Answer:
(ω₁ / ω₂) = 1.9079
Explanation:
Given
R₁ = 3.59 cm
R₂ = 7.22 cm
m₁ = m₂ = m
K₁ = K₂
We know that
K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²
if
v₁ = ω₁*R₁
and
I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²
∴ K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² <em>(I)</em>
then
K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²
if
v₂ = ω₂*R₂
and
I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²
∴ K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂² <em>(II)</em>
<em>∵ </em>K₁ = K₂
⇒ 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²
⇒ ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²
⇒ (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²
⇒ (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)
⇒ (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²
⇒ (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)
⇒ (ω₁ / ω₂) = 1.9079
Answer:
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Explanation:
The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.
The angle between the 2 forces and displacement is ∅ = 15°.
First we have to calculate the work done by the individual force and then we can calculate the total work.
The work done on a particle by a constant force F during a straight line displacement s is given by following formula:
W = F*s
W = F*s*cos∅
With ∅ = the angles between F and s
The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N
The total work done can be calculated as followed:
Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2
Wtotal = 2Fs*cos∅
Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°
Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>
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The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
