A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.20x10^4 rad/s to an angular speed of 3.14x10^4 rad/s. In the process, the bit turns through 1.92x10^4 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 7.85x10^4 rad/s, starting from rest?
To find the time it takes for the bit to reach its maximum speed, given a constant angular acceleration, we can use the equations of motion for rotational motion.
The first equation we'll use is:
ωf = ωi + αt
Where:
ωf = final angular speed (7.85x10^4 rad/s)
ωi = initial angular speed (0 rad/s, as the bit starts from rest)
α = angular acceleration (unknown)
t = time (unknown)
From the question, we know that the initial angular speed is 0 rad/s, and the final angular speed is 7.85x10^4 rad/s. So let's rearrange the equation to solve for α:
α = (ωf - ωi) / t
Substituting the given values, we have:
α = (7.85x10^4 rad/s - 0 rad/s) / t
Now, we can use the value of α to solve for the time it takes by rearranging the first equation:
t = (ωf - ωi) / α
Substituting the respective values:
t = (7.85x10^4 rad/s - 0 rad/s) / (7.85x10^4 rad/s - 1.20x10^4 rad/s)
Calculating the values:
t = 7.85x10^4 rad/s / (7.85x10^4 rad/s - 1.20x10^4 rad/s)
t = 7.85x10^4 rad/s / 6.65x10^4 rad/s
t ≈ 1.18 seconds
Therefore, it would take approximately 1.18 seconds for the bit to reach its maximum speed of 7.85x10^4 rad/s, starting from rest.
w = w0 + a*t
theta = w0*t + 1/2*a*t^2
where w is the angular speed, w0 is the initial angular speed, and a is the angular acceleration, and theta is the angle turned in radians, so
3.14*10^4 = 1.2*10^4 + a *t
1.92*10^4 = 1.2*10^4*t + .5*a*t^2
Solve this system of equations using algebra for a and t
Then, use
7.85*10^4 = 1.2*10^4 + a*t,
where you know a from before, and you now solve for time