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Naddik [55]
2 years ago
5

1. A flywheel is suspended by resting the inside of the rim on a horizontal knife edge so that the wheel can swing in a vertical

plane. The flywheel has a mass of 350Kg. The knife edge is parallel to and making 350mm from the axis of the wheel. The time for making one small oscillation is 1.77 seconds. Assuming that the centre of mass is in the axis of the wheel, determine:
(I).The radius of gyration of the wheel about the axis.
(II). The torque to increase the speed of the flywheel at uniform rate from 240 rev/min to 250 rev/min in 0.75seconds when the flywheel is revolving about its axis.
2. A solid metal cylinder 450mm diameter is suspended with its axis vertical by means of a wire coaxial with the cylinder and rigidly attached to it. The stiffness of the wire is 22Nm per radian of twist. Find the necessary mass of the cylinder so that when it is given a small angular displacement about its axis, it will make 40 vibrations per minute.​
Engineering
1 answer:
sammy [17]2 years ago
8 0

Answer: A fly wheel having a mass of 30kg was allowed to swing as pendulum about a knife edge at inner side of the rim as shown in figure.

Explanation:

You might be interested in
The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphe
Sonbull [250]

Answer:

a. The time required for the tank to empty halfway is presented as follows;

t_1   =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)

b. The time it takes for the tank to empty the remaining half is presented as follows;

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The total time 't', is presented as follows;

t =  \sqrt{2}  \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}

\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} =  \dfrac{v_2}{2 \cdot g}

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh

The time for the tank to drop halfway is given as follows;

\int\limits^{t_1}_0 {} \,  dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh

t_1  =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)

t_1   = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) =  { \dfrac{\sqrt{2}  \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)

t_1   =   { \dfrac{\sqrt{2}  \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)The time required for the tank to empty halfway, t₁, is given as follows;

t_1   =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

\int\limits^{t_2}_0 {} \,  dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh

t_2  =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The total time, t, to empty the tank is given as follows;

t = t_1 + t_2 =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} } =  \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}

t =  \sqrt{2}  \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }

3 0
2 years ago
Technician A says that 18 gauge AWG wire can carry more current flow that 12 gauge AWG wire. Technician B says that metric wire
denpristay [2]

Answer:

Technician B

Explanation:

Both AWG and metric are sized by cross-sectional area.

Technician A is wrong:  12 gauge wire is larger diameter rated for 20 amps in free air.  18 awg is smaller diameter and typically used for speaker wiring, Class II or low voltage and sub-circuits within appliances.

6 0
3 years ago
Where does Elizabeth want John to do and what does she want him to do there?​
ollegr [7]
What is the question?
8 0
2 years ago
The diffusion coefficients for species A in metal B are given at two temperatures:
Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

D = D_0e^{\frac{-Q_d}{RT} }

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

R = 8.314\ J/mol*K

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

And you should get D = 2.76*10^-16 m^/s as an answer for c)

5 0
3 years ago
Why will screws never replace nails​
cupoosta [38]

Answer:

because people have different opinions on nails and screws

Explanation:

3 0
3 years ago
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