Centripetal Acceleration HELP. Is it challenging? Well come find out

In summary, an athlete is whirling a 7.48 kg hammer tied to the end of a 1.4 m chain in a horizontal circle at a rate of 1.76 rev/s. The question is asking for the centripetal acceleration of the hammer, assuming the arm length is included in the chain length. The relevant equations are v = 2∏r/T and ac = v^2/r. The athlete tried finding the velocity first and then plugged it into the ac equation, but got the wrong answer. After further discussion, it was determined that the angular speed of 1.76 rev/s is not T and the correct approach is to find T by taking the inverse of 1.76,
  • #1
TwoTeez
4
0
1. An athlete whirls a 7.48 kg hammer tied to
the end of a 1.4 m chain in a horizontal circle.
The hammer moves at the rate of 1.76 rev/s.
What is the centripetal acceleration of the
hammer? Assume his arm length is included
in the length given for the chain.
Answer in units of m/s




2. Equations I am pretty sure I need are
v = 2∏r/T
and
ac= v2/r
not sure if I need any others.




3. I tried finding v first, and I assumed the 1.76 rev/s was T, so I did 2*∏*1.4(about 8.80) and divided it by 1.76, getting about 4.99 m/s
and I plugged it into the ac equation.
So I squared the 4.99, getting about 24.98, and dividng it by 1.4, which gets me about 17.843.
But that answer is wrong.
I don't know what the answer is.

I feel like what I am doing wrong is the 1.76 rev/s is not T, or it is, but I have to change it so it's in like m/s or something like that. If that's the case, then I need help finding that because I have no idea how to do that.


Thank you for any help!
 
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  • #2
in your equation for v, T is the time to complete one revolution (make sure you see why)

1.76 rev/s

is that what you want? or is it seconds per rev?
 
  • #3
I think I see where you are getting at.
For every second that passes, it makes 1.76 revolutions.
But to find the velocity, I need to put T as a time per revolution. So would I just do the inverse of T?
So if i do the inverse for T, wouldn't that just be multiplying T with the 2∏r?
But wouldn't I need to the the inverse of everything else as well? What you do to one side of the equation, you do to the other.

So 1/v = T/2∏r
And then do the inverse again to get the answer. But isn't that the same is the answer I would have gotten before? Am I even going in the right direction?
 
  • #4
well 1.76 is not T. Your messing around with thinking that it is. 1.76 is the angular speed, completely different from T.

the equation v = 2∏r/T is correct. Find T.
 
  • #5
If I want to turn the rev/s into common terms, would I multiply 1,76 by 2∏r?
 
  • #6
I have yet to learn about angular speed in my class, so I know practically nothing on it.
 
  • #7
TwoTeez said:
I think I see where you are getting at.
For every second that passes, it makes 1.76 revolutions.
But to find the velocity, I need to put T as a time per revolution. So would I just do the inverse of T?
So if i do the inverse for T, wouldn't that just be multiplying T with the 2∏r?
But wouldn't I need to the the inverse of everything else as well? What you do to one side of the equation, you do to the other.

So 1/v = T/2∏r
And then do the inverse again to get the answer. But isn't that the same is the answer I would have gotten before? Am I even going in the right direction?
With what you have now, you have 1.76 revolutions over one second. Just do the inverse of that: (1/1.76) to find your period, T

This is the relationship between frequency and period.

You are given a frequency of 1.76 rev/sec.

Frequency is (1/T) T is the period
 

Related to Centripetal Acceleration HELP. Is it challenging? Well come find out

1. What is centripetal acceleration and how is it different from regular acceleration?

Centripetal acceleration is the acceleration of an object moving in a circular path. It is different from regular acceleration because it is always directed towards the center of the circle, while regular acceleration can be in any direction.

2. How is centripetal acceleration calculated?

Centripetal acceleration can be calculated using the formula a = v²/r, where a is acceleration, v is the velocity of the object, and r is the radius of the circular path.

3. Is understanding centripetal acceleration important for everyday life?

While most people may not think about centripetal acceleration in their daily lives, it is actually very important in many areas such as amusement park rides, car and bicycle turns, and even the Earth's orbit around the sun.

4. Is learning about centripetal acceleration challenging?

For some people, understanding the concept and calculations of centripetal acceleration may be challenging at first. However, with practice and a solid understanding of the related concepts, it can become easier to grasp.

5. How can I apply centripetal acceleration in my own experiments or projects?

There are countless ways to apply centripetal acceleration in experiments or projects, such as designing a circular motion experiment or creating a device that utilizes centripetal acceleration, like a centrifuge. It all depends on your creativity and understanding of the concept.

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