Question

A uniform 7 kg rod with length 10 m has a frictionless pivot at one end. The rod is released from rest at an angle of 23 degrees beneath the horizontal.

What is the angular acceleration of the rod immediately after it is released?

The moment of inertia of a rod about the center of mass is 1/12 mL² , where m is the mass of the rod and L is the length of the rod. The moment of inertia of a rod about either end is 1/3 mL² , and the acceleration of gravity is 9.8 m/s² .
Answer in units of rad/s².

Answer 1

Mass of the Rod, \(m=7 \mathrm{~kg}\)

Lenght of the Rod, \(L=10 \mathrm{~m}\)

For uniform rod, the center of \(\mathrm{m}\) ass is at its

Geometric center.

The Position of Center of mass \(r=\frac{L}{2}\)

$$ =5 \mathrm{~m} $$

The angle made with the Horizontial, \(\theta=23^{\circ}\)

Now moment of inertia of the rod about an axis passing

through a point at one end of the rod is,

\(I=\frac{1}{3} m L^{2}\)

The torque acting on the rod,

$$ \begin{array}{l} \tau=\operatorname{mgr} \sin (90-\theta) \\ =\operatorname{mgr} \cos \theta \end{array} $$

But, torque, \(\tau=I \alpha\)

Thus, angual ar acceleration of the rod,

$$ \begin{aligned} \alpha &=\frac{\tau}{I} \\ &=\frac{m g \frac{L}{2} \cos \theta}{\frac{1}{3} m L^{2}} \\ &=\frac{3 g \cos \theta}{2 L} \end{aligned} $$

Hence, Angualar acceleration of the rod,

$$ \begin{aligned} \alpha &=\frac{3 g \cos \theta}{2 L} \\ &=\frac{3\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) \cos 23^{\circ}}{2(10 \mathrm{~m})} \\ &=1.353 \mathrm{rad} / \mathrm{s}^{2} \end{aligned} $$