The figure below (Figure 1) illustrates an Atwood\'s machine. Let the masses of

Question

The figure below (Figure 1) illustrates an Atwood's machine. Let the masses of blocks A and B be 4.00 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.300 kg?m2 and the radius of the wheel be 0.150 m .

Part A:

Find the linear acceleration of block A if there is no slipping between the cord and the surface of the wheel.

Express your answer in meters per second squared to three significant figures.

Part B:

Find the linear acceleration of block  B if there is no slipping between the cord and the surface of the wheel.

Express your answer in meters per second squared to three significant figures.

Part C:

Find the angular acceleration of the wheel C if there is no slipping between the cord and the surface of the wheel.

Express your answer in radians per second squared to three significant figures.

Part D:

Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel.

Express your answer in newtons to three significant figures.

Part E:

Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel.

Express your answer in newtons to three significant figures.

Explanation / Answer

here,

mA = 4 kg

mB = 2 kg

moment of inertia of pulley , I = 0.3 kg.m^2

radius , r = 0.15 m

A)

the linear acceleration of block A , a = the net force /effective mass

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

a = ( 4 - 2) * 9.81 /( 4 + 2 + 0.3 /0.15^2) m/s^2

a = 1.01 m/s^2

B)

the linear acceleration of block B is a = 1.01 m/s^2

C)

the angular accelration of wheel , alpha = a/r

alpha = 1.01 /0.15 rad/s^2

alpha = 6.76 rad/s^2

D)

for the left side chord

let the tension be Tl

for Block B

Tl - mB * g = mB * a

Tl = 2 * ( 9.81 + 1.01) N

Tl = 21.6 N

E)

for the right side chord

let the tension be Tr

for Block A

mA * g - Tr = mA * a

Tr = 4 * ( 9.81 - 1.01) N

Tr = 35.2 N

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