Question

using an equation to find angular acceleration, angular velocity and theta.

Answer 1

angular acceleration: \[\alpha = \frac{\alpha_t}{r}\]angular velocity \[\omega = \frac{v}{r}\]Theta (angular displacement)\[\theta = \frac{s}{r}\]

Answer 2

The angle theta through which a disk drive turns is given by\[\theta (t)=a+bt-ct^3\]where a, b, and c are constants. t is in seconds and theta is in radians. when t=0, theta= pi/4 rad and the angular velocity is 2.30 rad/s and when 1.60s, the angular acceleration is 1.40rad/s^2. find a, b, c, angular acceleration when theta is pi/4, theta when angular acceleration is 4rad/s^2, angular velocity when angular acceleration is 4rad/s^2/

Answer 3

i found a, b, c to be; a=pi/4, b=2.30, c= -0.146

Answer 4

so then i'm a bit confused at how to progress to the next part.

Answer 5

ok a, b and c are fine

Answer 6

write out your expression for theta subbing gin the values for a, b and c

Answer 7

theta(t)=(pi/4)+2.3t+0.146t^3

Answer 8

ok... so then i take the derivative of this twice right?

Answer 9

ok for...angular acceleration when theta = pi/4..
means

pi/4 = pi/4 + 2.3t + .146 t^3...agreed?

Answer 10

ok yeah i see

Answer 11

ok...so solve for t please

Answer 12

ok give me a second.

Answer 13

i... kinda forgot how to solve for variables to the 3rd power so i just went wolframalpha which i know i shouldn't but i feel like i'm holding everyone up here. it gave me the answer t= -15.7

Answer 14

lets see...
pi/4 = pi/4 + 2.3t + .146t^3 so
0 = 2.3t + .146t^3 so
0 = t(2.3 + .146t^2)
which means the only value that makes physical sense is t = 0...agreed (t cant be negative)

Answer 15

oh right. unless we time travel i suppose.

Answer 16

and that solution was painfully obvious i'm feeling bad since i didn't see it the first time. lol. i'm up too early. :P

Answer 17

now that you know t =0 for theta = pi/4...what is your general expression to calculate
alpha(t)?

Answer 18

that is ok...

Answer 19

here is:
theta(t)=(pi/4)+2.3t+0.146t^3
we need
alpha(t) = ?

Answer 20

ok the general expression would be... alpha(t)=1.314t ????

Answer 21

lets see take derivatives...
theta(t)=(pi/4)+2.3t+0.146t^3 so
omega(t) = 2.3 + 3(.146) t^2 = 2.3 + .438t^2 so
alphat(t) = 2(.438) t = .876t

agree?

Answer 22

ok. i see. i don't know how i got my result. after looking at my work it would come to the right conclusion.

Answer 23

so when omega = pi/4 we found t = 0 ...so alpha = .876t = .876(0) = 0

Answer 24

so then i just repeat this for the other values? for alpha i get 0.69

Answer 25

oh for theta i'm going to have to go backwards.

Answer 26

oh. cripes. ok.

Answer 27

alpha = .876t = .876(0) = 0 ...not .69

so you have these two left to answer:

theta when angular acceleration is 4rad/s^2
angular velocity when angular acceleration is 4rad/s^2/

Answer 28

you should be able to do the last two questions...yes?

Answer 29

here are the equations you need:

theta(t)=(pi/4)+2.3t+0.146t^3
omega(t) = 2.3 + 3(.146) t^2 = 2.3 + .438t^2
alphat(t) = 2(.438) t = .876t

Answer 30

ok i was just working on finding theta. thanks

Answer 31

your welcome...TTYL

Answer 32

so then i used 4rad/s^2 in the 3rd equation and got 4.57s for t and just plugged that into the 1st and 2nd equations to find theta and omega. for theta i got 25.23 rad and for omega i get 11.45 rad/s

Answer 33

looks good!