11.4 Precession of a gyroscope  (Page 3/6)

$a_{\text{CM}}=-\frac{3}{10}g$ ,
$v^2=v_0^2+2a_{\text{CM}}x\Rightarrow v^2=(7.0\text{m}\text{/}{\text{s})}^2-2\left(\frac{3}{10}g\right)x,$ $v^2=0\Rightarrow x=8.34\text{m;}$
b. $t=\frac{v-v_0}{a_{\text{CM}}},$ $v=v_0+a_{\text{CM}}t\Rightarrow t=2.38\text{s}$ ;
The hollow sphere has a larger moment of inertia, and therefore is harder to bring to a rest than the marble, or solid sphere. The distance travelled is larger and the time elapsed is longer.

a. $W=\mathrm{-500.0}\text{J}$ ;
b. $K+U_{\text{grav}}=\text{constant}$ ,
$500\text{J}+0=0+(6.0\text{kg})(9.8\text{m}\text{/}{\text{s}}^2)h$ ,
$h=8.5\text{m,}d=17.0\text{m}$ ;
The moment of inertia is less for the hollow sphere, therefore less work is required to stop it. Likewise it rolls up the incline a shorter distance than the hoop.

a. $\tau =34.0\text{N}·\text{m}$ ;
b. $l=m{r}^2\omega \Rightarrow \omega =3.6\text{rad}\text{/}\text{s}$

a. $d_{\text{M}}=3.85\text{×}{10}^8\text{m}$ average distance to the Moon; orbital period $27.32\text{d}=2.36\text{×}{10}^6\text{s}$ ; speed of the Moon $\frac{2\pi 3.85\text{×}{10}^8\text{m}}{2.36\text{×}{10}^6\text{s}}=1.0\text{×}{10}^3\text{m}\text{/}\text{s}$ ; mass of the Moon $7.35\text{×}{10}^{22}\text{kg}$ ,
$L=2.90\text{×}{10}^{34}\text{kg}{\text{m}}^2\text{/}\text{s}$ ;
b. radius of the Moon $1.74\text{×}{10}^6\text{m}$ ; the orbital period is the same as (a): $\omega =2.66\text{×}{10}^{\mathrm{-6}}\text{rad}\text{/}\text{s}$ ,
$L=2.37\text{×}{10}^{29}\text{kg}·{\text{m}}^2\text{/}\text{s}$ ;
The orbital angular momentum is $1.22\times {10}^5$ times larger than the rotational angular momentum for the Moon.

$I=0.135\text{kg}·{\text{m}}^2$ ,
$\alpha =4.19\text{rad}\text{/}{\text{s}}^2,$ $\omega ={\omega }_0+\alpha t$ ,
$\omega (5\text{s})=21.0\text{rad}\text{/}\text{s},$ $L=2.84\text{kg}·{\text{m}}^2\text{/}\text{s}$ ,
$\omega (10\text{s})=41.9\text{rad}\text{/}\text{s},$ $L=5.66\text{kg}·\text{m}\text{/}{\text{s}}^2$