Please help this is due today!!!!!

Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact di

Ilya [14]

Answer:

Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button, the motor that spins the disc slows down from 2000 rpm to 300 rpm and at 300 rpm a passive torsional spring-damper mechanism engages to decelerate and stop the disc. Here, your task is to design this spring-damper system such that the disc comes to rest without any oscillations. The rotational inertia of the disc (J) is 2.5 x 10-5kg m² and the torsional spring constant (k) is 5 × 10¬³NM. Calculate the critical damping coefficient cc for the system. choice of the damper, bear in mind that a good engineer stays at least a factor of In your 2 away from the danger zone (i.e., oscillations in this case). Use the Runge Kutta method to simulate the time dependent angular position of the disc, using the value of damping coefficient (c) that calculated. you Figure 1: Blue-ray disc and torsional spring-damper system.

5 0

3 years ago

Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kin

hammer [34]

Answer:

a. 149.74 KJ/KG

b. 97.9%

c. 0.81 kJ/kg K

Explanation:

8 0

4 years ago

A projectile is launched vertically with a velocity of 30 mxs . How long will it take to return to the original launch position?

rewona [7]

Answer: If the projectile is launched vertically, then you only aply velocity on the y axis.

The velocity 30 m/s is the initial velocity and if it is fired on the ground, then the initial position is 0m (this doesn't really matter), and now, let's analyze the forces on the projectile.

Once it is fired, the only force acting on the projectile is the force of gravity, and we know that the gravity acceleration is $-g = -9.8\frac{m}{s^{2} }$ , where the negative sign is there because this acceleration points downward.

so A(t) = $-9.8\frac{m}{s^{2} }$

For the velocity, we need to integrate the acceleration in the time, this is:

$v(t) = -9.8\frac{m}{s^{2} }*t + v0$

where v0 is the initial velocity, in this case 30m/s

And for the position, we need to integrate again, so:

$r(t) = -4.9\frac{m}{s^{2} }*t^{2} + 30\frac{m}{s} *t + r0$

where r0 is the initial position, in this case 0m.

now the question is "How long will it take to return to the original launch position?"

So now we need to find the time in which r(t) is zero again (so the projectile is in the ground again

so r(t) = 0 = $r(t) = -4.9\frac{m}{s^{2} }*t^{2} + 30\frac{m}{s} *t + r0$

this is: r(t) = t*(-4.9*t + 30) = 0

so is easy to see that t = 0 (because it is fired in the ground) is a solution, but is not the one that we are looking for,

so we only look inside the parentheses:

-4.9*t + 30 = 0

t = 30/4.9 = 6.12 s

So 6.12 seconds after the projectile is fired, it returns to the ground, or the original launch position.

6 0

4 years ago

A brittle failure has extensive plastic deformation in the vicinity of the advancing crack. This process proceeds relatively slo

Tomtit [17]

Answer:

False ( b )

Explanation:

In a brittle failure the cracks spreads rapidly without a significant deformation, and the cracks are very unstable with the cracks extending without an increase in the amount of applied stress.

Therefore the above description in the question is false.

3 0

3 years ago

Why is it a good idea to lock your doors while driving?<br> WRITER

inysia [295]

Answer: to avoid any accidents from happening, to ensure safety

Explanation:

3 0

4 years ago