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A pulley system has mass moment of inertia of 1.4kgm^2. If 8kg block is released from rest from S=1, by using energy approach what is angular velocity at S=2? Radius of pulley is 250mm. Note S is from the pulley centre.

This is what I did:

$$T_1+V_1=T_2+V_2$$ $$0+ 8(g)(1)= \frac{1}{2}(8)w^2(0.250)^2 + \frac{1}{2} Igw^2 - 8(g)1$$

Then rearranged for W and got 12.85rad/s?? Am I on the right tracks please? enter image description here

The mass is at the bottom and it is originally 1m away. The mass then moves down to 2m away from the pulley from this original position of 1m away.

Emmanuel
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1 Answers1

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This is in the right track but not entirely.

$$T_1+V_1=T_2+V_2$$

I found two problems

  • one with the initial position. Let's assume position is h=0 initially, and the mass travels by S=1m. So the initial energy can be thought of as zero.
  • the other is with an added g in the rotational energy of the pulley. It should be $\frac{1}{2} Iw^2 $ instead of $\frac{1}{2} Igw^2 $.

Below I am showing the differences.

$$0+ 8(g)(1)= \frac{1}{2}(8)w^2(0.250)^2 + \frac{1}{2} Igw^2 - 8(g)1$$

$$0+ \color{red}{0}= \frac{1}{2}(8)w^2(0.250)^2 + \frac{1}{2} \color{red}{I w^2 } + 8\cdot(g)\cdot \color{red}{(-1)}$$

The rest is exactly like you've done already.

NMech
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  • Hi Nmech, thank you much for looking into this. The Ig was meant to say it was mass at centre, sorry it was my fault I couldn’t write it here properly. From yours, wouldn’t I then get a negative number of the left and then square root of a negative number please? – Emmanuel Mar 23 '21 at 20:14
  • you are right I mistyped the $-$. Initially, I wanted to write it fancier. Update the page and you will see what I mean. But essentially you were right, I should have put -1, now its corrected. – NMech Mar 23 '21 at 20:25
  • Thank you Nmech! Youre a great mechanical engineer! Stackexchange should be proud to have you! – Emmanuel Mar 23 '21 at 20:34