Find acceleration of a Block within a simply pulley system.

AI Thread Summary
The discussion focuses on calculating the acceleration of block B in a pulley system where block B has twice the mass of block A. The initial approach involves using torque and moment of inertia equations, but it is pointed out that the assumptions made about tension forces are incorrect due to the system's acceleration. Participants emphasize the necessity of drawing free body diagrams for each block and the pulley to apply Newton's laws effectively. An alternative method suggested is to use energy conservation principles to find the acceleration. The correct approach is crucial for accurately determining the acceleration of block B.
cocoon
Messages
10
Reaction score
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, m_b = 2m_a = 2m. The mass of the pulley is equal to the mass of black A, m_p = m_a = m, the radius of the pulley is R. 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 \frac{m_pR^2}{2} 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



F_G = m * g (force of gravity equals mass times gravitational acceleration)
\tau_net = I * \alpha (torque equals moment of inertia times angular acceleration)

The Attempt at a Solution



\tau_{net} = I * \alpha = \frac{m * R^2}{2} * \alpha = \frac{m * R^3}{2} * a

furthermore,

\tau_{net} = F_{G on B} * R + F_{G on A} * R = 2 * m * g * R - m * g * R = m * g * R

so,

m * g * R = \frac{m * R^3}{2} * a

which leads to the answer:

a = \frac{2 * g}{R^2} m/s^2

Is this correct? Is there an easier way?
 
Last edited:
Physics news on Phys.org
bump... picture of problem lol
http://img340.imageshack.us/img340/690/18485819.jpg
 
Last edited by a moderator:
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, m_b = 2m_a = 2m. The mass of the pulley is equal to the mass of black A, m_p = m_a = m, the radius of the pulley is R. 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 \frac{m_pR^2}{2} 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



F_G = m * g (force of gravity equals mass times gravitational acceleration)
\tau_net = I * \alpha (torque equals moment of inertia times angular acceleration)

The Attempt at a Solution



\tau_{net} = I * \alpha = \frac{m * R^2}{2} * \alpha = \frac{m * R^3}{2} * a
oops, slight error in your equation there a = \alpha R, so \tau = MRa/2
furthermore,

\tau_{net} = F_{G on B} * R + F_{G on A} * R = 2 * m * g * R - m * g * R = m * g * R
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.
 
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..
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top