In the figure below, block 1 has mass m_1 = 460 g, block 2 has mass m_2 = 500 g,

Question

In the figure below, block 1 has mass m_1 = 460 g, block 2 has mass m_2 = 500 g, and the pulley, which is mounted on a horizontal axle with negligible friction, has radius R = 5.00 cm. When released from rest, block 2 falls 83.800 cm in 2.6100 s without the cord slipping on the pulley. (Do not treat the pulley as a uniform disk. Give your answers to three significant figures. Use these rounded values in subsequent calculations.) (a) What is the magnitude of the acceleration of the blocks? Block 2 falls at constant acceleration and so the constant-acceleration equations apply. Write Newton's second law separately for block 1 and for block 2. Write Newton's second law in angular form for the pulley, symbolizing the rotational inertia. The acceleration magnitude of a point on the rim must equal the acceleration magnitude of each block, m/s^2 (b) What is the tension T_2? (c) What is the tension T_1? (d) What is the magnitude of the pulley's angular acceleration? rad/s^2 (e) What is its rotational inertia? kg middot m^2

Explanation / Answer

Here ,

m1 = 0.460 Kg

m2 = 0.500 Kg

R = 0.05 m

a) let the acceleration is a

Using second equation of motion

d = 0.50 * a * t^2

0.838 = 0.50 * a * 2.61^2

a = 0.246 m/s^2

the magnitude of acceleration is 0.246 m/s^2

b)

Now, for the tension T2

Using second law of motion

0.50 * 9.8 - T2 = 0.50 * 0.246

T2 = 4.78 N

the tension T2 is 4.78 N

c)

for the tension T1

T1 = m1 * g + m1 * a

T1 = 0.46 * (9.8 + 0.246)

T1 = 4.62 N

the tension T1 is 4.62 N

d)

magnitude of angular acceleration = a/R

magnitude of angular acceleration = 0.246/0.05 rad/s^2

magnitude of angular acceleration =4.92 rad/s^2

e)

Now, using second law of motion

I * a = (T2 - T1) * R

I * 4.92 = (4.78 - 4.62) * 0.05

solving for I

I = 1.63 *10^-3 Kg.m^2

the rotational inertia of the pulley is 1.63 *10^-3 Kg.m^2

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