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Old May 7th 2017, 09:00 AM   #1
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Acceleration of a Block on an Atwood's Machine?

Hello. I'm stuck on a problem that has me determining the acceleration of a block on an Atwood's Machine. (The whole question and figure are shown in the attached screenshot).

Information
Tensions T1 and T2
M1 = 2.0kg, M2 = 4.0 kg, and Mp (pulley) = 2.0 kg
R = 6.0 cm
Frictional Torque = 0.5Nm

I started with 3 equations:
1) T1 - M1g = M1a1
2) T2 - M2g = M2a2
3) T2R - T1R - 0.5 = I (a/R) (Net Torque)

After plugging in I = 1/2mR^2, 3 simplifies to
T2-T1 = (1/2)Mpa + (0.5/R)

I'm not sure where to go from here. Please help!
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Acceleration of a Block on an Atwood's Machine?-capture.png  
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Old May 7th 2017, 01:02 PM   #2
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scalar equations for net torque & force ...

$T_1 r - T_2 r - \tau_f = \dfrac{1}{2}m_p r^2 \cdot \dfrac{a}{r}$

$T_1 - T_2 - \dfrac{\tau_f}{r} = \dfrac{1}{2}m_p \cdot a$

$\color{red}{T_1 - T_2 - \dfrac{25}{3} = a}$

$\color{red}{4g - T_1 = 4a}$

$\color{red}{T_2 - 2g = 2a}$

summing the last three equations yields ...

$2g - \dfrac{25}{3} = 7a \implies a = \dfrac{6g-25}{21} \, m/s^2$


$|\Delta y| = \dfrac{1}{2}at^2 \implies t = \sqrt{\dfrac{2\Delta y}{a}} = \sqrt{\dfrac{42}{6g-25}} \approx 1.1 \, sec$
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Old May 7th 2017, 06:39 PM   #3
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Thanks so much!
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Old May 10th 2017, 12:36 PM   #4
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Atwood was Nerdy!

Hi,

See what you think of this development....

Atwood's Machine | THERMO Spoken Here!

Good luck with your studies...

TSH
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acceleration, atwood, block, machine, torque



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