Consider the system shown in the diagram. the pulley is a uniform cylinder with

Question

Consider the system shown in the diagram. the pulley is a uniform cylinder with mass m_3 = 0.400 kg and radius R = 4.00 cm, the other two masses are m_1 = 2.00 kg and m_2 = 1.00 kg. and alpha = 25.0 degree. Assume the rope is massless. there is no slipping of the rope on the pulley, and there is no friction between m_1 and the incline. What is the acceleration of m_1 and m_2 (both magnitude and direction)? What is the angular acceleration of the pulley? What are the tensions in the rope (both to the left of the pulley and to the right of the pulley? If m_2 starts from rest 50.0 cm above the ground, where is it when it has speed 0.500 m/s?

Explanation / Answer

here,

m1 = 2 kg , m2 = 1 kg

m3 = 0.4 kg

R = 0.04 m

a)

the accelration of system , a= net force /effective mass

a = ( m2 * g - m1 * g * sin(25))/( m1 + m2 + 0.5 * m3 * r^2 /r^2)

a = ( 1 * 9.8 - 2 * 9.8 * sin(25))/( 2 + 1 + 0.5 * 0.4)

a = 0.47 m/s^2

the accelration of the system is 0.47 m/s^2

the angular accelration of the disc , alpha = a/r = 11.8 rad/s^2

let the tension in the string be t1 and t2

for m1

t1 - m1 * g * sin(theta) = m1 * a

t1 = 2 * ( 9.8 * sin(25) + 0.47 )

t1 = 9.22 N

for m2

t2 - m2 * g = m2 * a

t2 = 1 * ( 9.8 + 0.47)

t2 = 10.27 N

b)

when v = 0.5 m/s

h' = v^2 /2a

h' = 0.5^2 /( 2 * 0.47)

h' = 0.2659 m = 26.59 cm

so, the distance from the ground , d = h - h'

d = 23.4 cm

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