m and l are the mass and length of the rod, respectively.
\[ \tau_{net} = F_1 r_1 - F_2 r_2 \]
\[\text{Also, }\tau_{net} = I_{net} \times \alpha\]
On equating the value of \[\tau_{net}\] and putting the value of `l_("net"),` we get
\[F_1 r_1 - F_2 r_2 = \left( m_1 r_1^2 + m_2 r_2^2 + \frac{m l^2}{12} \right) \times \alpha\]
\[\left( - 2 \times 10 \times 0 . 5 \right) + \left( 5 \times 10 \times 0 . 5 \right) = \left[ 5 \left( \frac{1}{2} \right)^2 + 2 \left( \frac{1}{2} \right)^2 + \frac{\left( 1 \right)^2}{12} \right] \alpha\]
\[\Rightarrow 15 = \left( 1 . 75 + 0 . 084 \right) \alpha\]
\[ \Rightarrow \alpha = \frac{1500}{\left( 175 + 8 . 4 \right)} = \frac{1500}{183 . 4}\]
\[= 8 . 1\text{ rad/s}^2 ........\left( g = 10 \right)\]
\[= 8 . 01\text{ rad/s}^2 ......\left(\text{if }g = 9 . 8 \right)\]
(b) From the free body diagram of the block of mass 2 kg,
\[T_1 - m_1 g = m_1 a\]
\[ \Rightarrow T_1 = 2 \left( a + g \right)\]
\[ = 2\left( \alpha r + g \right)........ \left(\text{using, }a = \alpha r \right)\]
\[ = 2\left( 8 \times 0 . 5 + 9 . 8 \right)\]
\[ \Rightarrow T_1 = 27 . 6 N\]
From the free body diagram of the block of mass 5 kg,
\[m_2 g - T_2 = m_2 a\]
\[ \Rightarrow T_2 = m_2 \left( g - a \right)\]
\[ = 5 \left( g - a \right) = 5 \left( 9 . 8 - 8 \times 0 . 5 \right)............\left( a = \alpha r \right)\]
\[= 5 \times 5 . 8 = 29 N\]
