Physics, 23.11.2019 03:31 twhalon72

You are designing a rotating metal flywheel that will be used to store energy. the flywheel is to be a uniform disk with radius 23.0 cm. starting from rest at t = 0, the flywheel rotates with constant angular acceleration 3.00 rad/s2 about an axis perpendicular to the flywheel at its center. if the flywheel has a density (mass per unit volume) of 8600 kg/m3, what thickness must it have to store 800 j of kinetic energy at t = 8.00 s?

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We have that for the Question, it can be said that thickness must it have to store 800 J of kinetic energy at t = 8.00 s

h=0.0735m

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You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 23.0 cm. Starting from rest at t = 0, the flywheel rotates with constant angular acceleration 3.00 rad/s2 about an axis perpendicular to the flywheel at its center.If the flywheel has a density (mass per unit volume) of 8600 kg/m3,what thickness must it have to store 800 J of kinetic energy at t = 8.00 s?

Generally the equation for the is mathematically given as

$N=\pir^2h*P\\\\N=3.14*(23*10^{-2}^2)*h*8600\\\\N=1428.5h\\\\$

Therefore

$E=\frac{1}{2}Iw^2\\\\800=\frac{1}{2}*(\frac{NR^2}{2}w^2)\\\\800=\frac{1}{2}*(\frac{(1428.523*10^{-2})^2}{2}(2*8)^2)$

h=0.0735m

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Answer from: Bhoom1871

t = 0.0735 m

Explanation:

Angular acceleration of the flywheel is given as

$\alpha = 3 rad/s^2$

now after t = 8 s the speed of the flywheel is given as

$\omega = \alpha t$

$\omega = 3 \times 8$

$\omega = 24 rad/s$

now rotational kinetic energy of the wheel is given as

$K = \frac{1}{2}I\omega^2$

$K = \frac{1}{2}(\frac{1}{2}mR^2)(24^2)$

$800 = \frac{1}{4}m(0.23)^2(24^2)$

m = 105 kg

now we have

$m = \rho (\pi R^2) t$

$105 = 8600(\pi \times 0.23^2) t$

t = 0.0735 m

Answer from: Quest

pe = weightxheight = 2000x1000=2,000,000joules

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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