A cylinder has inner and outer radii of $16 c m$ $16 cm$ and $18 c m$ $18 cm$ , respectively, and a mass of $2 k g$ $2 kg$ . If the cylinder's frequency of rotation about its center changes from $4 H z$ $4 Hz$ to $9 H z$ $9 Hz$ , by how much does its angular momentum change?

Question

1579 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-27T05:44:58

The change in angular momentum is $= 1.82 k g m^{2} s^{- 1}$ $=1.82kgm^2s^-1$

The angular momentum is $L = I ω$ $L=Iomega$

where $I$ $I$ is the moment of inertia

and $ω$ $omega$ is the angular velocity

The mass of the cylinder is $m = 2 k g$ $m=2kg$

The radii of the cylinder are $r_{1} = 0.16 m$ $r_1=0.16m$ and $r_{2} = 0.18 m$ $r_2=0.18m$

For the cylinder, the moment of inertia is $I = m \frac{r_{1}^{2} + r_{2}^{2}}{2}$ $I=m(r_1^2+r_2^2)/2$

So, $I = 2 \cdot \frac{{0.16}^{2} + {0.18}^{2}}{2} = 0.058 k g m^{2}$ $I=2*(0.16^2+0.18^2)/2=0.058kgm^2$

The change in angular velocity is

$Delta omega=Deltaf*2pi=(9-4)*2pi=10pirads^-1$

The change in angular momentum is

$DeltaL=IDelta omega=0.058 xx10pi=1.82kgm^2s^-1$

A cylinder has inner and outer radii of 16 cm16cm16 cm and 18 cm18cm18 cm, respectively, and a mass of 2 kg2kg2 kg. If the cylinder's frequency of rotation about its center changes from 4 Hz4Hz4 Hz to 9 Hz9Hz9 Hz, by how much does its angular momentum change?