What is the moment of inertia and angular velocity of a released rod?

[RyRy19]

A uniform rod of mass M=5.0 kg and length ℓ=20 cm is pivoted on a frictionless hinge at the end
of it. The rod is held horizontally and then released.

a) Use the parallel-axis theorem to determine the moment of inertia of the rod about the hinge (ie
its end).

b) Determine the angular velocity of the rod when it reaches the vertical position and the speed of
the rod tip’s at this point.I=MR^2

Ok so what I did first was 1/12 ML^2 as "I" about the hinge is 1/12ML² + (M*(L/2)²) to work out A.

I don't really know how to work out b though, apart from possibly using that to work out the distance from the radius and then doing work done to then find out ω, I assume.

 

[SteamKing]

Draw a picture of the rod in the horizontal position before it is released. At the moment of release, what are the forces acting on the rod? How do these forces cause the rod to rotate about the hinged end? Draw a free body diagram.

Do you know the relationship between angular acceleration and the moment acting on a body? You know, something you might have picked up in a physics class about rotational motion.

 

[RyRy19]

Draw a picture of the rod in the horizontal position before it is released. At the moment of release, what are the forces acting on the rod? How do these forces cause the rod to rotate about the hinged end? Draw a free body diagram.

Do you know the relationship between angular acceleration and the moment acting on a body? You know, something you might have picked up in a physics class about rotational motion.

So If I work out I = (1/12)*M*R²
Then τ = MG* 10x10^-2 (as the length of the rod is 20x10^-2 - Only weight is
acting)

Then I use the solution to that and plug it into (1/2)ω²*I and solve for ω?

 

[haruspex]

So If I work out I = (1/12)*M*R²

Except that you want the moment about the hinge, as calculated in (a).

Then τ = MG* 10x10^-2

As you suggested, the easy way is to use work conservation. For that, you don't need the torque. What is the rotational KE gained? What PE has been lost?

 

[RyRy19]

Except that you want the moment about the hinge, as calculated in (a).

As you suggested, the easy way is to use work conservation. For that, you don't need the torque. What is the rotational KE gained? What PE has been lost?

So I would do mgh, for the GPE and then solve to work out v from 1/2mv^2 ? I'm guessing you take the height in regards to a triangle? Find the resultant vector or rather the hypotenuse?

If so I think I understand.

Thanks for the help.

 

[dauto]

So I would do mgh, for the GPE and then solve to work out v from 1/2mv^2 ? I'm guessing you take the height in regards to a triangle? Find the resultant vector or rather the hypotenuse?

If so I think I understand.

Thanks for the help.

No I don't think you understand. What's the height h? Why aren't you using the formula for the kinetic energy of rotational motion?

 

[haruspex]

So I would do mgh, for the GPE and then solve to work out v from 1/2mv^2 ?

No, $\frac 12 I\omega^2$, as you posted earlier. But make sure to use the right I.

I'm guessing you take the height in regards to a triangle? Find the resultant vector or rather the hypotenuse?

Triangle? You want the change in height of the mass centre of the rod.