Rotating rod, inertia and torque

  • Thread starter jl9999
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  • #1
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Homework Statement


A small ball with mass 1.70 kg is mounted on one end of a rod 0.750 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5030 rev/min.

(a) Calculate the rotational inertia of the system about the axis of rotation.
(b) There is an air drag of 2.20 10-2 N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?



Homework Equations


I=mr^2
T=FR
T=Ialpha



The Attempt at a Solution



So I got (a) which is simply I=mr^2, or I=(1.70)(.750)^2, I=.95625 kgm^2

I thought I had (b) but I was wrong. First I broke down the 5,030 rev/min to an angular acceleration which came out to 526 rad/s^2. Then I plugged that, along with the I into T=Ialpha which came out to, T=(.9562)(526), T=502.9Nm. I then added in the calculated torque from the air drag which was T=(.022)(.75), =.0165Nm to come out with a torque of 503Nm, which was wrong. I have absolutely no idea what detail(s) I'm missing because I was sure I had this one. Any help is greatly appreciated, thank you.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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hi jl9999! :smile:

(have an alpha: α and an omega: ω and a tau: τ and try using the X2 icon just above the Reply box :wink:)
I thought I had (b) but I was wrong. First I broke down the 5,030 rev/min to an angular acceleration which came out to 526 rad/s^2. Then I plugged that, along with the I into T=Ialpha …

i'm sorry, but both those are completely wrong :redface:

rev/min is a speed, not an acceleration, it converts to rad/s

and α is zero, so how will τ = Iα help?

try again :smile:
 
  • #3
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...actually (a) was right, according to the webassign(auto-grading homework website) Thanks for pointing out what I missed though. I often misread stuff. hopefully I can get it now.
 

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