Small Circular Acceleratioon Question

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In summary, the conversation is about determining the acceleration at the outer rim of a wheel with a 0.50 m radius rotating at 15 rev/s. There is confusion about which equation to use, with one person suggesting using the formula a=V^2/R and another pointing out that it is for radial acceleration, not tangential. Eventually, it is clarified that the correct formula for tangential acceleration is atangential = dV/dt and V can be found using the period equation T=2piR/V. The final answer is 4.4x103 m/s2, but there was confusion about the use of frequency and the equation for finding tangential acceleration.
  • #1
LadyMario
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A wheel of 0.50 m radius rotates at 15 rev/s. What is the acceleration at its outer rim in m/s2

They want tangential acceleration right? Is there anyway to solve this without angular velocity/speed as we haven't learned that yet. This is part of the practice questions for my midterm on Work/Energy and Kinematics, so I'm looking for one of those methods to solve it.

The answer is 4.4x103 m/s2 but I can't seem to get that. Help?
 
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  • #2
What did you get and how did you get it?
 
  • #3
I was using T=2piR/V to get V and then using that in a=V^2/R but a friend pointed out the 15 rev/s is frequency, and so I'm not sure of an equation for that.
 
  • #4
T = 1/f. If the frequency is 15 revolutions/s, how many seconds would it take for 1 revolution? Your formula for acceleration is for radial acceleration, not tangential. Since the answer is non-zero, question refers to radial.
 
  • #5
Yes I get that frequency means it will be 1/15 but what equation can I use to get tangential?
 
  • #6
atangential = dV/dt.
 
  • #7
lewando said:
atangential = dV/dt.

oh okay. I was thinking about how to get that equation, how do I get time and velocity though? From the Period equation: T=2piR/V but this time T= 1/15 and solve for V?
 
  • #8
Not sure if you are on the right track per your tangential acceleration inquiry. V can be found from the period equation. You are looking for aradial.
 

Related to Small Circular Acceleratioon Question

1. What is a small circular acceleration?

A small circular acceleration is a type of motion in which an object moves along a circular path with a constant change in direction. This means that the object is constantly accelerating towards the center of the circle, but its speed remains constant.

2. How is small circular acceleration different from linear acceleration?

Small circular acceleration is different from linear acceleration in that it involves a change in direction, while linear acceleration involves a change in speed or velocity. In small circular acceleration, the speed remains constant, but the direction of motion continuously changes.

3. What is the formula for calculating small circular acceleration?

The formula for calculating small circular acceleration is a = v^2/r, where a is the acceleration, v is the speed, and r is the radius of the circular path.

4. What are some real-life examples of small circular acceleration?

Some real-life examples of small circular acceleration include a car driving along a curved road, a satellite orbiting around the Earth, and a roller coaster moving along its track.

5. How is small circular acceleration used in scientific research?

Small circular acceleration is commonly used in scientific research to study the motion of objects in circular paths, such as planetary orbits or the movement of particles in a centrifuge. It is also used in experiments involving circular motion, such as the study of centripetal force and its effects on objects.

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