Working Out average angular velocity help please!?

Question

Hello, im having great difficulty on this question, ive done A - D but can't do the last 2 E and F, any help/advice or links would be great!

A flywheel has a mass of 225kg and rotates at 3000 revs per minute in a pair of 120mm diameter bearings, with a coefficient of friction of 0.032.
Calculate :
a)The load on each bearing.
b)The frictional force exerted on each bearing
c)The work done per minute in overcoming friction
d)The total power Consumed per minute in overcoming friction.
The flywheel takes 30 seconds to reach the angular velocity stated. Calculate
e) The average angular acceleration
f) The linear distance that a point on the flywheel would travel in that time


A)
F = mg = 225 x 9.81 = 2207.25N
2207.25/2 = 1103.625N

b) = m x mg (m = 0.032)
= 0.032 x 1103.625
= 35.316N

c) Work done = Force x Distance
In one revolution – 35 x πD
= 35.316 x π x 0.12
= 13.31
At 3000rpm = 13.31 x 3000
= 39941.45 n/m

d) Power = 39941.45/60
= 665.6pw

Answer 1

These seem easier than the work you've done already. I hope there isn't something unstated that makes them more difficult.
e) mean ang acc = omega/t = 3000 rpm / 30 s = 100 rpm/s = 2*pi*100 rad/s
f) since, from the data given, we don't know if the ang acc is constant or not, we'll assume it's constant. Also assume the point is located at radius R.:
s = 0.5 * ang acc * t^2 * R
BTW, the units in (c) are N-m, not N/m, and in (e) I assume the "p" in pw is a typo.

Answer 2

e) Maybe your problem is that you misread the question. It asks for the average angular acceleration, not the average angular velocity as in your title.

Angular acceleration is the rate of change of the angular velocity. We know it ended up with an angular velocity of 3000 rev/min, it started from zero and took 30 seconds to reach this velocity. So the average angular acceleration would be change in velocity divided by the change in time.

f) This is a little confusing because they don't tell you the diameter of the flywheel and I assume that by linear distance, they mean the distance traveled around in a circle. So I would interpret this to mean that they want you to calculate the distance the point on the wheel has traveled around in a circle during that 30 second acceleration. The equations that you have used for linear constant acceleration problems work just as well for rotary with the exception that you have to be careful to convert the revolutions into distance.

So d = vi*t + .5 *a*t^2 will work, but if a is in rev/min then d is in revolutions, not linear distance. Convert it to distance by determining the circumference of the circle that the point traces out. They did not give you a radius to the point of the flywheel but you could just use r and keep that in the answer.

Answer 3

Taking in simple terms over 9 days to trip 24,987 miles around the Earth over the two poles. locate the common speed and standard angular speed of the voyager. I honestly have the solutions: fifty one.7 m/s and eight.1x10^-6 rad/s standard speed = finished distance ÷ finished time finished distance = 24,987 mi * 5280 feet/mi * 12 in/feet * 2.fifty 4 cm/in * a million m/one hundred cm = 4.02 * 10^7 m finished time = 9 days * 24hr/day * 60 min/hr * 60 sec/min = 7.776 * 10^5 s standard speed = 4.02 * 10^7 m ÷ 7.776 * 10^5 = fifty one.7 m/s Angular speed = radians according to 2nd Angular speed = tangential speed ÷ length of radius tangential speed = fifty one.7 m/s Radius of earth = 6.38 * 10^6 m Angular speed = fifty one.7 ÷ 6.38 * 10^6 = 8.a million * 10^-6 rad/s