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pleasantresult
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- Homework Statement
- This is question 1.3 from Fetter and Walecka.
A massless string is placed over a massless pulley, and each end is wound around and fastened to a vertical hoop. The hoops have masses M1, M2 and radii R1, R2. The apparatus is placed in a uniform gravitational field g and released with each end of the string aligned along the field. Show that the tension in the string is tau = g*M1*M2*(M1+M2)^(-1).
- Relevant Equations
- tau = g*M1*M2*(M1+M2)^(-1)
I have seen the solution to this problem but I don't understand it.
Here is my drawing of the problem.
Apparently I should do the sum of the forces and sum of the torques for both hoops like this:
hoop 1:
(1) τ - M1g = M1(a1t-a1R)
(2) τR1 = ½M1R12a1r/R1
hoop 2:
(3) M2g - τ = M2(a1t+a2r)
(4) τR2 = ½M2R22a2r/R2
Where a1t and a2t are the translation accelerations up or down of hoops 1 and 2.
Where a1r and a2r are the rotational tangential accelerations of hoop 1 and hoop 2.
In the sum of the forces equations, the total linear accelerations are found by subtracting the tangential rotational acceleration for hoop 1 and adding the tangential rotational acceleration for hoop 2 based on the chosen orientations of positive linear and angular directions for the hoops.
In the third equation, you can sub in a1t where you would expect a2t because without rotation, the distance that hoop 2 drops will be the same as the distance that hoop 1 rises.
Then you have four equations with the four unknowns τ, a1t, a1r, and a2r which you can solve to get the result that τ = gM1M2/(M1+M2)-1
Everything about this setup makes sense to me except the moments of inertia that are used in equations 2 and 4. Isn't the moment of inertia of a thin hoop about its center MR2 not ½MR2?
Here is my drawing of the problem.
Apparently I should do the sum of the forces and sum of the torques for both hoops like this:
hoop 1:
(1) τ - M1g = M1(a1t-a1R)
(2) τR1 = ½M1R12a1r/R1
hoop 2:
(3) M2g - τ = M2(a1t+a2r)
(4) τR2 = ½M2R22a2r/R2
Where a1t and a2t are the translation accelerations up or down of hoops 1 and 2.
Where a1r and a2r are the rotational tangential accelerations of hoop 1 and hoop 2.
In the sum of the forces equations, the total linear accelerations are found by subtracting the tangential rotational acceleration for hoop 1 and adding the tangential rotational acceleration for hoop 2 based on the chosen orientations of positive linear and angular directions for the hoops.
In the third equation, you can sub in a1t where you would expect a2t because without rotation, the distance that hoop 2 drops will be the same as the distance that hoop 1 rises.
Then you have four equations with the four unknowns τ, a1t, a1r, and a2r which you can solve to get the result that τ = gM1M2/(M1+M2)-1
Everything about this setup makes sense to me except the moments of inertia that are used in equations 2 and 4. Isn't the moment of inertia of a thin hoop about its center MR2 not ½MR2?