The tail rotor of a helicopter has a mass of mr 50 kg...

Name: the tail rotor of a helicopter has a mass of mr 50 kg including a rotating unbalance of me 0075 kgm the tail section has a length of 4 m has a mass of mt 500 kg and can be simulated as a can 55293
Uploaded: 2023-08-25T19:35:36-08:00
Duration: 6 min 41 s
Channel: Madhur L
Description: the tail rotor of a helicopter has a mass of mr 50 kg including a rotating unbalance of me 0075 kgm the tail section has a length of 4 m has a mass of mt 500 kg and can be simulated as a can 55293

The tail rotor of a helicopter has a mass of MR = 50 kg, including a rotating unbalance of me = 0.075 kgm. The tail section has a length of 4 m, has a mass of MT = 500 kg, and can be simulated as a cantilever beam with an approximate constant circular cross-sectional area of diameter = 200 mm, made of aluminum (E = 73 GPa). The tail has a damping mechanism not shown that provides a damping coefficient of $\zeta = 0.15$. Determine the force transmitted to the base of the tail section when the blades rotate at 1500 rpm. Hint: Model the tail section as a cantilever beam of mass MT with an end mass MR, the Area moment of Inertia of a circular rod is I = (1/4)$\pi$R^4

Transcript

00:01 Hello students, referring to the given question, the given data are all given here.

00:05 We are asked to calculate the natural frequency of the system considering the effect of mass of the tail.

00:11 Therefore, for frequency the formula is 1 by 2 pi into root of k by m, where k is the stiffness of the system and m is the mass of the system, that is mass of the tail plus mass of the rotor, that is the total mass of the system.

00:33 Calculating frequency we need k, therefore k can be calculated using the formula 3 into e into i divided by l cube into pi by 4 into outer dia raised to the power 4 minus inner dia raised to the power 4.

00:54 So, again we need moment of inertia i, which can be calculated as pi by 64 into outer dia raised to the power 4 minus inner dia raised to the power 4.

01:08 Therefore, plugging the values that is 3 .14 divided by 64 into 30, that is diameter is 30 into 10 power minus 2 raised to the power 4 minus 25 into 10 to the power minus 2 raised to the power 4.

01:26 So, on calculating this value we will be getting i is equal to 2 .058 into 10 power minus 4 meter raised to the power 4.

01:39 Substituting the value of i, we can calculate k, where k is equal to 3 into that is 3 into e is 70 into 10 that is young's modulus pascal into i is 2 .058 into 10 power minus 4 meter power 4 divided by l cube is 4 raised to the power 4 meter cube into pi by 4 into again outer dia that is 30 into 10 power minus 2 raised to the power 4 minus 25 into 10 to the power minus 2 raised to the power 4.

02:28 On calculating this we will be getting k is equal to 2 .22 into 10 power 3 newton per meter.

02:37 Next we have to calculate the mass of the system.

02:50 Therefore, mass of the total mass of the system is equal to mass of tail plus mass of rotor.

02:58 Mass of rotor is given in data.

02:59 Therefore, we have to calculate mass of tail which is equal to rho into a into l.

03:06 For this we have to calculate area.

03:08 Area is equal to pi into r squared, r is pi into r naught squared minus r o outer radius the whole squared minus inner radius the whole squared.

03:21 Therefore, it is calculated as pi into outer radius is 15 into 10 power minus 2 raised to the power square minus 12 .5 into 10 power minus 2 raised to the power 2.

03:34 Therefore, the area is given as 0 .022 meter square...

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