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Abhisar

  • one year ago

A wheel having a radius of 10 cm is coupled by a belt to another wheel of radius 30 cm. 1st wheel increases its angular speed from rest at a uniform rate of 1.57 rad/s2. The time for the 2nd wheel to reach a rotational speed of 100 rev/min is (assume that the belt does not slip)

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  1. Compassionate
    • one year ago
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    There's actually a really simple equation for this. It's kind of like a system of equations. Since you have two variables, you'll need two equations. Did your book or course give you any equations to use? Do you have an idea of what you need to do?

  2. Abhisar
    • one year ago
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    Well, I know that linear velocity for both the wheels will be equal and I need to equate them to find the angular velocity of second.

  3. Compassionate
    • one year ago
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    Hmm. Marvelous. Give me a minute to write this down.

  4. Abhisar
    • one year ago
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    Sure..

  5. Compassionate
    • one year ago
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    I'm at McDonald's so my wifi is really slow. Bare with me.

  6. Abhisar
    • one year ago
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    ok..

  7. Compassionate
    • one year ago
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    I almost got it but my computer is freezing. I gotta use the restroom. Give me another minute. It's really freezing up and the wifi keeps going out. But I think I have it.

  8. Abhisar
    • one year ago
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    Sure, No problem

  9. Abhisar
    • one year ago
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    Well, I think I got it doe.....

  10. Compassionate
    • one year ago
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    Show me what you got.

  11. Abhisar
    • one year ago
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    \(\sf α_11=1.57rad/s^2\) \(ω_2=100rad/m\) =100×2π/60 =10π/3rad/s linear velocity of belt is equal ∴ω1r1=ω2r1 and α1r1=α2r2 ∴α2=(1.57)(1030) =1.573rad/s^2 \(\sf ω=ω_0+αt\) 10π/3=0+1.573×t t=20sec.

  12. Compassionate
    • one year ago
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    That's what I got.

  13. Compassionate
    • one year ago
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    I actually kept getting 21.666 repeating, but 20 looks good.

  14. Compassionate
    • one year ago
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    If I redo it, I got 20. My first two times I was going say 21.7, but decimals are... uncommon. I'll show my work

  15. Abhisar
    • one year ago
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    Sure..

  16. Compassionate
    • one year ago
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    \[\omega _{average} = \frac{ \Delta \theta }{ \Delta \tau } \] On the final step, I put in the known factors to isolate Wheel B, and simply solved for t, which gave me 21.67

  17. Compassionate
    • one year ago
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    I suspect I may be wrong.

  18. Abhisar
    • one year ago
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    I am not sure if that would be correct. I am myself confused >_<

  19. Compassionate
    • one year ago
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    You know what, I just realized w(a) is for average, but I am assumming you're looking for the Vf or Vi, even then, I think I royally messed up my average. I'm more of a newtonian kinda guy ;) 20 sounds good.

  20. Abhisar
    • one year ago
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    Thanks c:

  21. Abhisar
    • one year ago
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    I'll drop an explanation, may be some one else need it too. Speed of second disk will be \(\sf \Large {\frac{100 \times 2\pi}{60}=\frac{10 \pi}{3}}rad/s\) Since both the wheels or disks are connected, linear velocity and linear acceleration of each of them will be same

  22. Abhisar
    • one year ago
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    So, we can write that, \(\sf a_1=a_2\\\Rightarrow\alpha_1\times r_1=\alpha_2\times r_2\\ \therefore \alpha_2=0.52\) Now, using equations of motion, \(\sf \omega=\omega_0+\alpha t \\ 10\pi/3=0+0.52\times t\\ \therefore t=20.1 \approx 20sec\)

  23. IrishBoy123
    • one year ago
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    for connected wheels you can say \(r_1 \theta_1 = r_2 \theta _2\) provided the belt does not slip from there, it follows that \(r_1 \dot \theta_1 = r_2 \dot \theta _2\) and \(r_1 \ddot \theta_1 = r_2 \ddot \theta _2\) so the acceleration of the larger wheel is \(\ddot \theta = \frac{1.57}{3} rad/s^2\) using the acircular analogue for \(v = u + at\) ie \(\dot \theta_f = \dot \theta_i + \ddot \theta t\) with \(\dot \theta_i = 0\) gives \(t = \dfrac{\dot \theta_f}{\ddot \theta} = \dfrac{2\pi \frac{100}{60}}{\frac{1.57}{3}} \approx 20s\) so i guess i agree with you guys:p

  24. Astrophysics
    • one year ago
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    .

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