Moment of Inertia of a grinding wheel

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  • #1
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Homework Statement



A grinding wheel is initially at rest. A constant external torque of 52.5 N· m is applied to the wheel for 18.4 s, giving the wheel an angular speed of 605 rev/min. The external torque is then removed, and the wheel comes to rest 101 s later. Find the moment of inertia of the wheel.

Homework Equations



Torque = I * a

The Attempt at a Solution



Since Torque is equal to external torque + frictional torque, i decided to calculate what the acceleration would be without frictional torque to remove that variable. So I calculated the acceleration first, 605rev/min * 1 min / 60 sec = 10.08333 rev/s. 10.083333 rev/s divided by 18.4 sec is 0.548 rev/s^2, * 2pi = 3.44 rad/s^2. Then 10.083333 rev/s divided by 101 sec = 0.0998349835 rev/s^2. Multiplied that by 2 pi to get 0.62728 rad /s^2 and then added it to the original acceleration to get 4.0705 rad/s^2, the angular acceleration without friction. Then 52.5 = 4.0705 * I, to get 12.9 kg*m^2. This isn't right.

I'm assuming this strategy doesn't work. How else can I do it? Thanks for any help.
 

Answers and Replies

  • #2
cepheid
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This strategy should work fine. I can't find the error. How do you know the answer is wrong?
 
  • #3
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This strategy should work fine. I can't find the error. How do you know the answer is wrong?
Submission # Try Submitted Answer
1 Unable to interpret units. Computer reads units as "n*m".Help: Physical_Units 15.3 n*m
2 Unable to interpret units. Computer reads units as "n".Help: Physical_Units 15.3 n
3 Unable to interpret units. Computer reads units as "j*m".Help: Physical_Units 15.3 j*m
4 Unable to interpret units. Computer reads units as "n/m".Help: Physical_Units 15.3 n/m
5 Incompatible units. No conversion found between "N/m" and the required units.Help: Physical_Units 15.3 N/m
6 Incompatible units. No conversion found between "N*m" and the required units.Help: Physical_Units 15.3 N*m
7 Incorrect. (Try 1) 15.3 kg*m^2
8 Incorrect. (Try 2) 15.2 kg*m^2
9 Incorrect. (Try 3) 12.9 kg*m^2
10 Incorrect. (Try 4) 280.5 kg*m^2
11 Incorrect. (Try 5) 237 kg*m^2
12 Incorrect. (Try 6) 238 kg*m^2
13 Submission not graded. Use more digits. 240kg*m^2
14 Submission not graded. Use more digits. 13 kg*m^2
15 Incorrect. (Try 7) 13.1 kg*m^2
16 Incorrect. (Try 8) 12.0 kg*m^2
17 You have entered that answer before as submission # 9 12.9 kg*m^2
18 Incorrect. (Try 9) 11. 9kg*m^2
19 Incorrect. (Try 10) 11. 8 kg*m^2
20 Incorrect. (Try 11) 11.7 kg*m^2
21 Incorrect. (Try 12) 11.6 kg*m^2
22 Incorrect. (Try 13) 11.5 kg*m^2

Am I making a math error somewhere?
 
  • #4
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applied torque = T
friction torque = F
a= angular acceleration
I = moment of inertia
d = deaccleration after torque is removed


I * a = T + F ---- (1)

I * d = F ----- (2)

Put 2 in 1

I = T/(a-d)
 
  • #5
cepheid
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applied torque = T
friction torque = F
a= angular acceleration
I = moment of inertia
d = deaccleration after torque is removed


I * a = T + F ---- (1)

I * d = F ----- (2)

Put 2 in 1

I = T/(a-d)
Right, but since d is negative, that's exactly what the original poster did -- divided the applied torque by the sum of the angular accelerations. And you get 12.9 kg*m^2. So...I don't know what the issue is.
 
  • #6
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I got it, I must've been making a math error somewhere.

Thank you guys though!
 
  • #7
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Yes, the signs should be taken care. When I wrote down the equatons, I missed the signs.

