5

The angle through which rotating wheel has turned in time given by 0 = a t bt2+ct4 where H is in radians and t in seconds_Evaluate a at t 3.3 $Express your answer u...

Question

The angle through which rotating wheel has turned in time given by 0 = a t bt2+ct4 where H is in radians and t in seconds_Evaluate a at t 3.3 $Express your answer using two significant figuresAZdrad/s2SubmitRequest AnswerPart EWhat is the average angular velocity between t 2.0 and 33.3Express your answer using two significant figures_Azdrad/8SubmitRequest AnswerPart FWhat is the average angular acceleration between =2.0 and =3.3 8Wavk

The angle through which rotating wheel has turned in time given by 0 = a t bt2+ct4 where H is in radians and t in seconds_ Evaluate a at t 3.3 $ Express your answer using two significant figures AZd rad/s2 Submit Request Answer Part E What is the average angular velocity between t 2.0 and 33.3 Express your answer using two significant figures_ Azd rad/8 Submit Request Answer Part F What is the average angular acceleration between =2.0 and =3.3 8 Wavk



Answers

(II) The angle through which a rotating wheel has turned in time $t$ is given by $\theta=8.5 t-15.0 t^{2}+1.6 t^{4},$ where $\theta$ is in radians and $t$ in seconds. Determine an expression $(a)$ for the instantaneous angular velocity $\omega$ and $(b)$ for the instantaneous angular acceleration $\alpha .(c)$ Evaluate $\omega$ and $\alpha$ at $t=3.0 \mathrm{s}$ . (d) What is the average angular velocity, and $(e)$ the average angular acceleration between $t=2.0 \mathrm{s}$ and $t=3.0 \mathrm{s} ?$

All right. The position angular position is given by for t minus three T squared plus t cubed. What's the velocity at T equals zero. Where the velocity is going to be D data over D. T four minus 60 plus three T squared. Taking the derivative when t equals two or minus 12 plus four times three is 12 which is four um Not meters radiance per second. Okay, that's the answer to Part A B equals four. Six times four is 24. Four squared is 32. Four squared is 16. 16 times three. 48. 48 minus 24 is 24 plus two is 28. See was the average acceleration over that two second range? Well the average acceleration going to be the change in angular velocity should be 28 minus four over the amount of time which was two seconds. What are the instantaneous angular acceleration? So taking the derivative of the angular velocity I get where is it? It's up here. So at the beginning which is T equals two, six times two is 12. It's going to be six, and at the end four, six times four is 24 18. Yeah. Thank you for watching.

So we can say that the angular displacement in terms of time is equal to four T minus three t squared plus t cubed for party. We want to find the angular displacement in terms of time. This would simply be equal to derivative of the We want to find the angular velocity in terms of time that this would be equal to the derivative of the angular displacement. With respect to time, this would be equal to four minus six t plus three t squared. And we can say that the angular philosophy AT T equals 2.0 seconds. This would be equal to four minus six times 2.0 plus three times 2.0 squared. This is giving us 4.0 radiance for a second. It's also this would be measured in radiance For part B. The angular velocity AT T equals four seconds would be equal to four minus six times 4.0 plus three times 4.0 squared. This is equaling approximately 28 0.0 radiance for a second. Uh, for part C, we want to find the average angular acceleration between two and four seconds so we can say that the angular velocity at four seconds minus the angular velocity at two seconds, divided by four seconds minus two seconds. This would equal the average acceleration. Theat bridge, Angular acceleration between this time interval. This is giving us We referenced part A and part B. So we have 28 minus four, divided by four minus two. This is giving us approximately 12 radiance per second squared four part D. We can then say that the angular acceleration is gonna be the instantaneous, angular acceleration would be equal to the derivative of the angular velocity with respect to time. So this is going to be equal to negative six plus six T and so we can evaluate the angular acceleration at T equals two seconds. This is equaling negative. Six plus six times two. This is giving us 6.0 radiance per second squared radiance for second. Screw it, of course, being the units for the angular acceleration. And lastly for part E, we want to be about when we evaluate the expression in part D. At T equals four seconds. We can then say that the angular acceleration AT T equals four seconds with the equal to negative six plus six times four. This is giving us 18 radiance per second squared. And then we can say, essentially, we're noting that our answer for the, uh, hang for the average angular acceleration, uh, does turn out to be the earth arithmetic average of a sub to, uh, our other alphas up to an officer before, Um, But this you know, this, of course, won't always be the case. So we can say that essentially, um, 18 plus six divided by two. This is equal in 12 radiance per second squared. And we note that this is giving us the same answer as theano. Sir. However, just note that this isn't always the case. Um, this is just based on the equation that they gave us initially here. So in this case, it does. The average angular acceleration does equal the average angular acceleration when calculating it from the angular velocity is, however again, this is not always the case, and you do have to take the derivative in order to find the instantaneous, angular acceleration that is the end of the solution. Thank you for watching

