Answer:
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 %
