Question

A uniform slender rod(8 m length and 3 kg mass) rotates in a vertical plane about a horizontal axis 1 m from its end as shown in the figure. The magnitude of the angular acceleration (in $\frac{rad}{s^2})$ of the rod at the position shown is

Answer 1

$AP=1m,AC=4m,PC=3m$ $I_C=\frac{m{L}^2}{12}=\frac{3\times 8\times 8}{12}=16kg-m^2$ $I_P=I_C+m(PC{)}^2$ $=16+3\times 3^2=43kg-m^2$
Torque $=m_{2}g\times \frac{PB}{2}-m_{1}g\frac{AP}{2}$ $T=m_2\times 9.81\times \frac{7}{2}-m_1\times 9.81\times \frac{1}{2}$ $=34.335m_2-4.905m_1$ $m_1=\frac{3}{8}kg,m_2=\frac{3\times 7}{8}=2.625kg$ $T=34.335\times 2.625-4.905\times 0.375$ $=88.29N.m$
We can also calculate torque as $T=mg\times PC$ $=3\times 9.81\times 3=88.29N.m$
$T=I_P\alpha .$
$\alpha =\frac{T}{I_P}=\frac{88.29}{43}=2.053\frac{rad}{s^2}$