Mhm. This problem covers the concept of the can dramatic question for the traditional motion and this. To solve this problem first, we need to write the expression for the instantaneous velocity and this instantaneous acceleration. So by taking the time derivative, the instantaneous angular velocity, omega is for radiant four seconds plus six radiant All 2nd square into T plus three Radiance or 2nd Q into the square. That said, this is a question. And the expression for the analytic solution alpha is uh six gradients four, second square plus six radiant all second cube In two G. Let's say this is a question too. Now, for part a the angular velocity at Stephen equals from the equation one that we have derived for the uh and global ah city, you can substitute the value of Stephen into that equation. So by substituting four radian or second plus six radiant four seconds square Into T. and T. one equals two seconds less. Three radiant four second cube Into the square and there is two seconds square. So the travelocity two seconds comes out to me 28 radiance for second. No party. The anglo velocity at T. Two that is at four seconds equals from the same equation. We can write for radiance or second plus six radiance four seconds square into four seconds plus three radiance or 2nd square, sorry for a second. Q into four seconds square. So the angular velocity at the key to that is four seconds comes out to be uh scientist six radiance for a second. No party. The average acceleration there is the original acceleration is the changing uh angular velocity upon the changing tank. And between T. Two and T. Even the change in angular velocities 76 minus 28 radiance four seconds upon the changing time is full minus two seconds. So the average acceleration is uh 18. Sorry, every acceleration is 24, 24 radiance four seconds squared, no party the instantaneous acceleration alpha act even that is a two seconds equals from the second equation. We can write six radiant four seconds squared plus sixth radiant or second Q into no second. Are the angular instruments and good acceleration at even equals two secondaries 18 Radiance or 2nd square. Now part E. The angular acceleration alpha at T two equals six radiance four seconds squared plus six radiance four second cube Into that IMT two. And that is four seconds. So the instantaneous angular acceleration at T two equals 30 radiance for a second.

All right. The angular position is given by two plus. For he squared plus to t cubed. What's the angular position at? T. Equals zero. Well when T equals zero that's zero that's zero. You're just left with two. Oops radiant. Okay. Be with the angular velocity the angular velocity B. D. Theta over DT which is just going to be eight T. Just taking the derivative plus 60 square um At zero zero. See was the angular velocity when he is four. Just gonna be 32 eight times 4. 32 plus. Four times. I mean six times um four squared which is 16 six times 16 96 128 radiance per second. D. With the angular acceleration at T. Equals two. Well the acceleration is going to be the derivative of the angular velocity which is just gonna be eight plus. Well t when he is too going to be eight plus 12 times two is 24. 32. Radiance per second squared? Is its acceleration constant? No thank you for watching.


