Time required for disk to reach angular speed?

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Homework Help Overview

The problem involves a uniform disk being accelerated by a motor through a constant torque, with the goal of determining the time required to reach a specified angular speed. The subject area includes concepts of rotational dynamics, specifically torque, angular acceleration, and moment of inertia.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of angular acceleration and the moment of inertia, with some questioning the correctness of their units and the relationships between torque, moment of inertia, and angular acceleration.

Discussion Status

Participants are actively engaging with the problem, correcting misconceptions about the relationships between the variables involved. Some have identified errors in their calculations and are working towards clarifying their understanding of the units and equations needed to solve the problem.

Contextual Notes

There are indications of confusion regarding the conversion of angular speed from rpm to rad/s and the proper application of units throughout the calculations. Participants are encouraged to consistently apply units to avoid errors in their reasoning.

Rijad Hadzic
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Homework Statement


A uniform disk of mass M = 3 kg and radius r = .22 meters is mounted on a motor through its center. The motor accelerates the disk uniformly from rest by exerting a constant torque of 1 n * m

What is the time required for the disk to reach an angular speed of 800 rpm

Homework Equations


tor = i a
i = rotational intertia
a = acc

The Attempt at a Solution


So I convert 800 rotations / min to 80/6 rotations a second.

since angular acceleration = Torque / rotational interta, and I have torque is 1 n * m and rotational intertia = (1/2) m r^2

so I have (1) / (1/2)(3)(.22)^2 = 13.77

so that should be my angular acceleration right?

Now to find time all I have to do is use euqation
W_0 + ax t = W_x

w_o = 0 so

t = (W_x)/(a_x)

t = 80/(6)(13.77) = .9682 s

but my books answer was 6.08 seconds.

Does anyone know where I went wrong? I have a feeling I calculated acceleration wrong :/
 
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Rijad Hadzic said:
so I have (1) / (1/2)(3)(.22)^2 = 13.77

so that should be my angular acceleration right?
Wrong. That looks like the moment of inertia.
 
kuruman said:
Wrong. That looks like the moment of inertia.

Wow.. you are right.

So angular acc = I / Torque
so acc = (1/2)mr^2 = (1/2) (3) (.22)^2
But doing this I got angular acc = .0726 rad /s^2

Now using w_0 + ax t = wx

I get t = wx/ax = (80/6) (1/.0726) which = 183.65 which is still the wrong answer??
 
Rijad Hadzic said:
So angular acc = I / Torque
Incorrect. It would help if you got into the habit of appending units to any number you write down. What are your units in this case?
 
kuruman said:
Incorrect. It would help if you got into the habit of appending units to any number you write down. What are your units in this case?
Sorry I will do that from now on.

https://www.boundless.com/physics/textbooks/boundless-physics-textbook/static-equilibrium-elasticity-and-torque-8/torque-and-angular-acceleration-80/relationship-between-torque-and-angular-acceleration-319-6062/

This page makes the statement "Torque is equal to the moment of inertia times the angular acceleration."

moment of intertia = (1/2)MR^2 and torque = 1 N m

Why can't I set angular acceleration = Torque / moment of intertia?
 
Rijad Hadzic said:
Why can't I set angular acceleration = Torque / moment of intertia?
You can. What you had before is the inverse of that. You have to be careful because small things like that will trip you up. Now can you finish the problem?
 
kuruman said:
You can. What you had before is the inverse of that. You have to be careful because small things like that will trip you up. Now can you finish the problem?

I'm not sure :(

So I can state "acceleration = Torque / moment of intertia"

torque was given to be 1 N * m
moment of intertia given by my book is (1/2)MR^(2) = (1/2)(3kg)(.22m)^2 = .0736 kg m^2

so 1 N / .0736 kg m = 13.77 N/ (kgm) = acceleration

What am I missing? Why is this still the wrong acceleration value?? I'm literally plugging the values right into the equation I don't understand how it could be wrong..
 
You are missing the correct units for the acceleration. You list them as N/(kg m). Is that what they are? What does your textbook say they ought to be?

Edit: The number 13.77 is correct. It's the units you need to understand because that matters for finishing the problem correctly.
 
kuruman said:
You are missing the correct units for the acceleration. You list them as N/(kg m). Is that what they are? What does your textbook say they ought to be?

Edit: The number 13.77 is correct. It's the units you need to understand because that matters for finishing the problem correctly.

I see. I took the following steps and got 13.77 rad/s^2

( (1kg m^2)/(s^2) ) / ( (3kg/2)(.22m)^2 )

= (2) / ( 3 * (.22)^2 ) s^2

= 13.77rad s^(-2)
So is my value for acceleration correct??
 
  • #10
Rijad Hadzic said:
So is my value for acceleration correct??
Yes it is. Now proceed to find the required time.
 
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  • #11
kuruman said:
Yes it is. Now proceed to find the required time.
Okay now that I found angular acceleration = 13.77 rad s^(-2)

I use equation

W_o + (angacc)t = W_final

Well I know W_final = 800 rpm (I'm going to assume this to be rotations per minute not radians per minute) so 13.33 rotations per second, 1 rotation = 2pi so

2pi(13.33) rad/s

okay I'm just going to stop here.

I completely get the point you were trying to get to me here.

I got the answer that was written in my book... 6.08 seconds..

I see now. Write down your units. and follow them. Each. And. Every. Time.

Thank you for your patience with me.
 
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  • #12
Good job! :smile:
 

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