Rotational Motion Questions - Please Help

[porschedriver192]

I have been stuck on these forever...but here goes:

1st question:

A tire placed on a balancing machine in a service station starts from rest and turns through 5.5 revs in 1.0 s before reaching its final angular speed. Assuming that the angular acceleration of the wheel is constant, calculate the wheel's angular acceleration.

I know that you have to convert 5.5 rev into pi rad, but I keep getting the wrong answer (this is on an online HW assignment).

work:

formula aavg = (w2-w1)/(t2-t1)

5.5 x 2 pi = 34.5575 pi rad

(34.5575 - 0) / (1.0 s) = (I get it wrong here)
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2nd question (this one makes me angrier!)

The Emerald Suite, a revolving restaurant at the top of the Space Needle in Seattle, Washington, makes a complete turn once every hour. What is the tangential speed of a customer sitting 16.0 m from the restaurant's center?

Ok, I know the formula and setup, but am getting stuck on converting the rotation of the restaurant into what I need: is it rev/s, rev/min? And then do I multiply by 2 pi?

The formula is this:

at = ra

my work: 1 rev/hr = .0166 rev/min = 2.766e-4 rev/sec

2.766e-4 x 2 pi =

.0017 rad x 15 s = .0255 m/s (wrong answer it says)

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Any help would be greatly appreciated - thank you!

[Skomatth]

Question 1:

5.5 rev is the angular displacement the wheel goes through. You seem to have used it as angular velocity. Think of a different kinematics equation to use here.

Question 2:

You are using the equation for tangential acceleration but the restaurant isn't accelerating! The formula you need here is v=rw

[porschedriver192]

Thanks for the reply, I was eventually able to work through it and get my answer! It turns out that I had the formula wrong.
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This is similar to the second question I had, but they want the centripetal accel.

The Emerald Suite, a revolving restaurant at the top of the Space Needle in Seattle, Washington, makes a complete turn once every hour. What is the magnitude of the centripetal acceleration of the customer sitting 15.0 m from the restaurant's center?

I first found the avg disp: (1/60/60 x 2 pi) / (15) = 1.163e-4

then the avg angular speed: 1.163e-4/3600 seconds (this may be my problem)

= 3.232e-8

centripetal accel: a= rw^2

15 x 3.232e-8 = answer that it said is wrong
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Second question, going back to my orininal post, I cannot find the answer for the tire. (If you need the problem, it is in my first post)

The question asks for avg. angular accel,

so I did a avg = w2-w1 / t2-t1
5.5 (x 2pi) - 0 / 1 sec - 0 = 34.55 (which it said is wrong)

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Any help would again be appreciated. Thanks.

[Skomatth]

I think you have the right idea, but there's a much easier way.

Find the speed by considering the circumference and period in the form of distance=rateXtime

Put that into a=v^2/r

The problem in the first question:

5.5 rev is the angular displacement. 5.5 x 2pi is also displacement. Dividing this by time will just get you angular velocity. Recall some of your basic kinematic equations (list them here so I can see if you're missing one), and find one that uses displacement, acceleration and time.

[porschedriver192]

Thanks for the reply, but I am still unsure as what to do:

the first (here's the question again)

The Emerald Suite, a revolving restaurant at the top of the Space Needle in Seattle, Washington, makes a complete turn once every hour. What is the magnitude of the centripetal acceleration of the customer sitting 15.0 m from the restaurant's center?

Do I assume that the diameter is 15 m? And then do I do pi15^2 for the circumference? I am still unsure about the rate though.
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The second: these are my formulas so far (which would relate to this problem):

ang disp = change in arc lenth / distance from arc length

(I think this one is it!? but how do I get the disp from the given info?)
>>>avg angular speed = ang disp / time interval

avg angular accel = change in angular speed / time interval

(I also have the tangental ones, but they don't relate to this problem).
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Thanks for your continued help, I am looking forward to solving these problems!

[Skomatth]

The radius is 15m which means you can find the linear distance covered with C=2piR (formula for circumference). You also have the time it takes to cover this distance. Which should make it easy to find the speed of a person on the edge.

s=vt
v=vi + at
s= s_i + v_i*t + .5at^2
v^2=v_i^2 + 2as

Remember these equations from the kinematics you probably did at the beginning of the year? They can also be written for angular values.

[Skomatth]

s = v t



v = v_i + a t \ \


s = s_i + v_i t + \frac{1}{2} a t^2 \ \


v^2 = v^2_i + 2 a s

[porschedriver192]

Thanks again for sticking with me, but I am still pretty confused. I'll just ask my teacher to walk me through the steps to see where I messed up. Thanks though.