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Harlamova29_29 [7]
3 years ago
11

A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at its center to a long fiber. When the disk is turn

ed from the relaxed state through a small angle theta, the torque exerted by the fiber on the disk is proportional to theta. Find an expression for the torsional constant k in terms of the moment of inertia I of the disk and the angular frequency w of small, free oscillations. The disk, when twisted and released, oscillates with a period T of 1.00 s. Find the torsional constant k of the fiber.

Engineering
1 answer:
topjm [15]3 years ago
6 0

Answer:

k = 1.91 × 10^-5 N m rad^-1

Workings and Solution to this question can be viewed in the screenshot below:

You might be interested in
You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from
Aleksandr [31]

Answer:

a) \frac{Ws}{Es}  = \frac{200}{1+1.2s}

b) attached below

c) type zero system

d) k > \frac{g}{200}

e) The gain K increases above % error as the  steady state speed increases

Explanation:

Given data:

Motor voltage  = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; \frac{K1}{1+St} be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= \frac{Ws}{Es}  = \frac{200}{1+1.2s}

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > \frac{g}{200}

7 0
3 years ago
Four of the minterms of the completely specified function f(a, b, c, d) are m0, m1, m4, and m5.
Sveta_85 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms  for f so that f has eight primary implicants with two literals and no other prime implicant are m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14} and m_{15}

b) The essential prime implicant are c' d',a'b',ab and cd

c) The minimum sum-of-product expression for f are

                  a'b' +ab +c'd'+cd+a'c',\\ a'b'+ab+c'd'+cd+a'd,\\a'b'+ab+c'd'+cd+bc'  and  \\ a'b'+ab+c'd' +cd+bd

Explanation:

The explanation is shown on the second third and fourth image

8 0
2 years ago
The temperature of a system rises by 10 °C during a heating process. Express the rise in temperature of K, R, and °F.
Lorico [155]

Explanation:

Given T = 10 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (10 + 273.15) K = 283.15 K

<u>T = 283.15 K </u>

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32  

So,

T (°F) = (10 × 9/5) + 32 = 50 °F

<u>T = 50 °F</u>

The conversion of T( °C) to T(R) is shown below:

T (R) = (T (°C) × 9/5) + 491.67

So,

T (R) = (10 × 9/5) + 491.67 = 509.67 R

<u>T = 509.67 R</u>

6 0
3 years ago
What's the difference between accuracy and percision in measuring and gaging?
lidiya [134]

Answer:

The term Accuracy means that how close our result to the original result.

Suppose we do any experiment in laboratory and we calculate mass = 7 kg but answer is mass = 15 kg then our answer is not accurate.

And the term Precision means how likely we get result like this.

Suppose we do any experiment in laboratory and we calculate mass five times and each time we get mass = 7 kg then our answer is precised but not accurate.

4 0
3 years ago
Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
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