In an Atwood machine the two masses (m1 =5.0kg and m2 = 3.0 kg) are released from rest, with m1 at a height of h=0.75 m above the floor. When m1 hits the ground, its speed is v= 1.8 m/s. Assume that the pulley is a uniform disk with mass mp with radius r = 12 cm. Find a general expression of mp as a function of m1, m2, v and h. and then determine the pulleys mass. What % of total KE is stored in the KE of the pulley? (Icylinder = ½ mr2)

Explanation:

m₁ is heavy so it will go down and m₂ will go up. Let common acceleration be a .

For downward acceleration of m₁

m₁g - T₁ = m₁a

For upward acceleration of m₂

T₂- m₂g = m₂a

adding

m₁g - T₁ +T₂- m₂g = (m₁+m₂)a

m₁g- m₂g  +(T₂- T₁) = (m₁+m₂)a   ( 1 )

For rotatory motion of disc of mass mp and moment of inertia I

(T₁ -T₂ )R = I α , α is angular acceleration ( 2 )

From  1 and 2

m₁g- m₂g -  I α / R = (m₁+m₂)a

m₁g- m₂g -  I a / R² = (m₁+m₂)a

m₁g- m₂g = (m₁+m₂)a + I a / R²

m₁g- m₂g = [(m₁+m₂) + mp k² / R² ] a   ( k is radius of gyration of disk )

a = (m₁- m₂)g / [(m₁+m₂) +mp k² / R² ]

=  (m₁- m₂)g / [(m₁+m₂) + .5 mp  ]  ( for cylinder k² / R² = .5 )

If h be the height by which m₁ falls and its velocity becomes v

v² = 2 a h

a = v² / 2h

v² / 2h = (m₁- m₂)g / [(m₁+m₂) + .5 mp  ]

v²[(m₁+m₂) + .5 mp  ] = 2(m₁- m₂) gh

- v²[(m₁+m₂) +2(m₁- m₂) gh =  .5 mp v²

- v²[(m₁+m₂) +2(m₁- m₂) gh / .5 v ² = mp .

- 1.8² ( 5 + 3 ) + 2 ( 5 - 3 ) 9.8 x .75 / .5 x 1.8² = mp

- 25.92 + 29.4 / 1.62 = mp

2.15 kg = mp .

K.E  of masses = 1/2 x 5 x 1.8² + 1/2 x 3 x 1.8²

= 12.96 J

Rotational KE of disc

= 1/2 I ω²

= 1/4 mp r²ω²

= 1/4 x mp x v²

= .25 x 2.15 x 1.8²

= 1.74 J

in percent terms

(1.74 / 12.96) x 100

= 13.5 %

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.