A uniform solid disk with a mass of 22.8 kg and a radius of 0.633m is free to ro

Question

A uniform solid disk with a mass of 22.8 kg and a radius of 0.633m is free to rotate about a frictionless axle. Forces of 90.0 and 125 N are applied to the disk, as the figure illustrates (picture has the 125 N pointing to the right, on the horizontal, from the very top edge of the circle, at 90 degrees whereas the 90 N points to the right, on the horizontal, from the very bottom edge of the circle, at 90 degrees. Taking the clockwise direction to be the negative direction, what is (a) the net torque produced by the two forces and (b) the angular acceleration of the disk?

Explanation / Answer

To find the net torque, multiply the force times the radius and addthe two torques together, noting that one is negative since theclockwise direction is negative: Torque = - (125 * 0.633) + (90 * 0.633) = -22.15 Nm Note that this is negative because the torque is in the clockwisedirection. To find the angular acceleration, use the formula t = Ia,where t is the torque, I is the inertia of the disk, anda is the angular acceleration. The inertia of a uniform solid disk rotating through the center isgiven by (1/2)mR2 I = (1/2) * 22.8 * 0.633 I = 7.216 kg m2 Now we can solve for a: a = t / I a = -22.15 / 7.216 a = -3.069 rad / s2 The negative sign implies that the disk is accelerating in the clockwise direction.

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