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 0.310 m and moment of inertia I. The block on the frictionless incline is moving with a constant acceleration of magnitude a = 1.70 m/s2. (Let m1 = 15.5 kg, m2 = 21.5 kg, and = 37.0°.) From this information, we wish to find the moment of inertia of the pulley.

(a) What analysis model is appropriate for the blocks?

[]particle in equilibrium

[]particle under constant acceleration    

[]particle under constant speed

[]particle under constant angular acceleration

[]particle in simple harmonic motion

(b) What analysis model is appropriate for the pulley?

[]particle in equilibrium

[]particle under constant acceleration    

[]particle under constant speed

[]particle under constant angular acceleration

[]particle in simple harmonic motion

(c) From the analysis model in part (a), find the tension T1.

_______N

(d) From the analysis model in part (a), find the tension T2.
_______N

(e) From the analysis model in part (b), 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.
I = _________

(f) Find the numerical value of the moment of inertia of the pulley.
_________kg · m2

Explanation / Answer

here,

radius of pulley, r = 0.310 m
acceleration of block, a = 1.70 m/s^2

mass of block1, m1 = 15.5 kg
mass of block2, m2 = 21.5 kg

Angle of incline, A = 37.0 degrees

Part A:
The two blocks are moving with the same constant acceleration. Therefore, the model that's appropriate is that of a particle under constant acceleration

Part b:
The pulley is in rotation, its angular acceleration is constant because of the constant acceleration of the blocks. A particle under constant angular acceleration is the appropriate model.

Part C:
From newton Second Law, Fnet = 0
T1 - m1*g*Sin37 = m1*a

Solving for Tension in String, T1 = m1*a + m1*g*Sin37
T1 = 15.5*1.70 + 15.5*9.81*Sin37
T1 = 117.859 N

Part D:
From Newton Seocnd Law, SUM(F) = ma
m2*g - T2 = m2*a

Solving for Tension due to Block 2, T2
T2 = m2 ( g - a )
T2 = 21.5*( 9.81 - 1.70)
T2 = 174.365 N

Part E:
Newton's Law for the rotating pulley; taking the clockwise direction as positive :

T2*r - T1*r = I * alpha ( alpha is angular accelration)
Moment of inertia , I
I = r( t2 - t1) / alpha
I = r^2*( t2 - t1) / a ( alpha = a/r)

part f:
I = r^2*( t2 - t1) / a
I = 0.310^2*( 174.365 - 117.859) / 1.70
I = 3.194 Kg.m^2

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