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

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

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

$= <(0.3 k g) (38 m / s + 60 m / s) + (41 N - 26 N) (0.23 s), 0, 0> = <32.85 k g \cdot m / s, 0, 0>$

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Short Answer

Step by step solution

Definition of Force.

The given data is-

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

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

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