Question

A uniform solid disk is mounted on an axle n such a way that it
is free to rotate about a horizontal axis. The radius of the disk
is **0.450** m and its mass is **27.0**
kg. As shown in the diagram, two forces F_{1} =
**84.0** N and F_{2} = **130** N
applied to the disk sets the disk rotating with a constant angular
acceleration. Assume the axle is frictionless

(a) Calculate the magnitude and direction of the net torque produced by the two forces

(b) Determine the magnitude of the angular acceleration of the disk

Answer #1

the torque produced by F1 is r×F1 = (0.45 m) * (84 N) * sin(90°)
= 33.6 Nm and points counterclockwise (out of the page)

And by F2, (0.45 m) * (130 N) * sin(90°) = 58.5 Nm and points
clockwise (into the page)

The net torque is the vector sum of all the torques.

Which in this case is difference 58.5 - 33.6 Nm
= 24.9 N.m

And since torque from F2 is larger, the disk will accelerate
clockwise (net torque into page)

The moment of inertia of a solid, homogenous disk is
0.5mr^{2} where m is its mass and r is its radius.

I = 0.5x(27kg) x(.45)^{2} = 2.73 kg.m^{2}

Angular acceleration = torque / moment of inertia = 24.9/2.73 = 9.1
rad/s^{2}

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