Question

A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0 N and 125 N are applied to the disk, as the drawing illustrates. (a) What is the net torque produced by the two forces? (Assume counterclockwise is the positive direction.)(b) What is the angular acceleration of the disk? rad/s2

Answer 1

Answer:

a. $-12.7 Nm$

b. $-7.9 rad/s^2$

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

Forces applied to the solid disk include:

$F_1 = 90.0N\\F_2 = 125N$

Other parameters given include:

Mass of solid disk, $M = 24.3kg$

and radius of solid disk, $r = 0.364m$

a.) The formula for determining torque ($T$), is $T = r * F$

Hence the net torque produced by the two forces is given as a summation of both forces:

$T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm$

b.)  The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

$I_{disk} = {\frac{1}{2}}Mr^2$

where $M$ = Mass of solid disk

and $r$ = radius of solid disk

We then relate the torque and angular acceleration ($\alpha$) with the formula:

$T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha = -{\frac{12.7}{1.61}} = -7.9 rad/s^2$