Revolutions per minute, angular deceleration

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Homework Help Overview

The problem involves a grindstone's angular deceleration after being switched off, with initial and final rotational speeds provided. The task is to determine the additional revolutions made before the grindstone comes to a complete stop.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting rotational speeds from rpm to radians per second and calculating angular acceleration. Questions arise about the relationship between angular motion and linear distance, as well as the applicability of kinematic equations to circular motion.

Discussion Status

Some participants have suggested using kinematic equations related to angular motion, while others are exploring the implications of the angular deceleration and its effect on the total revolutions. There is an indication of progress as one participant identifies a relevant equation for the problem.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to find the time until the grindstone stops and how to relate angular displacement to revolutions. There is also a mention of a specific numerical answer that one participant has calculated, though consensus on this figure has not been reached.

Flucky
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Hi all,

Homework Statement



The motor driving a large grindstone is switched off when a rotational speed of 360 rpm has been achieved. After 15s the speed has decreased to 210rpm. If the angular deceleration remains constant, how many additional revolutions does the stone make before coming to rest?

The Attempt at a Solution



So to start with I converted rpm to rads-1 which was easy enough (I left it in pi to make life easier)

ω0 = 360/60 x 2∏ =12∏ rads-1
ω15 = 210/60 x 2∏ =7∏ rads-1

I then used v=u+at to find the acceleration

7∏ = 12∏ + 15a
a = -∏/3 rads-2

This is where I'm having a mind block as to what to do next, from 15s to when the stone comes to rest - how many additional revolutions does the stone make?

Would love some pointers (I'm sure it's a simple solution but my heads refusing to grasp it).
 
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If you have the angular velocity of the motor and the angular deceleration at t=15 s,
you should be able to easily determine how many additional revolutions occur before the motor stops.

Imagine this: you have a vehicle going a certain velocity when the brakes are applied. How would you calculate the stopping distance from the point where the brakes were applied?
 
SteamKing said:
If you have the angular velocity of the motor and the angular deceleration at t=15 s,
you should be able to easily determine how many additional revolutions occur before the motor stops.

Imagine this: you have a vehicle going a certain velocity when the brakes are applied. How would you calculate the stopping distance from the point where the brakes were applied?

Thanks for the reply.

For the car thing I'd just use a kinematic equation - but surely distance doesn't relate to revolutions?

Are there any equations that are cirular motion specific?
 
Right OK just found an equation θ = ωit + 0.5αt2 which looks very useful. With that I think all I need to do is find the time it takes to completely stop, use that in the equation I've just stated then convert θ into revolutions and Bob's your uncle.

I'll give that a go but I think I've got it from here.
 
Just as a check did anybody get the answer 363 revolutions?
 
Last edited:

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