How Do You Calculate Angular Acceleration of a Salad Spinner?

AI Thread Summary
To calculate the angular acceleration of a salad spinner, the initial angular velocity is determined to be 25.13272 radians/s after spinning 20 times in 5 seconds. Using the kinematic equation ω²f = ω²0 + 2α(θ), the calculated angular acceleration was found to be -8.37757 radians/s², which was marked incorrect. Participants noted that the problem asked for the magnitude of angular acceleration, which should be expressed as a positive value. There was confusion regarding the requirement for magnitude, as the original question did not specify this clearly.
Steelers72
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Homework Statement


Dario, a prep cook at an Italian restaurant, spins a salad spinner 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration.

What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in radians per second per second.

2. Homework Equations

ω2f = ω20 + 2α(θ)

The Attempt at a Solution



ω = θ / t
ω = 20 * 2 * π / 5.00
ω = 8π
ω = 25.13272 radians/s

now I can solve for α using a kinematic equation:

ω2f = ω20 + 2α(θ)
02 = 25.132722 + 2 * α * (2 * π * 6.00)
0 = 631.6536 + 75.39816α
75.39816α = -631.6536
α = -8.37757 radians/s2

apparently This answer is wrong...where did I go wrong?
 
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Significant figures??

Chet
 
Chestermiller said:
Significant figures??

Chet
-8.0 didnt work as well.
 
Steelers72 said:
-8.0 didnt work as well.
-8.38??

Chet
 
Steelers72 said:

Homework Statement


Dario, a prep cook at an Italian restaurant, spins a salad spinner 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration.

What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in radians per second per second.

2. Homework Equations

ω2f = ω20 + 2α(θ)

The Attempt at a Solution



ω = θ / t
ω = 20 * 2 * π / 5.00
ω = 8π
ω = 25.13272 radians/s

now I can solve for α using a kinematic equation:

ω2f = ω20 + 2α(θ)
02 = 25.132722 + 2 * α * (2 * π * 6.00)
0 = 631.6536 + 75.39816α
75.39816α = -631.6536
α = -8.37757 radians/s2

apparently This answer is wrong...where did I go wrong?
I used a somewhat different method and got -8π/3 rad/s2, which agrees with you.
 
α =
−8
radians/s2

ANSWER 2: Deduction: -3%
α =
−8π3
radians/s2

The question only asks for the magnitude of the angular acceleration.
ANSWER 3: Deduction: -3%
α =
−8.4
radians/s2

The question only asks for the magnitude of the angular acceleration.
ANSWER 4: Deduction: -3%
α =
−2.67π
radians/s2

The question only asks for the magnitude of the angular acceleration.These were my answers. All marked wrong.
 
Steelers72 said:
...

These were my answers. All marked wrong.
After the first message stating that only the magnitude was asked for, it seems reasonable to give the magnitude. (The statement in the OP clearly asks for angular acceleration, no mention of magnitude.)
 
SammyS said:
After the first message stating that only the magnitude was asked for, it seems reasonable to give the magnitude. (The statement in the OP clearly asks for angular acceleration, no mention of magnitude.)

What would be the magnitude though? I am confused as to why my answer of -8.38 radians/s^2 is wrong
 
Steelers72 said:
What would be the magnitude though? I am confused as to why my answer of -8.38 radians/s^2 is wrong
When they ask for the magnitude of a vector, they want it without the sign. The magnitude is always taken to be positive, and the sign is absorbed into the direction of the unit vector. Pretty silly to require that in the answer to this problem, however.

Chet
 
  • #10
Chestermiller said:
... Pretty silly to require that in the answer to this problem, however.​

Chet
I agree with Chet on this point, especially since there was no mention of magnitude in the problem statement.
 
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