Posted by **Papito** on Monday, May 7, 2007 at 1:18pm.

A uniform solid disk with a mass of 40.3 kg and a radius of 0.454 m is free to rotate about a frictionless axle. Forces of 90.0 N and 125 N are applied to the disk

(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?

This is how i solved it.

Torque = Fl

net torque = (125N x 0.454m) + (90N x 0.454N)=97.81

Angular acceleration = torque/ inertia

Inertia = mass x radius ^2

= 40.3 kg x 0.454 ^2 =8.31

Angular acceleration = 97.81/ 8.31 =11.77 rads/s^2

Please check solution

Isnt the moment of inertia 1/2 m r^2 ?

So Moment of Inertia = 0.5x 40.3kg x 0.454^2 = 4.15

Angular acceleration =97.81/4.15

=23.55

## Answer This Question

## Related Questions

- Physics - Can you please help with this problem. I don't know how to solve it at...
- physics - A uniform disk with a mass of 27.3 kg and a radius of 0.309 m is free ...
- physics - A uniform disk with a mass of 27.3 kg and a radius of 0.309 m is free ...
- Physics please check - A uniform solid disk with a mass of 32.3 kg and a radius ...
- Physics - A uniform disk of radius 0.485 m and unknown mass is constrained to ...
- Physics - A solid cylindrical disk has a radius of 0.18 m. It is mounted to an ...
- physics - A solid cylindrical disk has a radius of 0.17 m. It is mounted to an ...
- Physics - A flat uniform circular disk (radius = 2.10 m, mass = 1.00 102 kg) is ...
- Physics - A uniform disk with mass 6 kg and radius 1.8 m is pivoted at its ...
- physics - A solid cylindrical disk has a radius of 0.16 m. It is mounted to an ...

More Related Questions