Chapter 10

$\omega =0\text{.}\text{737 rev/s}$

(a) $-0\text{.}{\text{26 rad/s}}^2$

(b) $\text{27}\text{rev}$

(a) $80{\text{rad/s}}^2$

(b) 1.0 rev

(a) 45.7 s

(b) 116 rev

a) $6{\text{00 rad/s}}^2$

b) 450 rad/s^s

c) 21.0 m/s^s

(a) 0.338 s

(b) 0.0403 rev

(c) 0.313 s

$\text{0.50 kg}\cdot {\text{m}}^2$

(a) $\mathrm{50.4\; N}\cdot \text{m}$

(b) $\text{17.1}{\text{rad/s}}^2$

(c) $17.0{\text{rad/s}}^2$

$3\text{.}\text{96}\times {\text{10}}^{\text{18}}\text{s}$

or $1.26\times {\text{10}}^{\text{11}}\text{y}$

$\begin{array}{c}{I}_{\text{end}}=I_{\text{center}}+m{\left(\frac{l}{2}\right)}^2\\ \text{Thus,}{I}_{\text{center}}=I_{\text{end}}-\frac{1}{4}{\text{ml}}^2=\frac{1}{3}{\text{ml}}^2-\frac{1}{4}{\text{ml}}^2=\frac{1}{\text{12}}{\text{ml}}^2\end{array}$

(a) 2.0 ms

(b) The time interval is too short.

(c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of $\text{500 N}\cdot \text{m}$ is reasonable.

(a) 17,500 rpm

(b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is > 50,000 gs.

(c) Flywheel mass and radius should both be much greater, allowing for a lower spin rate (angular velocity).

(a) 185 J

(b) 0.0785 rev

(c) $W=9\text{.}\text{81 N}$

(a) $2.57\times {\text{10}}^{\text{29}}\text{J}$

(b) ${\text{KE}}_{\text{rot}}=2\text{.}\text{65}\times {\text{10}}^{\text{33}}\text{J}$

(a) $\text{128 rad/s}$

(b) $\text{19.9 m}$

(a) $\text{10.}{\text{4 rad/s}}^2$

(b) $\text{net}W=6.\text{11 J}$

(a) 1.49 kJ

(b) $2.52\times {\text{10}}^4\text{N}$

(a) $2\text{.}\text{66}\times {\text{10}}^{\text{40}}\text{kg}\cdot {\text{m}}^2\text{/s}$

(b) $7\text{.}\text{07}\times {\text{10}}^{\text{33}}\text{kg}\cdot {\text{m}}^2\text{/s}$

The angular momentum of the Earth in its orbit around the Sun is $3\text{.}\text{77}\times {\text{10}}^6$ times larger than the angular momentum of the Earth around its axis.

$\text{22}\text{.}\text{5 kg}\cdot {\text{m}}^2\text{/s}$

25.3 rpm

(a) $\mathrm{0.156\; rad/s}$

(b) $1\text{.}\text{17}\times {\text{10}}^{-2}\text{J}$

(c) $0\text{.}\text{188 kg}\cdot \text{m/s}$

(a) 3.13 rad/s

(b) Initial KE = 438 J, final KE = 438 J

(a) 1.70 rad/s

(b) Initial KE = 22.5 J, final KE = 2.04 J

(c) $1\text{.}\text{50 kg}\cdot \text{m/s}$

(a) $5\text{.}\text{64}\times {\text{10}}^{\text{33}}\text{kg}\cdot {\text{m}}^2\text{/s}$

(b) $1\text{.}\text{39}\times {\text{10}}^{\text{22}}\text{N}\cdot \text{m}$

(c) $2\text{.}\text{17}\times {\text{10}}^{\text{15}}\text{N}$

(b)

(d)

(d)

You are given a thin rod of length 1.0 m and mass 2.0 kg, a small lead weight of 0.50 kg, and a not-so-small lead weight of 1.0 kg. The rod has three holes, one in each end and one through the middle, which may either hold a pivot point or one of the small lead weights.

(a)

(c)

(a)

(a)

(b)

(c)

(b)

(b)

(c)

(d)

A door on hinges is a rotational system. When you push or pull on the door handle, the angular momentum of the system changes. If a weight is hung on the door handle, then pushing on the door with the same force will cause a different increase in angular momentum. If you push or pull near the hinges with the same force, the resulting angular momentum of the system will also be different.

Since the globe is stationary to start with,

$\tau =\frac{\text{Δ}L}{\text{Δ}t}$

$\tau \cdot \text{Δ}t=\text{Δ}L$

By substituting,

120 N•m • 1.2 s = 144 N•m•s.

The angular momentum of the globe after 1.2 s is 144 N•m•s.