angular velocity and revolutions problem?

Question

At t=0 the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by theta(t) = (A)t - (B)t^2 - (C)t^3.

(a) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero?

(b) Calculate the average angular velocity for the time period from t=0 to the time when the angular velocity of the motor shaft zero.

Thanks for the help, sorry it's such a long problem!

Answer 1

theta(t) = At - Bt^2 - Ct^3 -------(1)

angular velocity =rate of change of angle (theta)
w(t) = d(theta)/dt differentiating (1)
w(t) = A - 2Bt - 3Ct^2 -------(2)

(a) In given time interval if "delta (theta)" angle is covered by motor then its ROVOLUTIONs equivalent can be obtained by :

revolutions (R) = (1/2pi) delta (theta) -----(3)

delta (theta): it is to be calculated for time from t=0 (I reversed) to t=t (when w=o in (2)). The former (t=0) is Zero from (1).

delta(theta) = theta(t=t, w=0) - 0 = theta(t=t)

we find out "t" >> w(t) =0= A - 2Bt - 3Ct^2

3Ct^2 +2Bt - A = 0 ----(4)

t = (B/3C)[SQRT{1+(3AC/B^2)}] Negative time left out ---(5)

Also t^2 =(B^2/9C^2) [2+(3AC/B^2) - SQRT{1+(3AC/B^2)}] (6)
put in (1)

theta(t) = At - Bt^2 - Ct^3
= (2A/3) t - (B/3) t^2 using (4)

put t and t^2 from (5) and (6)

theta = [6ABC+2B^3) / 27C^2] SQRT{1+(3AC/B^2)} -

..........- [9ABC+2B^3) / 27C^2] .......(7)

ROVOLUTIONS = EQU(7) * (1/2pi)
answer
-------------------
b) average angular velocity is

w(av) = [w(t=0) + w(t=t)] /2,

for t = 0, w(t) = A
for t = t (w=o). we calculated t=t (w=0) as in (5), if we again try to work out w (t) at that very instant then w=o

or w(t)=o >> t=t >> w(t) =o

so average angular velocity is

w(av) = [A - 0] /2 = (A/2)
answer

Answer 2

Open the e book (or google it, or in basic terms open wikipedia) and you will see what you like. i think which you like angular velocity i.e. angular frequency and the appropriate formula is a normal multiplication v=rw the place v is linear speed (m/s) and w (genuinely it particularly is omega) and it represents frequency (a million/s), or radians according to 2nd (rad/s), see on your e book. So, the 2nd activity you ask approximately is the hassle-free conversion from w in revolutions according to 2nd to v in cm/s, (no longer m/s). One revolution is two*pi radians. in the 1st activity you're given the speed of the truck this is genuinely v of the tires. you have the radius, so which you will get w in rad/s. considering the fact which you like revolutions according to 2nd, in basic terms divide the end result by technique of two*pi. base line, for the 1st activity you calculate w=v/(2*pi*r), v=60miles/h, r=26 inches, and for the 2nd activity you calculate v=2*pi*r*w, r=4.6 cm, w=3.

Answer 3

For part a) :

theta(t) = At - Bt^2 - Ct^3

d(theta) / dt = w(t) >>> angular velocity

w(t) = A - 2Bt - 3Ct^2

for t = 0, theta = 0

for t = 0, w(t) = A

w(t) = 0, for A - 2Bt - 3Ct^2 = 0

3CT^2 +2Bt - A = 0

t = - 2B + sqr(4B^2 + 12A) / 6C

that's the instant when w = 0

replace "t" in theta, so well, just operate it.

the result will be :

theta(t) = A[ - B +sqr( 4B^2 + 12A) / 6C] - B[ - B +sqr( 4B^2 + 12A) / 6C]^2 - C[- B +sqr( 4B^2 + 12A) / 6C]^3

b) w(t), for t = 0, w(t) = A

the average angular velocity, I think it's A/2.

A/2 rad/s