Converting centrifugal acceleration to frequency

[NODARman]

Homework Statement

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Relevant Equations

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I've calculated this. Is it correct?

a=ω²R => ω=√(a/R)
T(period)=1/ω=1 / (√(a/R))
f = 1/T = √(a/R)
[ if a=10G=100 m/s² ; R=1m ]
then: f = √(a/R) = √(100/1) = 10 hertz

In angular velocity, can I just convert a=ω²R to ω=√(a/R) and write instead of ω=(2π)/T (in the first line)?

 

[topsquark]

You have a mistake in line 2. $T = \dfrac{2 \pi}{\omega}$. In line 3 you are also saying that T = 1/f (this is correct), which means you are saying that $\omega = f$, which is not true. The definition is $\omega = 2 \pi f$.

As to the rest, clearly you are doing more than deriving an expression for frequency. Please post the whole problem statement.

-Dan

 

[NODARman]

You have a mistake in line 2. $T = \dfrac{2 \pi}{\omega}$. In line 3 you are also saying that T = 1/f (this is correct), which means you are saying that $\omega = f$, which is not true. The definition is $\omega = 2 \pi f$.

As to the rest, clearly you are doing more than deriving an expression for frequency. Please post the whole problem statement.

-Dan

Is this now correct?

a=ω²R => ω=√(a/R)
ω=2πf => f = ω/2π

if a=10G=100 m/s² ; R=1m

then:
ω = √(a/R) = 10 rad/s
f = ω/2π = 10/2π = 5/π = 1.59154943 hz

 

[topsquark]

Is this now correct?

a=ω²R => ω=√(a/R)
ω=2πf => f = ω/2π

if a=10G=100 m/s² ; R=1m

then:
ω = √(a/R) = 10 rad/s
f = ω/2π = 10/2π = 5/π = 1.59154943 hz

Yes and no.
Yes: Your derived equation is correct. I have no idea where you got your numbers so I can't verify if you are putting them in right. Again, please post the whole problem.

No: The number you calculated for f is correct. But if you are going to be using g = 10 m/s^2 then your answer shouldn't be given to anywhere near that number of significant digits. Technically your answer should only be to one sigfig but you could probably get away with two. No more.

No: The unit for frequency is written as "Hz" not "hz."

Otherwise, good work.

-Dan

 

[BvU]

If you take $g$ = 10 m/s², there is no point in specifying your answer to ten digits ...

$\$

 

[NODARman]

Yes and no.
Yes: Your derived equation is correct. I have no idea where you got your numbers so I can't verify if you are putting them in right. Again, please post the whole problem.

No: The number you calculated for f is correct. But if you are going to be using g = 10 m/s^2 then your answer shouldn't be given to anywhere near that number of significant digits. Technically your answer should only be to one sigfig but you could probably get away with two. No more.

No: The unit for frequency is written as "Hz" not "hz."

Otherwise, good work.

-Dan

So, here's the problem:

R = 1-meter block of wood rotating in space. Calculate the frequency of rotation, if the centripetal acceleration on the edge of the block is equal to 10Gs or ~100 m/s^2

 

[topsquark]

So, here's the problem:

View attachment 313551

R = 1-meter block of wood rotating in space. Calculate the frequency of rotation, if the centripetal acceleration on the edge of the block is equal to 10Gs or ~100 m/s^2

Yep, it looks good, then. Along with the comments made above, that is.

-Dan

 

[kuruman]

R = 1-meter block of wood rotating in space. Calculate the frequency of rotation, if the centripetal acceleration on the edge of the block is equal to 10Gs or ~100 m/s^2

Is this a 1-meter block that rotates about its middle? If so, the radius of rotation is 0.5 m. The statement "R = 1-meter block" is ambiguous.

 

[NODARman]

Yep, it looks good, then. Along with the comments made above, that is.

-Dan

Thanks for helping.

 

[NODARman]

Is this a 1-meter block that rotates about its middle? If so, the radius of rotation is 0.5 m. The statement "R = 1-meter block" is ambiguous.

Well, I meant radius.

 

[kuruman]

Well, I meant radius.

OK, then.