An Atwood machine consists of two masses, m A= 7.0 kg and m B = 8.0 kg , connect

Question

An Atwood machine consists of two masses, mA= 7.0 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to rotate about a fixed axis. The pulley is a solid cylinder of radius R0 = 0.40 m and mass 0.80 kg .

Part A

Determine the magnitude of the acceleration a of each mass.

Express your answer using two significant figures.

Part B

What percentage of error in a would be made if the moment of inertia of the pulley were ignored? Ignore friction in the pulley bearings.

Express your answer using two significant figures.

Explanation / Answer

The moment of inertia of a solid cylinder of radius R and mass M is
I = 1/2 M R^2
Denote the tension in the left side if the cord by Ta and the tension in the right side of the cord by Tb. Then the equations of motion are:

mA a = TA - mA g
mB a = mB g - Tb
I a/R = (Tb - Ta) R (I times angular acceleration = applied torque)

The first is: Ta = mA (a + g)
The second is: Tb = mB (g - a)

Substituting this in the third

I a = mB (g - a) R^2 - mA (a + g) R^2

a * ( I + mB R^2 + mA R^2) = (mB - mA) g R^2

a = g ( mB-mA) / ( 1/2 M + mA + mB)

Therefore the acceleration is
a = 9.81 m/s^2 * 1 kg /(0.40 kg + 7 kg + 8 kg)
= 0.64 m/s^2

Ignoring the inertia of pulley

T = mA (a + g)
T = mB (g - a)

so,

mA (a + g)=mB (g - a)

(mA+mB)a=g(mB-mA)

a=g(mB-mA)/(mA+mB)

a=9.8(8-7)/(8+7)=9.8/15=.65 m/sec^2

%error=.01*100/.64=1.56%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.