- #1
tandoorichicken
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A cylinder of mass m and radius R is rigidly mounted to the same shaft as a lightweight cylinder of radius R/2. The shaft is free to turn with negligible friction. Cords are wound in opposite directions about the cylinders, and identical masses m are hung from the cords. a) Which way do the cylinders turn? (clockwise) b) what is the acceleration of the mass hung on the cylinder of radius R?
The actual problem comes with a picture that has a cord coming off the right side of the big disk and a cord coming off the left side of the small disk. I know the assembly spins clockwise because the mass on the big cylinder has more torque than the one on the small cylinder.
Part B I'm not so sure about. Basically I equated the tension in the wire to the tangental force from the pulley, and wrote the torque equation: [itex] \tau = R F [/itex]. From there I went to [itex] R F = I\alpha = \frac{1}{2} mR^2\alpha [/itex] and solved for [itex] \alpha [/itex]. The final result was [itex] \alpha = \frac{2F}{mR} [/itex]. My only question is did I do it right and if not what other factors do I need to include?
The actual problem comes with a picture that has a cord coming off the right side of the big disk and a cord coming off the left side of the small disk. I know the assembly spins clockwise because the mass on the big cylinder has more torque than the one on the small cylinder.
Part B I'm not so sure about. Basically I equated the tension in the wire to the tangental force from the pulley, and wrote the torque equation: [itex] \tau = R F [/itex]. From there I went to [itex] R F = I\alpha = \frac{1}{2} mR^2\alpha [/itex] and solved for [itex] \alpha [/itex]. The final result was [itex] \alpha = \frac{2F}{mR} [/itex]. My only question is did I do it right and if not what other factors do I need to include?