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cricket20 [7]
3 years ago
13

A torsional pendulum consists of a 5 kg uniform disk with a diameter of 50 cm attached at its center to a rod 1.5 m in length. T

he torsional spring constant is 0.625 N-m/rad. Disregarding the mass of the rod, what is the natural frequency of the torsional pendulum?
Engineering
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

natural frequency of the torsional pendulum =  1.4 rad/s

Explanation:

given data

mass = 5 kg

diameter = 50 cm = 0.50 m

so radius R = 0.25 m

length =  1.5 m

torsional spring constant = 0.625 N-m/rad

solution

first we get here moment of inertia that is

moment of inertia = mR²  .........1

moment of inertia = 5 × 0.25²

moment of inertia  = 0.3125 kg-m²

so now we get here natural frequency of the torsional pendulum that is

natural frequency of the torsional pendulum = \sqrt{\frac{Kt}{I} }    ..................2

here Kt is torsional spring constant  and I is moment of inertia  

natural frequency of the torsional pendulum = \sqrt{\frac{0.625}{3125} }  

natural frequency of the torsional pendulum =  1.4 rad/s

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A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end
hjlf

Answer:

a) v=19.6 m/s

b) H=19.58 m

c) v_{f}=29.57 m/s  

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J

But we know the work is equal to change of kinetic energy, so:

W=\Delta K=\frac{1}{2}mv^{2}

v=\sqrt{\frac{2W}{m}}=19.6 m/s

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

v_{f}^{2}=v_{i}^{2}-2gh

At the maximum height the velocity is 0, so v(f) = 0.

0=v_{i}^{2}-2gH

H=\frac{19.6^{2}}{2*9.81}=19.58 m  

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

v_{f}^{2}=v_{i}^{2}-2gh

v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0
3 years ago
You have a 12-inch PVC water main that is 850 feet long flowing at 5.6 cfs. Point A is at an elevation of 750 ft. Point B is at
alex41 [277]

Known :

D = 12 in = 1 ft

L = 850 ft

Q = 5.6 cfs

hA = 750 ft

hB = 765 ft

PA = 85 psi = 12240 lb/ft²

Solution :

A = πD² / 4 = π(1²) / 4

A = 0.785 ft²

<u>Velocity of water :</u>

U = Q / A = 5.6 / 0.785

U = 7.134 ft/s

<u>Friction loss due to pipe length :</u>

Re = UD / v = (7.134)(1) / (0.511 × 10^(-5))

Re = 1.4 × 10⁶

(From Moody Chart, We Get f = 0.015)

hf = f(L / d)(U² / 2g) = 0.015(850 / 1)((7.134²) / 2(32.2))

hf = 10 ft

PA + γhA = PB + γhB + γhf

PB = PA + γ(hA - hB - hf)

PB = 12240 + (62.4)(750 - 765 - 10)

PB = 10680 lb/ft²

PB = 74.167 psi

8 0
3 years ago
Consider an ideal gas undergoing a constant pressure process from state 1 to state
Radda [10]

Answer:

s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy <em>at constant pressure</em>:

ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,

We obtain the expression to compute the specific entropy change:

s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})

Best regards.

6 0
4 years ago
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balu736 [363]

Answer: You're question is very vague... be more specific in the problem, what is it asking you to do?

Explanation: This shouldn't be under "Engineering" you should put it under "Mathematics" for a better result. Sorry for the mix up! Hope this helps! ^^

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3 years ago
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