As shown in the figure below, two blocks are connected by a string of negligible

Question

As shown in the figure below, two blocks are connected by a string of negligible mass passing over a pulley of radius 0.330 m and moment of inertia I. The block on the frictionless incline is moving with a constant acceleration of magnitude a = 1.40 m/s2. (Let m1 = 13.0 kg, m2 = 17.0 kg, and ? = 37.0°.) From this information, we wish to find the moment of inertia of the pulley.

(a), find the tension T1.

(b), find the tension T2.

(c) find a symbolic expression for the moment of inertia of the pulley in terms of the tensions T1 and T2, the pulley radius r, and the acceleration a.

(d) Find the numerical value of the moment of inertia of the pulley.

Explanation / Answer

here,

radius of pulley, r = 0.330 m

acceleration fo block, m1 = m2 , a = 1.40 m/s^2

mass of block1, m1 = 13 kg
mass of block2. m2 = 17 kg
mass of pulley = mp

Angle of inclined, A = 37 degrees

Part A :
From newton Second Law, Sum(F) = 0
T1 - m1*g*SinA = m1*a

t1 = m1*a + m1*g*SinA

t1 = 13*1.40 + 17*9.81*Sin37

t1 = 118.565 N

Part b:
From ewton Second Law,
T2 - m2*g = m2*a
T2 = m2*g + m2*a
T2 = 17*9.81 + 17.*1.40
T2 = 190.57 N

Part C and part D
From newton law for rotating pullley, taking clockwise as positive direction

t2*r - t1*r = I*alpha ( Torque = moment of inertia(I) * Angular Acceleration)

Since alpha = a / r

Solving for moment of inertia I,
I = (t2 - t1)*r^2 / a
I = (190.57 - 118.565)*(0.330)^2 / 1.40
I = 5.601 N.m^2

Part D:

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