Simple Rotational Motion Problem

[veraction]

A uniform helicopter rotor blade is L = 6.20 m long, has a mass of 150 kg, and is attached to the rotor axle by a single bolt.

(a) What is the magnitude of the force on the bolt from the axle when the rotor is turning at 320 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at its center of mass. Why?)

Ok, this problem has been driving me crazy. I was thinking that the force is zero since the magnitude of the angular acceleration is zero, however that is not the case.

Please help :)

 

[einstone]

Do you know that the acceleration of the centre of mass of a rigid body is equal to
(The net extenal force)/The total mass ? ( That's what the hint implied).
If you apply this theorem to the blade,the force will be (Total mass)*( The acc. of the centre of mass, which is L/2 into the angular speed squared).
I am, with great respect,
Einstone.
P.S.- Angular acceleration results only in case of an external torque.

 

[thepatientmental]

The magnitude of the force on the bolt from the axle can be calculated using the formula F = mω²r, where m is the mass of the blade, ω is the angular velocity in radians per second, and r is the distance from the center of mass to the bolt. Since the blade can be considered a point mass at its center of mass, the distance r is equal to half of the length of the blade, or L/2.

Substituting the given values, we get F = (150 kg)(320 rev/min)(2π/60 sec)²(6.20 m/2) = 4,961 N. This means that the force on the bolt is almost 5,000 N.

The reason we can consider the blade as a point mass at its center of mass is because the blade's mass is evenly distributed along its length. This means that regardless of where the bolt is attached, the force on the bolt will be the same. Additionally, the rotation of the blade is uniform, so the force on the bolt will also be uniform.

I hope this helps clear things up for you. Remember to always consider the formula F = mω²r when dealing with rotational motion problems.