rotational motion

a uniform helicopter rotor blade is 7.8m long , has a mass of 110kg, 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 at its com . why ? please answer this as well ). B) calculate the torque that must be applied to the rotor to bring it to full speed from rest in 6.7 s . ignore air resistance . ( the blade cannot be considered to a point mass for this calculation . why not ? assume the mass distribution of a thin uniform rod . ) (C) how much work does the torque do on the blade in order for the blade to reach a speed of 320 rev / min . ?? any help is appreciated .

Easy Math Editor

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Comments

rotational inertia of a point mass concentrated at center of mass is I=mrr = 110kg7.87.83

ayesha nawab - 7 years, 4 months ago

320rev/min means 640 rad/min means 10.67rad/sec so omega or angular velocity w= 10.67pie rad/sec the calculate rotational k.e by formula Iwsquare then calcalate angular acceleration by

ayesha nawab - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$