Chapter 11: Q27 P (page 460)

In figure two small objects each of mass $m = 0.3 k g$ are connected by a lightweight rod of length $d = 1.5 m .$ At a particular instant they have velocities whose magnitude are $v_{1} = 38 m / s$ and $v_{2} = 60 m / s$ and are subjected to external forces whose magnitudes are $F_{1} = 41 N$ and $F_{2} = 26 N$ . The distance role="math" localid="1668661918159" $h = 0.3 m,$ and the distance $w = 0.7 m .$ The system is moving in outer space. Assuming the usual coordinate system with $+ x$ to the right, $+ y$ toward the top of the page, and $+ z$ out of the page toward you, calculated these quantities for this system:
(a) ${\vec{p}}_{t o t a l,}$ (b) ${\vec{v}}_{C M,}$ (c) ${\vec{L}}_{t o t, A,}$ (d) ${\vec{L}}_{r o t,}$ (e) ${\vec{L}}_{t r a n s A,}$ (f) ${\vec{P}}_{t o t a l}$ at a time $0.23 s$ after the initial time.

Short Answer

Expert verified

The total momentum of the system ${\vec{p}}_{t o t a l,}$ - $29.4 k g \cdot m / s$ .

the velocity of center of mass is $(49 m / s, 0, 0)$ .

The total angular momentum ${\vec{L}}_{A}$ of the system relative to point $A$ is $- 25.92 k g \cdot m^{2} / s$ .

The rotational angular momentum of the system is $<0, 0, 4.95 k g \cdot m^{2} / s>$ .

The translational angular momentum of the system is $<0, 0 l - 30.87 k g \cdot m^{2} / s>$ .

Step by step solution

Definition of Force.

A force is a push or pull on an object caused by the interaction of the object with another object. Every time two things interact, a force is exerted on each of them. The two items no longer feel the force after the interaction ends.

The given data is-

Mass of each object, $m = 0.3 k g$

Length of the limit weight rod, $d = 1.5 m$

Velocity of the first object, $v_{1} = 38 m / s$

Velocity of second object, $v_{2} = 60 m / s$

Magnitude of external force on first object, $F_{1} = 41 N$

Magnitude of external force on second object, $F_{2} = 26 N$

Height of second object about point $A, h = 0.3 m$

Distance of the second object about point $A, w = 0.7 m$

Find the total momentum, the velocity of center of mass and the total angular momentum of the system.

Figure show the two small objects connected by a lightweight rod and are subject to external forces.

Consider two objects of mass $m$ are connected by a lightweight rod of length $d$ about a point $A$ .

These objects are moving with velocities $v_{1}$ and $v_{2}$ when subjected to external forces $F_{1}$ and $F_{2}$ respectively.

(a) The direction of velocities of the two objects are along $x -$ axis and both are in the same direction towards right.

Thus, the total momentum of this system is

${\vec{p}}_{t o t a l} = m \vec{v_{1}} + m \vec{v_{2}} = m (v_{1} + v_{2}) = (0.3 k g) (38 m / s + 60 m / s) = 29.4 k g \cdot m / s$

(b) The velocity of the center of mass is given by

$v_{C M} = \frac{{\vec{p}}_{C M}}{M_{t o t a l}} = \frac{m_{1} \vec{v_{1}} + m_{2} \vec{v_{2}}}{m_{1} + m_{2}} = \frac{29.4 k g \cdot m / s}{2 (0.3 k g)}$

$= 49 m / s$

Therefore, the velocity of center of mass is $(49 m / s, 0, 0)$ .

(c ) The angular momentum of the system is given by

$L = r \times p$

The total angular momentum ${\vec{L}}_{A}$ of the system relative to point $A$ is $L_{A =} \vec{r_{A}} \times P_{t o t a l} = <- w, h + d, 0> \times (m_{1} v_{1}, 0, 0) + <- w, h + d, 0> \times (m_{1} v_{1}, 0, 0)$

Therefore, the total angular momentum ${\vec{L}}_{A}$ of the system relative to point A is $- 25.92 k g \cdot m^{2} / s$ .

Find the rotational angular momentum of the system and the translational angular momentum of the system.

(d)The total angular momentum of the system is

${\vec{L}}_{t o t a l} = {\vec{L}}_{r o t} + {\vec{L}}_{t r a n s}$

$Here, the translational momentum of the system is$

${\vec{L}}_{t r a n s} = {\vec{r}}_{c m} + p_{t o t a l} = <- w, (h + \frac{d}{2}), 0> \times <(m_{1} v_{1} + m_{2} v_{2}), 0, 0>$

$= <[(h + \frac{d}{2}) (0) - (0) (0)], [(0) (m_{1} v_{1} + m_{2} v_{2}) - (- w) (0)], [(- w) (0) - (h + \frac{d}{2}) (m_{1} v_{1} + m_{2} v_{2})]>$

$= <0, 0 - (h + \frac{d}{2}) (m_{1} v_{1} + m_{2} v_{2})>$

Therefore, the rotational angular momentum of the system is

${\vec{L}}_{r o t} = {\vec{L}}_{t o t a l} - {\vec{L}}_{t r a n s}$

$= <0, 0, (- m v_{1} (h + d) - m v_{2} h)> - <0, 0 - (h + \frac{d}{2}) (m v_{1} + m v_{2})>$

$= (- m v_{1} (h + d) - m v_{2} h) + (h + \frac{d}{2}) (m v_{1} + m v_{2})$

$= - m v_{1} (h + d - h - \frac{d}{2}) - m v_{2} (h - h - \frac{d}{2})$

$= - m v_{1} (\frac{d}{2}) - m v_{2} (- \frac{d}{2})$

$= <0, 0 - m v_{1} (\frac{d}{2}) + m v_{2} (\frac{d}{2})>$

$= <0, 0, [(0.3 k g) (- (38 m / s) + 60 m / s) (\frac{1.5 m}{2})]>$

$= <0, 0, 4.95 k g \cdot m^{2} / s>$

(e)The translational angular momentum of the system relative to a point A is

$\vec{L_{t r a n s}} = \vec{r_{C M}} \times p_{t o t a l}$

$= <- w (h + \frac{d}{2}), 0> \times <(m_{1} v_{1} + m_{2} v_{2}), 0, 0>$

$= <[(h + \frac{d}{2}) (0) - (0) (0)], [(0) (m_{1} v_{1} + m_{2} v_{2}) - (- w) (0)], [(- w) (0) - (h + \frac{d}{2}) (m_{1} v_{1} + m_{2} v_{2})]>$

$= <0, 0 - (h + \frac{d}{2}) (m_{1} v_{1} + m_{2} v_{2})>$

$= <0, 0, [- (0.3 k g) ((38 m / s) + 60 m / s) (0.3 m + \frac{1.5 m}{2})]>$

$= <0, 0 l - 30.87 k g \cdot m^{2} / s>$

(f) The change in momentum of the system after a short interval of time $Δ t$ is

$Δ \vec{p} = {\vec{F}}_{n e t} Δ t$

Therefore, the final momentum of the system after a short interval of time $Δ t$ is

${\vec{p}}_{f} = {\vec{p}}_{i}_{n e t} Δ \vec{p}$

$= <(m v_{1} + m v_{2}), 0, 0> + <({\vec{F}}_{1} Δ t + (- {\vec{F}}_{2}) Δ t), 0, 0> = <(m v_{1} + {\vec{F}}_{1} Δ t + m v_{2} - {\vec{F}}_{2} Δ t), 0, 0> = <(m (v_{1} + v_{2}) + (F_{1} + F_{2}) Δ t), 0, 0>$