using sinusoidal functions and radians to solve the following

A sine wave has a maximum value of 100v and a frequency of 50hz. Determine the instantaneous voltage present at
(a) 2.5ms and (b) 15ms from the start of the cycle. You must use the wave function v=Vmax sin(2 pi ft)

An aircraft sits on a runway ready for take off. It has 1.4m diameter wheels and accelerates uniformly from rest to 225 kmh (take off speed) in 40 seconds. Determine the angular acceleration of the undercarige wheels and the number of revolutions made by each wheel during take off run.

can anyone help me with these questions please

thanks

Andy

Did you try the first one? You have the formula given to you:



You can complete it from here.

For the second, use the relationship that the linear velocity at the rim of the wheel is related to its angular velocity by:



You are given v at take off, so you can calculate the angular velocity of the wheel at that point. Then you can calculate the average angular acceleration :



and the number of rotations is determined by dividing the distance the plane covers in those 40 seconds by the circumference of the wheel. Post back and tell us what you find for an answer.

thanks for your help

1a.
100 x sin(2 pi x (50 x o.oo25)) = 1.370

1b.
100 x sin(2 pi x (50 x 0.015)) = 8.215

I'm not sure if these are write.

I still don't understand the second question though could you do an example using difernt values.

thanks andy

also how are you using the symbols like pi are they on your keyboard or have you downloaded something.

Hello Bayley86.

For both 1a and 1b, check your calculator - I think you calculated the sine values using degress rather than radians! You should see that sin(2 pi*50*.0025) = sin (.25* pi) = sin(pi/4), which you should recognize as sqrt(2)/2. And here's a hint - the answer to 1b is a whole number.

As for the second problem, I'll start you out on an analogous problem. Consider a car with wheels 0.5 meters in diameter, that accelerates from a standing start to 100 m/s in 20 seconds. OK - it's linear acceleration is 100m/s/(20 sec) = 5 m/s^2. The angular aceleration of the wheel is 5m/s^2 divided by 0.5 m = 10 rad/s^2. You could also get that same answer by noting that the angular velocity of the wheel when the car reaches 100 m/s is 100 m/s divided by the radius of the wheel , or 200 rad/sec. Since it takes 20 seconds to spin up to that velocity, its angular acceleration is 200 rad/sec divided by 20 sec = 10 rad/ sec^2.

Hope this helps. I suggest you try these same principals with problem 2, and post back what you get. If you have difficulty please show us what you tried and how far you got.

ebaines:

Slap on the wrist. Don't teach bad habits.. Don't EVER write .005. Always write 0.005. Countless times I lost points for that.

Units abbreviation for seconds is s; In this problem since it involves trig, best not to throw in sec for the Secant function. I like sec. too.

OP:
I gueess you realize that the first problem should have been cake. Frequency is in cycles/s and 2*PI is the number of radians in 360 degrees. So, the only real issues are units.

ms = s/1000 and the radians and degree problem. It's always wise to cleck your calculator, because you should know, for instance the cos(180 deg) and the cos(PI) being near 1. Use this as a method of checking the calculator, or Excel for that matter.

Quote:

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Originally Posted by KeepItSimpleStupid

ebaines:

Slap on the wrist. Don't teach bad habits.. Don't EVER write .005. Always write 0.005. Countless times I lost points for that.

Units abreviation for seconds is s; In this problem since it involves trig, best not to throw in sec for the Secant function. I like sec. too.

OP:
I gueess you realize that the first problem should have been cake. Frequency is in cycles/s and 2*PI is the number of radians in 360 degrees. So, the only real issues are units.

ms = s/1000 and the radians and degree problem. It's always wise to cleck your calculator, because you should know, for instance the cos(180 deg) and the cos(PI) being near 1. Use this as a method of checking the calculator, or Excel for that matter.

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And for a second, I thought that ebaines introduced the cosecant function in there :eek: Then, reading on the second time, I realised it was seconds.

EDIT: oops, sorry, I meant secant :o And as I just read your second post here, you initially said 0.5 m in diameter and then used 0.5 as a radius.

