Posted by **seth** on Saturday, May 2, 2009 at 9:16am.

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

- physics -
**drwls**, Saturday, May 2, 2009 at 11:03am
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

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