Period/Frequency of a Rotational Object

  • Thread starter BBallman_08
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In summary, the object moves uniformly around a circular path of radius 0.20 meters, completing one revolution every 2.00 seconds. The translational speed of the object is 2*PI*0.20/2.00 meters per second, the frequency is 1/2.00 revolutions per second, and the angular speed is 2*PI/2.00 radians per second.
  • #1
BBallman_08
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Homework Statement



An object moves uniformly around a circular path of radius .20 meters, making one complete revolution every 2.00 sec. What are (a) the translational speed of the object, (b) the frequency of the object, and (C) the angular speed of the object?



Homework Equations



x = A Cos (angular speed * t)

Angular speed = 2 pi frequency

T = 1/f


The Attempt at a Solution



I am lost as to how to approach this one... I think I might need to solve for the angular speed first?
 
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  • #2
Looks to me like you could easily find the frequency from the given info.

making one complete revolution every 2.00 sec...

Casey
 
  • #3
Frequency = 1 / period correct? So, 1 / 2 sec = .5 rev/sec?
 
  • #4
Anyone?
 
  • #5
Would this be suffice to solve it?

r = .2m and T = 2s
v = 2pi(r) / T
a) so just plug in.

b) f = 1/T so just plug in again

c) omega = v/r so just plug in.

On part c) how can I get the final answer to be in rad/sec?
 
  • #6
BBallman_08 said:
c) omega = v/r so just plug in.
On part c) how can I get the final answer to be in rad/sec?
[itex]\omega = v/r[/itex] is in radians per second.
 
  • #7
Oh okay, so in part a) the r is reffering to 2 Pi Radians or not 2 pi radius?
 
  • #8
You have the right answer for part (a), [itex]v = 2\pi\, r/T[/itex]. Here is how this expression arises. The circumference of the circle is [itex]2\pi r[/itex]. This is the arc length the object travels in In one revolution. Dividing by the time gives the average speed. Since the speed is constant in uniform circular motion, the average speed is the speed, period. Given [itex]v = 2\pi\, r/T[/itex], what is [itex]\omega=v/r[/itex]?
 

1. What is the period of a rotational object?

The period of a rotational object is the time it takes for the object to make one complete rotation around its axis.

2. How is the period of a rotational object calculated?

The period of a rotational object is calculated by dividing the time it takes for one rotation by the number of rotations. It can also be calculated by dividing the circumference of the object by its linear velocity.

3. How does the mass of a rotational object affect its period?

The mass of a rotational object does not affect its period. The period is solely determined by the distance from the axis of rotation and the linear velocity.

4. What is the relationship between the period and frequency of a rotational object?

The period and frequency of a rotational object are inversely related. This means that as the period increases, the frequency decreases, and vice versa.

5. Can the period/frequency of a rotational object change over time?

Yes, the period/frequency of a rotational object can change over time if there is a change in the object's angular velocity. This can be caused by external forces such as friction or a change in the object's shape or mass.

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