Answer:
(A) 0.54 kg.m^{2}
(B) 0.0156 N
Explanation:
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here is the complete question:
A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5100 rev/min. (a) Calculate the rotational inertia of the system about the axis of rotation. (b) There is an air drag of 2.60 x 10^{-2} N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?.
solution
mass of the ball (m) = 1.5 kg
length of the rod (L) = 0.6 m
angular velocity (ω) = 4900 rpm
air drag (F) = 2.60 x 10^{-2} N = 0.026 N
(take note that values from the original question are used, with the exception of the air drag which was not in the original question)
(A) because the rod is mass less, the rotational inertia of the system is the rotational inertia of the rod about the other end, hence rotational inertia = where m = mass of ball and L = length of rod
= = 0.54 kg.m^{2}
(B) The torque that must be applied to keep the ball in motion at constant speed = FLsin90
= 0.026 x 0.6 x sin 90 = 0.0156 N
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I trust the appropriate response will help you.
Answer:
The error in tapping is ±0.02828 ft.
Explanation:
Given that,
Distance = 200 ft
Standard deviation = ±0.04 ft
Length = 100 ft
We need to calculate the number of observation
Using formula of number of observation
Put the value into the formula
We need to calculate the error in tapping
Using formula of error
Put the value into the formula
Hence, The error in tapping is ±0.02828 ft.
Answer:
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