Abhisar
  • Abhisar
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)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Compassionate
  • Compassionate
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?
Abhisar
  • Abhisar
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.
Compassionate
  • Compassionate
Hmm. Marvelous. Give me a minute to write this down.

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Abhisar
  • Abhisar
Sure..
Compassionate
  • Compassionate
I'm at McDonald's so my wifi is really slow. Bare with me.
Abhisar
  • Abhisar
ok..
Compassionate
  • Compassionate
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.
Abhisar
  • Abhisar
Sure, No problem
Abhisar
  • Abhisar
Well, I think I got it doe.....
Compassionate
  • Compassionate
Show me what you got.
Abhisar
  • Abhisar
\(\sf α_11=1.57rad/s^2\) \(ω_2=100rad/m\) =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.
Compassionate
  • Compassionate
That's what I got.
Compassionate
  • Compassionate
I actually kept getting 21.666 repeating, but 20 looks good.
Compassionate
  • Compassionate
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
Abhisar
  • Abhisar
Sure..
Compassionate
  • Compassionate
\[\omega _{average} = \frac{ \Delta \theta }{ \Delta \tau } \] On the final step, I put in the known factors to isolate Wheel B, and simply solved for t, which gave me 21.67
Compassionate
  • Compassionate
I suspect I may be wrong.
Abhisar
  • Abhisar
I am not sure if that would be correct. I am myself confused >_<
Compassionate
  • Compassionate
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.
Abhisar
  • Abhisar
Thanks c:
Abhisar
  • Abhisar
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
Abhisar
  • Abhisar
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, \(\sf \omega=\omega_0+\alpha t \\ 10\pi/3=0+0.52\times t\\ \therefore t=20.1 \approx 20sec\)
IrishBoy123
  • IrishBoy123
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
Astrophysics
  • Astrophysics
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