Please tell whether my method is correct or not.
Problem statement: Two blocks are connected over pulley with light thread. The pulley is rough and does’t rotate.
I have taken the angular acceleration of pulley as 0.
you forgot about the friction in the pulley
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how will it effect? could you provide your solution?
there is no rotation so there shouldn’t be any change in tension forces… i think??
There’s a well known formula that when the rope is wrapped at angle \alpha, then
T_2 = T_1 e^{\displaystyle\pm\mu\alpha}
plus-minus depends on the direction of motion
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but how can we derive it here? could you explain please?
The derivation is very similar to the derivation of Laplace pressure formula. We take a very small part of the rope which makes a small angle d\alpha.
The normal reaction force is N = 2\cdot T\cdot d\alpha/2, and dT = \mu N = \mu Td\alpha. Then you simply integrate it
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Thanks!
I have got the answer as this :
If we put in values, m1 = 2/3 kg, m2 = 8 kg, µ = ln(2)/π. I am getting 50/7 m/s^2. is this correct?
yes, it’s right
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Oh thanks!