- #1

Zyxer22

- 16

- 0

## Homework Statement

A uniform helicopter rotor blade is 8.82 m long, has a mass of 108 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 302 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at its center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in 7.14 s. Ignore air resistance. (The blade cannot be considered to be a point mass for this calculation. Why not? Assume the mass distribution of a uniform thin rod.) (c) How much work does the torque do on the blade for the blade to reach a speed of 302 rev/min?

## Homework Equations

F=mv[itex]^{2}[/itex]/r

F= mr[itex]\omega^{2}[/itex]

Since it's a uniform rod;

I=[itex]\frac{1mL^{2}}{12}[/itex]

[itex]T[/itex] = I[itex]\alpha[/itex]

[itex]\alpha[/itex] = [itex]\omega[/itex]/t

W=.5I[itex]\omega^{2}[/itex]

## The Attempt at a Solution

Using F= mr[itex]\omega^{2}[/itex]

I changed 302rev/min to 31.625 rad/s

plugging in m= 108kg r = 4.41m

F= 4.76 * 10^5

-->This answer is wrong.

[itex]\alpha[/itex] = [itex]\omega[/itex]/t

[itex]\alpha[/itex] ~ 4.429 rad/s[itex]^{2}[/itex]

I=[itex]\frac{mL^{2}}{12}[/itex]

[itex]T[/itex] = I[itex]\alpha[/itex]

[itex]T[/itex] = 775.58 Nm

--> This answer is also wrong

W=.5I[itex]\omega^{2}[/itex]

W=87530 J

--> Also wrong

Any help would be greatly appreciated, thanks :)