A uniform helicopter rotor blade is 7.80 m long, has a mass...

A uniform helicopter rotor blade is 7.80 $\mathrm{m}$ long, has a mass of 110 $\mathrm{kg}$ 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 ro- tor 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? (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in 6.70 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 in order for the blade to reach a speed of 320 rev/min?

Transcript

00:01 Hello students, in this question the following data is given.

00:04 The length of the helicopter rotor blade l is given as 7 .8 meters and the mass of the helicopter rotor blade m is given as 110 kilogram.

00:15 So, in the part of the question we have to find the magnitude of the force.

00:18 The magnitude of the force on the bolt from the axle when the rotor is turning at 320 revolutions per minute.

00:27 So, we can use a formula fc is equal to mv square by r and that is equal to m omega square by 2l.

00:40 So, now fc is equal to mass that is 110 kilogram into omega which is 2 pi into f which is 320 revolutions per minute.

00:54 We have to convert it into seconds.

00:56 So, we divide it by 60 the whole square into length l which is 7 .8 divided by 2.

01:04 This we get the centripetal force is equal to 481741 .97 newton.

01:15 So, this is the centripetal force.

01:19 In the b part of the question, in the b part of the question we have to calculate the torque that must be applied to the rotor to bring it to the full speed.

01:28 So, we know the initial angular velocity, initial angular velocity omega i is equal to 0 radian per second.

01:41 So, in the question it is mentioned that we have to find the torque at time t is equal to 0 .7 seconds.

01:49 So, we can write that the after 6 .7 seconds the angular velocity omega f will be equal to omega will be equal to 320 revolutions per minute.

02:14 So, the angular acceleration of the rotor is calculated using the equation of the angular motion for constant angular velocity that is omega is equal to omega i plus alpha t that is alpha is equal to 320 into 2 pi by 60 divided by 6 .7 seconds and that is equal to 5 .0015.

02:48 Therefore, alpha is equal to 5 .0015 radian per second square...

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