An Atwood\'s machine consists of two masses, m A and m B, which are connected by

Question

An Atwood's machine consists of two masses, mA and mB, which are connected by a massless inelastic cord that passes over a pulley, see the figure(Figure 1) .

Part A

If the pulley has radius R and moment of inertia I about its axle, determine the acceleration of the masses mA and mB. [Hint: The tensions FTA and FTBare not equal. We discussed the Atwood machine in Example 4-13 in the textbook, assuming I=0 for the pulley.]

Express your answer in terms of the variables mA, mB, I, and appropriate constants.

Part B

Compare to the situation in which the moment of inertia of the pulley is ignored.

Express your answer in terms of the variables mA, mB, and appropriate constants.

a=

Part B

Compare to the situation in which the moment of inertia of the pulley is ignored.

Express your answer in terms of the variables mA, mB, and appropriate constants.

aI=0 = Problem 10.51 An Atwood's machine consists of two masses, m A and mB which are connected by a massless inelastic cord that passes over a pulley, see the figure(Figure 1) Figure 1 v of 1 OR TA TB mB (m ma Submit My Answers Give U Incorrect: Try Again: 9 attempts remaining; no points deducted The correct answer does not depend on: a, man mb, r. Part B Compare to the situation in which the moment of inertia of the pulley is ignored. Express your answer in terms of the variables mA,mB, and appropriate constants (m ini )g m (m m1) s Submit My Answers Give U Incorrect; Try Again: 9 attempts remaining The correct answer does not depend on: m, m 1, m1, m2, s.

Explanation / Answer

(A)

For the first mass

mBg - FTB = mBa

FTB = mBg- mBa

For the second mass...

FTA - mAg = mAa

FTA = mAg+ mAa

For the pulley

(FTB - FTA)R = Ia/R

mBg- mBa - mAg- mAa = Ia/R2

mBg - mAg = Ia/R2 + mAa + mBa

mBg - mAg = a(I/R2 + mA + mB)

a = {mBg - mAg} / {(I/R2 )+ mA + mB}

(B)

The tensions are the same if we ignore the moment of inertia of the pulley

For the first mass

mBg - FT = mBa

FT = mBg- mBa

For the second mass...

FT - mAg = mAa

FT = mAg+ mAa

mBg- mBa = mAg+ mAa

mBg- mAg = mBa+ mAa

a = (mBg- mAg) / (mA + mB)

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