1. The problem statement, all variables and given/known data A high-speed drill reaches 2760 rpm in 0.260 s. Through how many revolutions does the drill turn during this first 0.260 s? 2. The attempt at a solution UPDATED: Here's what I have right now 2760 rpm * (2n/1 rev) * (60 s / 1 min) = 1040495.49 rad/s 1040495.49 rad/s * 0.260 s = 270,528.83 radians 270,528.83 radians * (1 rev / 2n) = 43,056 revolutions Is that right? I haven't put the answer in because I have a limited amount of tries but I want to make sure I did it right.
when you converted rpm you got rad/s. So multiplying that by 0.26s will give you the radians it moved. now you know that 2π rad = 1 rev. You need to do another conversion to get the revolutions.
I divided 11.96 by 2pi and got 1.90 revolutions, but the computer program says that's wrong. What's my mistake?
You know the final (and initial) angular velocity and the time it took to get there. With this you can get the angular acceleration. Given that, you can find how many revolutions it traversed in the given time.
Hi smpolisetti, welcome to PF. in the problem. initial angular velocity is zero and final angular velocity = 2760*2π/60 rad./s. Find the angular acceleration using ω = ωο + α*t. Then find the angular displacement using θ = ωο*t + 1/2*α*t^2
I know that the acceleration is 1110 rad/s/s but I don't know how to get the amount of revolutions from that
no no 2760 rpm you have. 1 rpm = 2π/60 rad/s you do not need angular acceleration. Convert the rpm to rad/s and then multiply by the 0.26 sec.
Here's what I have 2760 rpm * (2n/1 rev) * (60 s / 1 min) = 1040495.49 rad/s 1040495.49 rad/s * 0.260 s = 270,528.83 radians 270,528.83 radians * (1 rev / 2n) = 43,056 revolutions Is that right? I haven't put the answer in because I have a limited amount of tries but I want to make sure I did it right.
Angular acceleration = 1100 rad/s/s. θ = ωο*t + 1/2*α*t^2 θ = 1/2*1100*(0.26)^2 find θ and then find n.