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.
is my work correct
7500 rpm * 2pi* (1/60sec) = 785.4 rad/s
angular accel = (785.4-0)/5 = 157rad/s^2
F = mass*angular acceleration*radius
F =0.04Kg * 157rad/s^2 * (0.0635m)
F =0.4N
The angular acceleration is correct.
The Greek symbol "alpha" is usually used for that.
For the force F, you need to use the angular acceleration equation:
Torque L = F*r = I*(alpha)
= (Moment of Inertia)*(Angular acceleration) = (1/2) M R^2 * alpha
F = (1/2)(R^2/r)*M*(alpha)
R is the disc radius and r is the radial distance to the point of application of the force
Your work is almost correct, but there is a small error. Let's go through the steps again:
First, convert the rotational speed from RPM to rad/s. You correctly calculated this as 7500 rpm * 2π * (1/60 s) = 785.4 rad/s.
Next, calculate the angular acceleration. You divided the change in angular speed (785.4 rad/s) by the time (5 s) to get 157 rad/s^2. This step is correct.
Now, let's calculate the force applied. You correctly used the formula F = mass * angular acceleration * radius, but there is a slight mistake in the value of the radius.
The problem states that the radius is 2.5 in, but the force is applied at a point 0.25 in from the center. So, the actual radius to be used in the calculation is the sum of these two values: 2.5 in + 0.25 in = 2.75 in.
To convert the radius to meters, you can use the conversion factor 1 in = 0.0254 m. Therefore, the radius in meters is 2.75 in * 0.0254 m/in = 0.06985 m (rounded to five decimal places).
Now, plug in the correct values into the formula:
F = mass * angular acceleration * radius
F = 0.04 kg * 157 rad/s^2 * 0.06985 m
F ≈ 0.43 N (rounded to two decimal places)
Therefore, the correct answer is approximately 0.43 N.