Work done dynamics question

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  • #1
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Dear All, would really appreciate some help with the following question - am struggling with dynamics at the minute!
1. Homework Statement

An 8kg mass is suspended from a grooved drum at radius r1=200mm and with a speed of 0.3m/s it drops 1.5m.
The outer radius of the drum is r2=300mm. Assume that the drum has radius of gyration k=210mm and mass 12kg and a constant frictional torque in its bearings of 3Nm. Determine the speed once it has dropped the 1.5m.

Homework Equations


So in a hint I have been told to utilise s=r x theta and also v=wr

The Attempt at a Solution


I attempted the following:
mgh + 1/2m(mass)v^2 + 1/2 Iw(drum)^2 = 1/2m(mass)v^2 + 1/2 Iw(drum)^2

With the left hand side being the initial velocities with the 0.3m/s.

This threw me well out (the answer is supposed to calculate to 3.01m/s.

Anyone able to help, it would greatly appreciated!

Thanks!
 

Answers and Replies

  • #2
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Sorry, forgot to add - I had a -3 in the left hand side as the friction.....
 
  • #3
BvU
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Well, did you work out your attempt ? How did you wrangle your -3 into the left hand side ? Is the friction there only initially ?
 
  • #4
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Hi BvU,
I was using the theory that,

T1 + ∑U1-2 = T2

Hence the 3Nm frictional torque was placed as a minus from the mgh on the left hand side. The frictional torque is stated as constant in the question hence I understand that it has to included as part the energy equation.
I did this problem earlier in the day and I belive I got something like 2 x10^-4 (!) so you can see I was well out....
I'm going to try it again as possibly tonight or tomorrow morning as I am flats out with assignments and revision at the moment so will stick up the answer I get the 2nd time around - never kn ow, I might get it right!
 
  • #5
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So to be a bit clearer on this and as sort of an answer to BvU, the following is the equation I have come up with to solve:
https://pod51046.outlook.com/owa/service.svc/s/GetFileAttachment?id=AAMkAGY5NTlkMjQ3LTdkOTUtNGRhMS05YjFlLTVhNzkzMTliNzUzOQBGAAAAAABNCMhttiqcQLg6VZ%2FUAKBeBwCA0zEXai05TZTyoTdiil0LAAAAefBKAADPKniALmOrSrBCx4TcTWeaAAE1I%2B0BAAABEgAQALuE7DSqlMpKlAAKu8VK5ek%3D&X-OWA-CANARY=OHs_oKPiqEyBerIAxjKWY6DFHxa79NEIHWIw4x0FUc0LBuIuZ27LjwRQCZZZe69f5KNq6ky2dbw.
However, I cannot solve as I cannot work out the linear acceleration to calculate M; (M = F*s and the F is calculated from the tension in the rope which is T=m(g-a).
This as well as the unknown V^2 I (which is what I'm looking for) is really acting as a 'show stopper' for me.
Anyone??
 
  • #7
Stephen Tashi
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Why don't you type out a list of the numerical values you used in the equation?
 
  • #8
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m(block)=8kg
g=9.81m/s^2
r1=0.2m
r2=0.3m
k=0.21m
m(drum)-12kg
Torque(tf)=3Nm
u=0.3 m/s

I=mk^2 was used to calculate I of the drum
Theta was calculated from = s/r
 
  • #9
BvU
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[edit] I started this yesterday but apparently forgot to post, sorry.

Although it is tempting to consider the dimension of a torque to be equal to the dimension of an energy, this is not the case: a torque is a (pseudo) vector and an energy is a scalar. So you can't just subtract the 3 from Δmgh.

Could you post your calculations in detail ? It enables more helpful assistance.

[edit] Well, you did some of that, so that deserves a bit more help:

I suppose the initial 1/2 mu2 gave you 0.36 J ? That's what I get.
And the initial 1/2 I u2/r22 another 0.265 J ?

As I said, the ##\tau_f## can't just be added to ##mgh##, which is a substantial 118 J . Strange you should get such low final speed !

And what is ##M\theta## ? Same story as ##\tau_f## ? Didn't ##M\theta## change into ##{1\over 2} I\; \omega^2 \ ##?

Now comes the bummer: (since you dropped the supposed answer) I can't reconstruct the 3 m/s from the energies (0.5 J + 118 J - Wf =? about 60 J from 3 m/s).

Perhaps it's necessary to find the actual acceleration of the mass and use v = at with t from v0+at2 ?
 
  • #10
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θHi BvU, thanks for the input.
The question was handed in a tutorial on work done from uni with the answer given as 3.01m/s and I have since found this question in a Meriam textbook with the same answer but no worked example or one that is anything like it to draw from.

I included the Mθ as I assumed this was a rotational energy, so yes , same as the torque - this is incorrect I take it?
 
  • #11
BvU
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Rotational energy is already accounted for in ##{1\over 2} I\; \omega^2 ##, so this way you count it twice !

Any ideas how to deal with the friction already ?

And: -- just to be sure -- is the mass initially moving upward at 0.3 m/s or is it moving downward ?
 
Last edited:

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