Converting centrifugal acceleration to frequency

• NODARman
In summary: The frequency is given as 5/π hertz. In summary, the block of wood is rotating at a frequency of 5/π hertz.
NODARman
Homework Statement
.
Relevant Equations
.
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

Steve4Physics and NODARman
topsquark said:
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

NODARman said:
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

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

##\ ##

topsquark
topsquark said:
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

NODARman said:
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

NODARman
NODARman said:
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.

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

-Dan
Thanks for helping.

kuruman said:
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 said:
OK, then.

NODARman

1. What is centrifugal acceleration?

Centrifugal acceleration is the acceleration experienced by an object moving in a circular path. It is caused by the force of inertia acting on the object, which pulls it away from the center of rotation.

2. How is centrifugal acceleration different from centripetal acceleration?

Centrifugal acceleration is the outward acceleration experienced by an object in circular motion, while centripetal acceleration is the inward acceleration that keeps the object moving in a circular path. They are equal in magnitude but opposite in direction.

3. Why is it important to convert centrifugal acceleration to frequency?

Converting centrifugal acceleration to frequency allows us to understand the relationship between the two and make calculations or predictions based on this relationship. It is particularly useful in fields such as engineering, physics, and aviation.

4. What is the formula for converting centrifugal acceleration to frequency?

The formula for converting centrifugal acceleration to frequency is f = √(a/2πr), where f is the frequency in Hertz, a is the centrifugal acceleration in meters per second squared (m/s²), and r is the radius of the circular path in meters (m).

5. Can centrifugal acceleration be converted to frequency in any unit of measurement?

Yes, as long as the units are consistent, centrifugal acceleration can be converted to frequency in any unit of measurement. For example, if the acceleration is given in G-forces, the resulting frequency will be in Gs (1 G = 9.8 m/s²).

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