Question

The figure below (Figure 1) illustrates an Atwood's machine. Let the masses of blocks A and B be 3.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.400 kg⋅m2 and the radius of the wheel be 0.100 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.

Answer 1

Let, tension in left cable is T1 , tension in right cable is T2 and linear acceleration is a.

(a)

From equilibrium of torque,

T2*R - T1*R = l

For no slipping between the cord and the surface of the wheel,

= a / R

so, T2 - T1 = l*(a / R^2) ......(1)

equation for block B,

T1 - m_Bg = m_Ba ........(2)

Equation for block A,

m_Ag - T2 = m_Aa ..........(3)

T1 = m_B(a + g) ..........(4)

T2 = m_A(g - a) ...........(5)

From eq (1), (4) and (5),

m_A(g - a) - m_B(a + g) =  l*(a / R^2)

(m_A - m_B)g = ((m_A + m_B) + l / R^2) a

linear acceleration a = (3.5 - 2)*9.81 / [(3.5 + 2) + 0.40 / 0.10^2]

a = 0.323 m/s^2

linear acceleration of block A = 0.323 m/s^2

(b)

linear acceleration of block B = 0.323 m/s^2

(c)

Angular acceleration,

= a / R = 0.323 / 0.10

= 3.23 rad/s^2

(d)

Tension in left side of the cord,

T1 = m_B(a + g) = 2*(0.323 + 9.81)

T1 = 20.26 N

(e)

Tension in right side of the cord,

T2 = m_A(g - a) = 3.5 (9.81 - 0.323)

T2 = 33.20 N