The spring-mounted 0.84-kg collar A oscillates along the horizontal rod, which is rotating at the constant angular rate rad/s. A
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Answer:
T = 5.36 s
Explanation:
given,
depth of the mine shaft = 122.5 m
speed of the sound = 340 m/s
time taken = ?
time taken by the stone to reach at the bottom
using equation of motion
initial speed , u = 0 m/s
t = 5 s
time taken by the sound to travel
d =v x t
t = 0.36 s
total time taken for the sound to reach carol after dropping the stone
T = 5 + 0.36
T = 5.36 s
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