Physics Help Forum Acceleration of a Block on an Atwood's Machine?

 May 7th 2017, 09:00 AM #1 Junior Member   Join Date: May 2017 Posts: 2 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! Attached Thumbnails
 May 7th 2017, 01:02 PM #2 Member     Join Date: Aug 2008 Posts: 46 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 ... $2g - \dfrac{25}{3} = 7a \implies a = \dfrac{6g-25}{21} \, m/s^2$ $|\Delta y| = \dfrac{1}{2}at^2 \implies t = \sqrt{\dfrac{2\Delta y}{a}} = \sqrt{\dfrac{42}{6g-25}} \approx 1.1 \, sec$ TheNerdyGinger likes this.
 May 7th 2017, 06:39 PM #3 Junior Member   Join Date: May 2017 Posts: 2 Thanks so much!
 May 10th 2017, 12:36 PM #4 Senior Member   Join Date: Jun 2010 Location: NC Posts: 341 Atwood was Nerdy! Hi, See what you think of this development.... Atwood's Machine | THERMO Spoken Here! Good luck with your studies... TSH

 Tags acceleration, atwood, block, machine, torque

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