Find acceleration of a Block within a simply pulley system.

In summary, the figure shows two blocks suspended by a cord over a pulley, with the mass of black B being twice the mass of black A. The mass of the pulley is equal to the mass of black A, and the blocks are let free to move without slipping or stretching. The rotational inertia of the pulley is given as \frac{m_pR^2}{2}. Using the equations F_G = m * g and \tau_net = I * \alpha, and considering the free body diagrams of each block and the pulley, the magnitude of acceleration of block B can be solved for by writing 3 equations with 3 unknowns: T_A, T_B, and a. Alternatively, the principle
  • #1
cocoon
10
0

Homework Statement



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.

Homework 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)

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?
 
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  • #2
bump... picture of problem lol
http://img340.imageshack.us/img340/690/18485819.jpg
 
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  • #3
cocoon said:

Homework Statement



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.

Homework 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)

The Attempt at a Solution



[tex]\tau_{net} = I * \alpha = \frac{m * R^2}{2} * \alpha = \frac{m * R^3}{2} * a[/tex]
oops, slight error in your equation there [tex]a = \alpha R, [/tex] so [tex] \tau = MRa/2 [/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]
no, that's not right, you've assumed the respective tension forces are equal to the respective weights, but the system is accelerating, so the tensions cannot equal the weights
Is this correct? Is there an easier way?
No, this is incorrect, you need to write 3 equations with 3 unknowns, T_A, T_B, and a, so draw Free Body diagrams (FBD's ) of each block and the pulley to solve for them... separte FBD's and application of Newton's laws to each are essential in these type problems. Trying to take shortcuts often results in incorerct methods and solutions.
 
  • #4
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..
 

1. How do you determine the acceleration of a block within a simple pulley system?

The acceleration of a block within a simple pulley system can be determined by using the equation a = (m2 - m1)/(m1 + m2) * g, where m1 is the mass of the lighter block, m2 is the mass of the heavier block, and g is the acceleration due to gravity.

2. What factors affect the acceleration of a block in a simple pulley system?

The acceleration of a block in a simple pulley system is affected by the mass of the blocks, the friction in the pulley system, and the angle at which the rope is pulled.

3. Can the acceleration of a block in a simple pulley system be negative?

Yes, the acceleration of a block in a simple pulley system can be negative if the heavier block is moving downwards with a greater force than the lighter block is moving upwards.

4. How does the acceleration of a block change if the angle of the rope is changed?

If the angle of the rope is increased, the acceleration of the block will decrease as the force pulling the block upwards decreases. If the angle of the rope is decreased, the acceleration of the block will increase as the force pulling the block upwards increases.

5. What is the relationship between the acceleration of the block and the tension in the rope?

The acceleration of the block is directly proportional to the tension in the rope. This means that if the tension in the rope increases, the acceleration of the block will also increase.

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