# A solid disk, spinning counter-clockwise, has a mass of 7 kg and a radius of 3 m. If a point on the edge of the disk is moving at 16 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Jun 2, 2016

For a disc rotating with its axis through the centre and perpendicular to its plane, the moment of inertia , $I = \frac{1}{2} M {R}^{2}$

So, the Moment of Inertia for our case, $I = \frac{1}{2} M {R}^{2} = \frac{1}{2} \times \left(7 \setminus k g\right) \times {\left(3 \setminus m\right)}^{2} = 31.5 \setminus k g {m}^{2}$

where, $M$ is the total mass of the disc and $R$ is the radius.

the angular velocity ($\omega$) of the disc, is given as: $\omega = \frac{v}{r}$ where $v$ is the linear velocity at some distance $r$ from the centre.

So, the Angular velocity ($\omega$), in our case, = $\frac{v}{r} = \frac{16 m {s}^{-} 1}{3 m} \approx 5.33 \setminus r a d \text{/} s$

Hence, the Angular Momentum = $I \omega \approx 31.5 \times 5.33 \setminus r a d \setminus k g \setminus {m}^{2} \setminus {s}^{-} 1 = 167.895 \setminus r a d \setminus k g \setminus {m}^{2} \setminus {s}^{-} 1$