Question

Bro why cant I post this T-T

Answer 1

Answer: 12.0 m/s^2

Explanation:

Let $\alpha$ be the angular acceleration of the end of the rod

Taking torque about the link, we have:

$\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)$

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

$\tau = I_{rod}\ \alpha......(ii)$

From equations (i) and (ii) we have:

$mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }$

The acceleration of the end of the rod farthest from the link is given by:

$a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}$