Boy - you guys are harsh! Cosecant? Glad you didn't go down that path. And I thought I was being pretty smart understanding that when the OP wrote "sin(2 pi ft)" that he wasn't talking about "pi-feet"!

hi

1a. =70.71
1b. =-100

2a.
first I converted 225 kmh into meters a second

225 x (5/18) =62.5

62.5/0.7 =89.285
89.285/40 =2.232rad/s^2

2b.
circumference of wheel = 1.4 x 3.142 =4.3988

62.5 x 40 =2500
2500/4.3988 =795.672 revolutions

these are the answers I got

thanks again

andy

Hmm... all is good except the last one.

The speed of the plane varies as it accelerates. So, you cannot use the formula here but the distance formula with acceleration included, where s is the displacement, a the acceleration and t the time.

Care to try again? :)

The question dose say it accelerates uniformly though or is that different

The formula you used is applicable only when speed is constant. Since you have acceleration, the distance is altered.

Say, I travel for 10s at 10 m/s. I cover 100 m.
Now, I travel with constant acceleration of 1 m/s, for 10 s and thus reaching a maximum speed of 10 m/s. I do not cover 100 m because, I started slow, at 1 m/s at the first second, then 2 m/s at the second second, etc. Hence, in 10 seconds, I cover less distance. In fact, I cover

Is that OK?

is the constant acceleration
40/62.5=0.46
s=1/2(0.64)(62.5)^2=1250
1250/4.3988=284.168 revolutions

Quote:

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Originally Posted by bayley86

is the constant acceleration
40/62.5=0.46
s=1/2(0.64)(62.5)^2=1250
1250/4.3988=284.168 revolutions

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That's the right answer, but please, please show units as you do your work! It took me several minutes to understand why you started with 40/62.5. I finally figured out that you used the formula v^2 = 2as which you rearranged into s = 1/2(1/a)v^2. So the 40/62.5 is really 1/a.

is that definitely write or should it be
s=1/2(0.64)(40)^2=512
512/4.3988=116.395 revolutions

if the formula is s=1/2at^2 is the t time or speed

Quote:

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Originally Posted by bayley86

or
s=1/2(0.64)(40)^2=512
512/4.3988=116.395 revolutions

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Wrong - this is why I implore you to write out the units! Please tell us what the number 0.64 represents in your equation. Hint - it does NOT have units of m/s^2 !

To be honest I don't no what the units are I've confused myself now

OK. Consider the equation s = 1/2 a t^2. The value of s (position) is in meters, a (acceleration) is in meters per second squared, and t (time) is in seconds. So to properly calculate s you need to make sure that you have a in terms of m/s^2. Now, you before you used the fact that acceleration can be calculated by dividing the final velocity by the amount of time it takes to get from a standing start to that velocity: a = v/t. the units for v are m/s. So you have:



Now put this into s = 1/2 a t^2:



Do you see how the units work out so that s is ultimately expressed as meters?

Now divide 1250 meters by the circumference of the wheel to find out how many times it revolves:

Thank you that makes sense now I was confused because I put the speed in the formula by accident instead of the time but the answer was still write

An English correction for you: "answer was right" not wriite. See Dictionary.com | Find the Meanings and Definitions of Words at Dictionary.com

Quote:

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Originally Posted by bayley86

thankyou that makes sense now i was confused because i put the speed in the formula by accident insted of the time but the answer was still write

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Well, you made two mistakes at once that coincidentally corrected the first mistake.

1st mistake : you didn't find acceleration, but the inverse of acceleration
2nd mistake : you substituted the speed instead of the time.

Ok, that's how I do it:

a) Circumference of wheel =

Final linear speed = 225 km/h = = 62.5 m/s

Now, rotational speed in rev/s = = 14.2 rev/s

Rotational acceleration = = 0.355 rev/s^2

In radians (1 rev = 2pi rad) =

b) Now, I'll do some 'transposition' of the formula .

I'll substitute distance by the total amount of revolution and acceleration into rotational acceleration in revolutions.

Thus; Number of rotations =