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Mumz [18]
3 years ago
5

Gear A has a mass of 1 kg and a radius of gyration of 30 mm; gear B has a mass of 4 kg and a radius of gyration of 75 mm; gear C

has a mass of 9 kg and a radius of gyration of 100 mm. The system is at rest when a couple M0 of constant magnitude 4 N, m is applied to gear C. Assuming that no slipping occurs between the gears, determine the number of revolutions required for disk A to reach an angular velocity of 288 rpm. (Round the final answer to three decimal places.)
The number of revolutions required for disk A to reach an angular velocity of 288 rpm is ________ rev.
Engineering
1 answer:
Kruka [31]3 years ago
6 0

Answer:

(4.5125 * 10^-3 kg.m^2)ω_A^2

Explanation:

solution:

Moments of inertia:

I = mk^2

Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2

Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2

Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2

Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.

r_A=50 mm

r_1 =100 mm

r_2 =50 mm

r_C=150 mm

At the contact point between gears A and B,

r_1*ω_b = r_A*ω_A

ω_b = r_A/r_1*ω_A

       = 0.5ω_A

At the contact point between gear B and C.

At the contact point between gears A and B,

r_C*ω_C = r_2*ω_B

ω_C = r_2/r_C*ω_B

       = 0.1667ω_A

kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2

                           =(4.5125 * 10^-3 kg.m^2)ω_A^2

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Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?
lianna [129]

Answer:

Both B and G ( Hexagonal and Tetragonal )

Explanation:

The crystals system listed below has the following relationship for the unit cell edge lengths; a = b ≠ c ( hexagonal and Tetragonal )

hexagonal ; represents  a crystal system  which has three equal axes that have an angle of 60⁰ between them while Tetragonal denotes crystals that have  three axes which have only two of its axes equal in length.

5 0
3 years ago
An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s
nignag [31]

Given:

X_{L} = 50.0 \ohm

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

V_rms} = 85.0 V

Solution:

To calculate max current in inductor, I_{L(max):

At f = 60.0 Hz

X_{L} = 2\pi fL

50.0 = 2\pi\times 60.0\times L

L = 0.1326 H

Now, reactance X_{L} at f' = 45.0 Hz:

X'_{L} = 2\pi f'L

X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm

Now, I_{L(max) is given by:

I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}

I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A

Therefore,  max current in the inductor, I_{L(max) = 2.13 A

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I need ideas of usernames for a 2021 Jeep Wrangler Rubicon!!
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Answer:

2021 super star wagon master

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3 years ago
Compare the use of a low-strength, ductile material (1018 CD) in which the stress-concentration factor can be ignored to a high-
kicyunya [14]

Answer:

Step 1 of 3

Case A:

AISI 1018 CD steel,

Fillet radius at wall=0.1 in,

Diameter of bar

From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel

Tensile strength

Yield strength

The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.

At the critical stress elements on the top and bottom surfaces transverse shear is zero

Explanation:

See the next steps in the attached image

5 0
3 years ago
Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm
Simora [160]

Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow.  You can also see that R_{th} represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

  1. Wall Thickness.  More thickness, more thermal resistance.
  2. Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
  3. Cross-section Area. More cross-section area, less thermal resistance.

A expression to define  the thermal resistance for the wall is as follows:  R_{th} =\frac{l}{Ak}, where  l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.

5 0
3 years ago
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