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calculating angular speed in revolutions per second?
A constant torque of 26.3N*m is applied to a grindstone whose moment of inertia is 0.101kgm2. Using energy principles, calculate the angular speed (in revolutions per second) after the grindstone has made 12.0 revolutions. (Neglect friction.)
1 Answer
- RealProLv 73 years ago
Torque = moment of inertia * angular acceleration.
T = Iα
α = 260.39 rad/s^2
Angular acceleration is like regular acceleration, but with angles. It tells you how fast angular speed is changing, how many more radians the thing rotates through each second.
Each second angular velocity is 260.39 rad/s greater.
Given an initial angular velocity ω0, some final angular velocity ω, and the angle (or number of revolutions) Φ it made between the end and the start, we can write
ω^2 = (ω0)^2 + 2αΦ
In this case I think it's not spinning at the start, so ω0 = 0.
Since 1 revolution = 2pi radians, we turn α to rev/s^2 by dividing with 2pi.
α = 41.44 rev/s^2
Plug into equation
ω^2 = 2αΦ
ω^2 = 2(41.44 rev/s^2)(12.0 rev)
ω = 31.54 rev/s
Multiply with 60.
31.54 rev/s * (60 s)/(1 min) = 1.89×10^3 rev/s