Question
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
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
Answers
drwls
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
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