Angular Acceleration of a washer

In summary, the conversation discusses the problem of a washer's spin-dry cycle being interrupted by a safety switch and the calculation of the number of revolutions the tub makes during this time. The person asking for help is unsure of how to use the kinematic equations for this problem and receives assistance in understanding and solving it. The final answer is determined to be 94.63 revolutions.
  • #1
ryan838
3
0
I don't really know what to do on this problem. So if someone could get me pointed in the right direction I would appreciate it.

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 7.0 rev/s in 13.0 s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 14.0 s. Through how many revolutions does the tub turn during this 27 s interval? Assume constant angular acceleration while it is starting and stopping.
 
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  • #2
You have to apply the kinematic equations for constant acceleration:
x = x0 + v0t + 1/2 a t2
vf2 = vi2 + 2aΔx

Of course, instead of distance, velocity, and acceleration, you must use angular displacement, angular velocity, and angular acceleration. Make sense?
 
  • #3
No that doesn't really make to much sense. Here is what I did so maybe you can tell me where I went wrong.

I took the 7 rev/s and divided it by 13s to get the acceleration to be .54 rev/s. Then to get the deacceleration I took 7 rev/s divided by 14s to get -.5 rev/s. So for the first 13s it is speeding up by .54 rev/s right? Then the last 14s it is deacclerating at a rate of -.5 rev/s? From there I don't get how to put it into the equation you posted. Thanks for your help though.
 
  • #4
Originally posted by ryan838
I took the 7 rev/s and divided it by 13s to get the acceleration to be .54 rev/s.
Right. The units should be rev/s2. (I should have given you the equation Δv = at .)
Then to get the deacceleration I took 7 rev/s divided by 14s to get -.5 rev/s.
Right. Same comment about units.
So for the first 13s it is speeding up by .54 rev/s right? Then the last 14s it is deacclerating at a rate of -.5 rev/s? From there I don't get how to put it into the equation you posted.
You know the times and the accelerations. Now you need to find the angle (distance in revs). Which of the two equations I gave give the distance? (The other one you don't need!)
 
  • #5
Thank you for your help. I got the right answer which turned out to be 94.63 revolutions.
 

1. What is angular acceleration?

Angular acceleration is the rate of change of angular velocity, which is the rate at which an object rotates around an axis. It is measured in radians per second squared (rad/s²).

2. How is angular acceleration different from linear acceleration?

Angular acceleration is the change in rotational motion, while linear acceleration is the change in straight-line motion. Linear acceleration is measured in meters per second squared (m/s²), while angular acceleration is measured in radians per second squared (rad/s²).

3. How is angular acceleration calculated?

Angular acceleration can be calculated by dividing the change in angular velocity by the change in time. In other words, it is the change in angular velocity divided by the change in time. The formula for angular acceleration is: α = (ω₂ - ω₁) / (t₂ - t₁), where α is the angular acceleration, ω is the angular velocity, and t is the time.

4. What factors affect angular acceleration?

The factors that affect angular acceleration include the force applied, the mass and distribution of the object, and the distance from the axis of rotation. In general, a larger force applied further from the axis of rotation will result in a greater angular acceleration.

5. How is angular acceleration related to torque?

Angular acceleration is directly proportional to torque, which is the twisting force that causes an object to rotate. This means that the greater the torque, the greater the angular acceleration. The relationship between torque and angular acceleration is given by the formula: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

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