A uniform rod with a mass of m = 1.94 kg and a length of l = 2.10 m is attached to a horizontal surface with a hi=nge. The rod can rotate around the hi=nge without friction. (See figure.)
Initially the rod is held at rest at an angle of θ = 70.4° with respect to the horizontal surface. Then the rod is released.
What is the angular speed of the rod, when it lands on the horizontal surface?
1.61 rad/s is incorrect.
What is the angular acceleration of the rod, just before it touches the horizontal surface?
1.06 rad/s^2 is incorrect.

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okpalawalter8 okpalawalter8 · Answer 1 · 2022-07-18T19:52:53+02:00

The angular acceleration of the rod, just before it touches the horizontal surface is mathematically given as

w = 3.6 rad/s

Generally, the equation for Torque is mathematically given as

[tex]T = F r[/tex]

Therefore

[tex]T = (mg Cos\theta) (L/2)\\\\T = I \alpha[/tex]

[tex]\alpha = (1.5) (9.8 Cos65.2) (2.10)[/tex]

[tex]\alpha = 12.95 rad/s2[/tex]

Generally, the equation for conservation of energy is mathematically given as

P.E=K.E

Therefore

[tex]mg (L/2) sin65.2 = (0.5) I w2[/tex]

[tex](0.5) (9.8) Sin65.2 = (0.5) (2.1/3) w2[/tex]

w = 3.6 rad/s

In conclusion, the angular acceleration

w = 3.6 rad/s

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