A computer hard drive rotates at 5400 revolutions per minute. Starting from rest, what torque will get the hard drive up to that speed in just 150 revolutions? The moment of inertia of the hard drive is 1.3×10^-2 kg·m^2.

I know the answer is 2.2 N.m, but I'm not sure what the steps are. Thank you in advance!

To find the torque required to accelerate the hard drive to the desired speed, we can use the principle of rotational dynamics.

The principle of rotational dynamics states that the torque (τ) applied to an object is equal to the moment of inertia (I) of the object multiplied by its angular acceleration (α).

In this case, we want to find the torque (τ) required to accelerate the hard drive to a specific speed, given the moment of inertia (I) and the number of revolutions (N). The angular acceleration (α) can be calculated using the following formula:

α = (final angular velocity - initial angular velocity) / time

First, let's convert the number of revolutions (N) to radians by multiplying it by 2π:

θ = N * 2π

Next, let's find the final angular velocity (ω) using the formula:

ω = (2π * N) / time

The initial angular velocity (ω_0) is 0 since the hard drive starts from rest.

Now, we can calculate the angular acceleration (α):

α = (ω - ω_0) / time

Finally, we can calculate the torque (τ):

τ = I * α

Let's plug in the values given:

Moment of inertia (I) = 1.3×10^-2 kg·m^2
Number of revolutions (N) = 150
Time = 60 seconds (since rpm is given)

θ = 150 * 2π
ω = (2π * 150) / 60
ω_0 = 0
α = (ω - ω_0) / time
τ = I * α

Now we can calculate the values step by step:

θ = 150 * 2π = 300π
ω = (2π * 150) / 60 = 5π
α = (5π - 0) / 60 = π/12
τ = (1.3×10^-2 kg·m^2) * (π/12)

Now we can simplify and calculate the final torque value:

τ = 1.3×10^-2 * π / 12 ≈ 2.2 N.m

Therefore, the torque required to accelerate the hard drive to the desired speed in just 150 revolutions is approximately 2.2 N.m.