Angular Acceleration/Velocity

  • #1

Homework Statement



A laboratory apparatus consists of a low friction, low mass turntable which carries a large, flat, round, solid disk. The disk is 10.0 inches or 25.4 cm in diameter and has a mass of 4.34 kg. The turntable base turns on a cylinder 1 7/8 inches (4.76 cm) in diameter. A string is wrapped around the base cylinder (hub) of the turntable. The string passes over pulley of negligible friction and mass to a hanging mass of 200 g. The hanging mass is released from rest and falls 80.0 cm in 7.05 s. (a) What is the acceleration of the falling mass assuming constant acceleration? (b) What is the angular acceleration of the turntable? Through how many radians does the turntable turn in the 7.05 seconds? How many revolutions is this? What is the angular velocity and the tangential velocity of an object attached to the edge of the large disk at the end of the 7.05 s?


Homework Equations





The Attempt at a Solution



Well I'll write down every variable I have

DISK:
Diameter = 25.4cm
Mass = 4.34 kg


CYLINDER:
Diameter = 4.76 cm


HANGING MASS:
200 kg
Velocity_i = 0
Δy = 80.0cm
Time = 7.05s


(a) What is the acceleration of the falling mass assuming constant acceleration?

So for this, knowing that acceleration is constant and this is a linear acceleration I can use

Δy = v_i * t + 1/2 * a * t²

we want a so

a = 2(Δy - (v_i * t)) / t²
0
a = 2Δy / t²
a = 2(.80m) / 7.05s² = .032m/s²



(b) What is the angular acceleration of the turntable? Through how many radians does the turntable turn in the 7.05 seconds? How many revolutions is this? What is the angular velocity and the tangential velocity of an object attached to the edge of the large disk at the end of the 7.05 s?


This is where I get stuck...I know that I can use

ω = ω_o + α * t
Δθ = ω_o * t + 1/2 * α * t²
ω² = ω_o ² + 2 * α * t²


I mean I can rearrange each one for α

α = (ω - ω_o) / t
α = 2(Δθ - (ω_o * t)) / t²
α = (ω² - ω_o²) / 2t²

But I don't know the final angular velocity, nor the Δθ because I don't know how many times it will spin.

Can anyone give me a hint here?
 

Answers and Replies

  • #2
gneill
Mentor
20,875
2,838
Your linear acceleration looks okay. Do you happen to know an expression that relates linear acceleration to angular acceleration? (hint: it involves the radius)
 
  • #3
Your linear acceleration looks okay. Do you happen to know an expression that relates linear acceleration to angular acceleration? (hint: it involves the radius)
I think I saw somewhere that Angular acceleration is = to linear acceleration / the radius? Is that correct?
 
  • #4
gneill
Mentor
20,875
2,838
I think I saw somewhere that Angular acceleration is = to linear acceleration / the radius? Is that correct?
Yes.
 
  • #5
Yes.
Perfect....so then if

α = a/R
α = .032m/s² / R


But now.....it asks for the angular acceleration of the turntable.... is this the disk on TOP of the turntable...or the cylinder that the string is wrapped around and the turntable base turns on?
 
  • #6
gneill
Mentor
20,875
2,838
Perfect....so then if

α = a/R
α = .032m/s² / R


But now.....it asks for the angular acceleration of the turntable.... is this the disk on TOP of the turntable...or the cylinder that the string is wrapped around and the turntable base turns on?
Yes :wink:

They are the same, as they are joined together and rotate together. But the string is tangential to only one of them...
 
  • #7
Yes :wink:

They are the same, as they are joined together and rotate together. But the string is tangential to only one of them...
ahh okay so the cylinder is what I'll be using in this example...

so then

α = a / R
α = .032m/s² / .0476m = .672



so now

Through how many radians does the turntable turn in the 7.05 seconds

Wouldn't I just multiply that number by 2∏?

If that is true then this next part of:
How many revolutions is this?

seems redundant since it would be right back to .672

So I'm thinking my hunch is incorrect lol
 
  • #8
gneill
Mentor
20,875
2,838
ahh okay so the cylinder is what I'll be using in this example...

so then

α = a / R
α = .032m/s² / .0476m = .672
Check your radius value; re-read the problem statement.
so now

Through how many radians does the turntable turn in the 7.05 seconds

Wouldn't I just multiply that number by 2∏?
No, acceleration is not a distance. How do you find distance given acceleration and time?
If that is true then this next part of:
How many revolutions is this?

seems redundant since it would be right back to .672

So I'm thinking my hunch is incorrect lol
Your hunch is incorrect :smile:
 
  • #9
Check your radius value; re-read the problem statement.
Figures I would use the diameter rather than the radius -_- lol okay lets try that again

α = .032 m/s² / .0238m
α = 1.34

No, acceleration is not a distance. How do you find distance given acceleration and time?
Δθ = ω_o * t + 1/2 * α * t²
Δθ = 1/2 * α * t²
Δθ = 1/2 * (1.34) * (7.05)²
Δθ = 33.3 radians



How many revolutions is this?

NOW I would DIVIDE that number by 2∏ to get the revolutions correct?

so

33.3 / 2∏ = 5.30 revolutions

good so far?
 
  • #10
gneill
Mentor
20,875
2,838
Figures I would use the diameter rather than the radius -_- lol okay lets try that again

α = .032 m/s² / .0238m
α = 1.34



Δθ = ω_o * t + 1/2 * α * t²
Δθ = 1/2 * α * t²
Δθ = 1/2 * (1.34) * (7.05)²
Δθ = 33.3 radians



How many revolutions is this?

NOW I would DIVIDE that number by 2∏ to get the revolutions correct?

so

33.3 / 2∏ = 5.30 revolutions

good so far?
The method's fine so far.

You might want to hang onto a few more decimal digits for intermediate results. Rounding errors are starting to creep into your significant digits. In general, when a problem has many steps where results build in previous results, you want to keep extra "guard digits" for the intermediate values. You still report individual results rounded to the required significant figures, but use the extra digits for following calculations.
 

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