The applied torque must overcome frictional and inertial torque. So
T = F + I*a

During deceleration. the frictional torque supplies the inertial torque, so
I * d = F

So I = T/(a + d)
 
  • #8
cepheid
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So what was the answer? I couldn't find any arithmetic errors. Here is what I was doing:

According to Newton's 2nd law for rotational motion the net torque is equal to the moment of inertia times the angular acceleration:

[tex] \sum \vec{\tau} = I \vec{\alpha} [/tex]

Call the first time interval (while speeding up) [itex]\Delta t_1 [/itex] and the second time interval (while slowing down) [itex]\Delta t_2 [/itex]. During the first time interval, the equation for Newton's 2nd becomes;

[tex] \tau_\mathrm{ext} + \tau_\mathrm{fric} = I \alpha_1 [/tex]

and during the second time interval, it is just:

[tex] \tau_\mathrm{fric} = I \alpha_2 [/tex]

In the above equations, I've dropped the vector notation, since this is a 1D problem, and so tau and alpha can each either be positive or negative scalars, representing the two possible directions they can have. Now we can compute the angular accelerations during each of the two time intervals. From basic kinematics, we know that IF the acceleration is constant, then it is equal to the total change in velocity (over a time interval) divided by that time interval:

[tex] \alpha_1 = \frac{\Delta \omega_1}{\Delta t_1} = \frac{\omega_{1f} - \omega_{1i}}{\Delta t_1} = \frac{\omega_{1f} }{\Delta t_1} [/tex]

where the last part is because the initial angular velocity is zero (the wheel starts from rest). This works out to:

[tex] \frac{605~\frac{\mathrm{rev}}{\mathrm{min}} \cdot 2\pi~\frac{\mathrm{rad}}{\mathrm{rev}} \cdot \frac{1~\mathrm{min}}{60~\mathrm{s}} }{18.4~\mathrm{s}} = 3.443~\mathrm{rad}/\mathrm{s}^2 [/tex]

Similarly, for the second time interval:

[tex] \alpha_2 = \frac{\Delta \omega_2}{\Delta t_2} = \frac{\omega_{2f} - \omega_{2i}}{\Delta t_2} = \frac{0- \omega_{1f} }{\Delta t_2} = \frac{-605~\frac{\mathrm{rev}}{\mathrm{min}} \cdot 2\pi~\frac{\mathrm{rad}}{\mathrm{rev}} \cdot \frac{1~\mathrm{min}}{60~\mathrm{s}} }{101~\mathrm{s}} = -0.627281701~\mathrm{rad}/\mathrm{s}^2 [/tex]

So, the net torque equation during time interval 1 becomes (when you substitute into it the expression for the frictional torque from the equation for time interval 2):

[tex] \tau_\mathrm{ext} + I\alpha_2 = I \alpha_1 [/tex]

[tex] \tau_\mathrm{ext} = I (\alpha_1 - \alpha_2) [/tex]

[tex] I = \frac{\tau_\mathrm{ext}}{\alpha_1 - \alpha_2} = \frac{52.5~\mathrm{N}\cdot\mathrm{m}}{3.443~\frac{\mathrm{rad}}{\mathrm{s}^2} - (-0.627281701~\frac{\mathrm{rad}}{\mathrm{s}^2})} = \frac{52.5~\mathrm{N}\cdot\mathrm{m}}{4.0702817~\frac{\mathrm{rad}}{\mathrm{s}^2}} = 12.9~\mathrm{kg}\cdot\mathrm{m}^2 [/tex]
 
Last edited:
  • #9
60
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I dont see anything wrong with 12.9kgm^2
 
  • #10
I have the same problem but with different numbers. When I use the equations you guys are using, though, I keep getting 13.4 kg*m^2, which is not correct. Could you help me?

A grinding wheel is initially at rest. A constant external torque of 51.9 N· m is applied to the wheel for 18.8 s, giving the wheel an angular speed of 589 rev/min. The external torque is then removed, and the wheel comes to rest 106 s later. Find the moment of inertia of the wheel.
 
  • #11
60
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I think the time interval is not right. Probably the time intervals are not cumulative.
 

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