Your hard drive can spin up to 7500 rpm from rest in 5 s. If your hard drive's
mass is 40 g and its radius is 2.5 in, what is the force applied to the drive if the force is
applied at a point 0.25 in from the center? Assume the drive is a solid, uniform disk with
I = (1/2)mr2
, and assume the force is applied tangentially. (1 kg = 1000 g, 1 in = 2.54 cm,
1 m = 100 cm)
To find the force applied to the hard drive, we need to use the equation for rotational motion:
τ = Iα
Where τ represents the torque applied to the object, I represents the moment of inertia, and α represents the angular acceleration.
To find the moment of inertia (I), we can use the formula for a solid disk rotating about its axis:
I = (1/2)mr^2
Where m is the mass of the disk and r is the radius.
Given that the mass (m) is 40 g and the radius (r) is 2.5 in, we need to convert the units to kg and meters. Since 1 kg is equal to 1000 g and 1 in is equal to 2.54 cm, we can make the conversions as follows:
m = 40 g / (1000 g/kg) = 0.04 kg
r = 2.5 in * (2.54 cm/in) * (1 m/100 cm) = 0.0635 m
Now we can plug the values of m and r into the moment of inertia equation:
I = (1/2) * 0.04 kg * (0.0635 m)^2 = 0.0000509425 kg·m^2
Next, we can calculate the angular acceleration (α) using the formula:
α = Δω / Δt
Where Δω is the change in angular velocity and Δt is the change in time.
The problem states that the hard drive spins up from rest, so the initial angular velocity (ω_initial) is 0.
ω_initial = 0 rad/s
ω_final = (7500 rpm) * (2π rad/1 min) * (1 min/60 s) = 785.398 rad/s
Δt = 5 s
Δω = ω_final - ω_initial = 785.398 rad/s
Now we can plug the values of Δω and Δt into the equation for α:
α = (785.398 rad/s) / (5 s) = 157.0796 rad/s^2
Finally, we can substitute the values of I and α into the equation for torque:
τ = Iα = (0.0000509425 kg·m^2) * (157.0796 rad/s^2) = 0.008 kg·m^2/s^2
The force applied to the drive at a point 0.25 in from the center is equal to the torque divided by the radial distance (r):
F = τ/r
Given that the radial distance (r) is 0.25 in * (2.54 cm/in) * (1 m/100 cm) = 0.00635 m, we can calculate the force:
F = (0.008 kg·m^2/s^2) / 0.00635 m = 1.2598 N
Therefore, the force applied to the hard drive is approximately 1.2598 Newtons.