Question

A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The moment of inertia of the rod about this axis is ML2/3.) The rod is released when it is being held up above the level of horizontal and when it makes an angle of 37° with the horizontal. What is the angular acceleration of the rod at the instant it is released? (in units of rad/s/s )

Answer 1

Answer:

$a=9.8 rad/s^{2}$

Explanation:

Torque, $\tau$ is given by

$\tau=Fr$ where F is force and r is perpendicular distance

$R=0.5Lcos\theta$ where $\theta$ is the angle of inclination

Torque, $\tau$ can also be found by

$\tau=Ia$ where I is moment of inertia and a is angular acceleration

Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity

Making a the subject, $a=\frac {Fr}{I}=\frac {mgr}{I}$ and already I is given as  

$I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta$ hence  

$a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}$

$a=\frac {3gcos\theta}{2L}$

Taking g as 9.81, $\theta$ is given as 37 and L is 1.2

$a=\frac {3*9.81cos37}{2*1.2}=9.7932679419$

$a=9.8 rad/s^{2}$