A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil
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Answer:
N₂=20.05 rpm
Explanation:
Given that
R= 19 cm
I=0.13 kg.m²
N₁ = 24.2 rpm
ω₁= 2.5 rad/s
m= 173 g = 0.173 kg
v=1.2 m
Initial angular momentum L₁
L₁ = Iω₁ - m v r ( negative sign because bird coming opposite to motion of the wire motion)
Final linear momentum L₂
L₂= I₂ ω₂
I₂ = I + m r²
The is no any external torque that is why angular momentum will be conserve
L₁ = L₂
Iω₁ - m v r = I₂ ω₂
Iω₁ - m v r = ( I + m r²) ω₂
Now by putting the all values
Iω₁ - m v r = ( I + m r²) ω₂
0.13 x 2.5 - 0.173 x 1.2 x 0.19 = ( 0.13 + 0.173 x 0.19²) ω₂
0.325 - 0.0394 = 0.136 ω₂
ω₂ = 2.1 rad/s
N₂=20.05 rpm
Answer:
Explanation:
From the Question we are told that:
Mass
Coefficient of kinetic friction
Generally the equation for Frictional force is mathematically given by
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
Therefore