Similar Solved Questions

5 answers
Determine Polar ncutral whether cach 1 amino acids Hydrophobic polar neutral 1 V chargel N 0
Determine Polar ncutral whether cach 1 amino acids Hydrophobic polar neutral 1 V chargel N 0...
5 answers
Given acceleration vector of an object is a(t) = 6t i + (2t + 1)j - 12k, with initial velocity vector at time equals zero is v(0) = -2i + 4k and the initial position vector is r(0) = ~i + 2j, determine the velocity vector as function of ii_ speed of the object at time t = 2 iii. position vector at time t
Given acceleration vector of an object is a(t) = 6t i + (2t + 1)j - 12k, with initial velocity vector at time equals zero is v(0) = -2i + 4k and the initial position vector is r(0) = ~i + 2j, determine the velocity vector as function of ii_ speed of the object at time t = 2 iii. position vector at ...
5 answers
+y -25Describe and sketch the domain of the function f6y)=
+y -25 Describe and sketch the domain of the function f6y)=...
5 answers
The following position vs. time graph for a simple pendulum near the Earth'$ surface is (in SI units): s(t) = 0.182 cos(2.85t + 1.66)pts) Find the length of the pendulum. pts) Find the maximum acceleration of the pendulum_
The following position vs. time graph for a simple pendulum near the Earth'$ surface is (in SI units): s(t) = 0.182 cos(2.85t + 1.66) pts) Find the length of the pendulum. pts) Find the maximum acceleration of the pendulum_...
5 answers
5 L 2 ue} n 1.9 n= Mm n
5 L 2 ue} n 1.9 n= Mm n...
5 answers
6. Let f (x,y) = er ~ Find all the second partial derivatives.
6. Let f (x,y) = er ~ Find all the second partial derivatives....
5 answers
Make up a table with columns headed by values of $mathrm{kx}$ nunning from $x=-lambda 2$ to $x=+lambda$ in intervals of $x$ of $lambda / 4$ of course, $k=$ 2m A. In each column place the corresponding values of $cos (k x-$ m 4 ) and beneath that the values of $cos (k x+3 pi / 4)$. Next plot the functions $15 cos (k x-pi) 4}$ and $25 cos (k x+3 pi .4)$
Make up a table with columns headed by values of $mathrm{kx}$ nunning from $x=-lambda 2$ to $x=+lambda$ in intervals of $x$ of $lambda / 4$ of course, $k=$ 2m A. In each column place the corresponding values of $cos (k x-$ m 4 ) and beneath that the values of $cos (k x+3 pi / 4)$. Next plot the func...
5 answers
Solve each equation.$$ rac{r}{r^{2}+8 r+15}- rac{2}{r^{2}+r-6}= rac{2}{r^{2}+3 r-10}$$
Solve each equation. $$ \frac{r}{r^{2}+8 r+15}-\frac{2}{r^{2}+r-6}=\frac{2}{r^{2}+3 r-10} $$...
5 answers
1 1 1 Mecharnt 8 Mechanism 8 Mechanism sm anism 2020-09 2020-09 2020-092 3 Bud WdJdd (HUOIDEuJ
1 1 1 Mecharnt 8 Mechanism 8 Mechanism sm anism 2020-09 2020-09 2020-09 2 3 Bud Wd Jdd (HUOIDEuJ...
1 answers
Find the values of $a$ and $b$ that make the following function differentiable for all $x$ -values. $$f(x)=\left\{\begin{array}{ll}a x+b, & x>-1 \\b x^{2}-3, & x \leq-1\end{array}\right.$$
Find the values of $a$ and $b$ that make the following function differentiable for all $x$ -values. $$f(x)=\left\{\begin{array}{ll}a x+b, & x>-1 \\b x^{2}-3, & x \leq-1\end{array}\right.$$...
5 answers
Simplify.$$i^{42}$$
Simplify. $$i^{42}$$...
5 answers
Suppose that two carts, one twice as massive as the other, fly apart when the compressed spring that joins them is released. What is the acceleration of the heavier cart relative to that of the lighter cart as they start to move apart?
Suppose that two carts, one twice as massive as the other, fly apart when the compressed spring that joins them is released. What is the acceleration of the heavier cart relative to that of the lighter cart as they start to move apart?...
1 answers
For the following exercises, the rectangular coordinates of a point are given. Find two sets of polar coordinates for the point in $(0,2 \pi] .$ Round to three decimal places. $(-6,8)$
For the following exercises, the rectangular coordinates of a point are given. Find two sets of polar coordinates for the point in $(0,2 \pi] .$ Round to three decimal places. $(-6,8)$...
5 answers
Using the Bohr quantization condition, considering only circularorbits, find the energy levels of a particle of mass m moving inthe potential energy field U(r) = 6 r^4 (b is a positive realconstant). Express your answer in terms of m, b n and h.
Using the Bohr quantization condition, considering only circular orbits, find the energy levels of a particle of mass m moving in the potential energy field U(r) = 6 r^4 (b is a positive real constant). Express your answer in terms of m, b n and h....
5 answers
The equlpotentla Ilnes reqion of electric fleld are shown the dlagram below; along- that path? Here Vo +190each path Indlcated below, what the work done bY the eCtric fleldmovinacharge+9,7* 10-73Vo ZVo3Vot(a) from_ WAB(b} from Wac(c} from WAD(d) from Wpc
The equlpotentla Ilnes reqion of electric fleld are shown the dlagram below; along- that path? Here Vo +190 each path Indlcated below, what the work done bY the eCtric fleld movina charge +9,7* 10-7 3Vo ZVo 3Vot (a) from_ WAB (b} from Wac (c} from WAD (d) from Wpc...
5 answers
Y (ucte?} eler farmey lans nce p 6 Postr audlyacen} Jo Fivr Con-kzun 245 000 08 Jc Thc pasuvc 7usy Fuve Tov #h< herd Povide 2mo Ugh pesWkay chmrnions dcl li rn |mun 7mov ' { ~fancng /5 na 0f fzncimg) reen € nec(sSarY lon) Khe ritr
Y (ucte?} eler farmey lans nce p 6 Postr audlyacen} Jo Fivr Con-kzun 245 000 08 Jc Thc pasuvc 7usy Fuve Tov #h< herd Povide 2mo Ugh pes Wkay chmrnions dcl li rn |mun 7mov ' { ~fancng /5 na 0f fzncimg) reen € nec(sSarY lon) Khe ritr...
5 answers
3 One mole of an ideal mono atomic gas absorbs 5 x 103 J of heat while doing 2 x 10' J of work on the environment What is the change in the temp of the gas in Kelvin?
3 One mole of an ideal mono atomic gas absorbs 5 x 103 J of heat while doing 2 x 10' J of work on the environment What is the change in the temp of the gas in Kelvin?...

-- 0.018959--