s and ca : 6 colli airs of 2. The Atwood mashine shown in figure below ver a M-5

Question

s and ca : 6 colli airs of 2. The Atwood mashine shown in figure below ver a M-500.g and a mass m 300.g connected by pulley in the shape of a disk of mass 750.g does not slip on th that passes o The string 10.0cm. and radius ra e edge of the pulley, which turns without friction about its center. (a)Find the tensions in th system. (b)Use th after it has fallen 1.00m if system starts from rest mentum glider e string segments and the acceleration a of the e conservation of energy to calculate the velocity of the 500.g mass

Explanation / Answer

here,

m = 300 g =0.3 kg

M = 500 g = 0.5 kg

mass of disk , mp = 750 g = 0.75 kg

r = 10 cm = 0.1 m

a)

the accelration , a = net force /effective mass

a = ( M - m) * g /( m + M + mp)

a = ( 0.5 - 0.3) * 9.81 /( 0.5 + 0.3 + 0.5 * 0.75)

a = 1.67 m/s^2

the tension in left side string be Tl and right side loop be Tr

for m

Tl - m * g = m * a

Tl = 0.3 * ( 9.81 + 1.67) N

Tl = 3.44 N

for M

M * g - Tr = M * a

Tr = 0.5 * ( 9.81 - 1.67) N

Tr = 4.07 N

b)

let the final speed be v

using conservation of energy

0.5 * (m + M) * v^2 + 0.5 * I * w^2 = g * h * ( M - m)

0.5 * (m + M) * v^2 + 0.5 * (0.5 * mp * r^2) * (v/r)^2 = g * h * ( M - m)

0.5 * ( 0.3 + 0.5) * v^2 + 0.5 * 0.5 * 0.75 * v^2 = 9.81 * 1 * ( 0.5 - 0.3)

solving for v

v = 1.83 m/s

the speed of 500 g mass is 1.83 m/s

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