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Find acceleration of a Block within a simply pulley system. |
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| Dec11-09, 02:57 AM | #1 |
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Find acceleration of a Block within a simply pulley system.
1. The problem statement, all variables and given/known data
The figure shows two blocks suspended by a cord over a pulley. The mass of black B is twice the mass of black A, [tex]m_b = 2m_a = 2m[/tex]. The mass of the pulley is equal to the mass of black A, [tex]m_p = m_a = m[/tex], the radius of the pulley is [tex]R[/tex]. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cords are massless. The rotational inertia of the pulley is [tex]\frac{m_pR^2}{2}[/tex] about a perpendicular axis through the center. Find the magnitude of acceleration of the block B. Express your results in the simplest possible form in terms of any or all the following: m, g, R, and universal physical or mathematical constants. 2. Relevant equations [tex]F_G = m * g[/tex] (force of gravity equals mass times gravitational acceleration) [tex]\tau_net = I * \alpha[/tex] (torque equals moment of inertia times angular acceleration) 3. The attempt at a solution [tex]\tau_{net} = I * \alpha = \frac{m * R^2}{2} * \alpha = \frac{m * R^3}{2} * a[/tex] furthermore, [tex]\tau_{net} = F_{G on B} * R + F_{G on A} * R = 2 * m * g * R - m * g * R = m * g * R[/tex] so, [tex]m * g * R = \frac{m * R^3}{2} * a[/tex] which leads to the answer: [tex]a = \frac{2 * g}{R^2} m/s^2[/tex] Is this correct? Is there an easier way? |
| Dec11-09, 11:33 AM | #2 |
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bump... picture of problem lol
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| Dec11-09, 01:09 PM | #3 |
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| Dec11-09, 03:37 PM | #4 |
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Find acceleration of a Block within a simply pulley system.
this accn.. u solved is for t he pulley not for the block B
so use the principle of energy conversion to solve the problem then it will give the acceleration for block B.. |
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