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 )

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opudodennis opudodennis · Answer 1 · 2019-10-29T19:25:40+01:00

Answer:

[tex]a=9.8 rad/s^{2}[/tex]

Explanation:

Torque, [tex]\tau[/tex] is given by

[tex]\tau=Fr[/tex] where F is force and r is perpendicular distance

[tex]R=0.5Lcos\theta[/tex] where [tex]\theta[/tex] is the angle of inclination

Torque, [tex]\tau[/tex] can also be found by

[tex]\tau=Ia[/tex] 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, [tex]a=\frac {Fr}{I}=\frac {mgr}{I}[/tex] and already I is given as

[tex]I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta[/tex] hence

[tex]a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}[/tex]

[tex]a=\frac {3gcos\theta}{2L}[/tex]

Taking g as 9.81, [tex]\theta[/tex] is given as 37 and L is 1.2

[tex]a=\frac {3*9.81cos37}{2*1.2}=9.7932679419[/tex]

[tex]a=9.8 rad/s^{2}[/tex]