Physics, 25.08.2019 09:00 LJOrozco

Problem 10.32
a very thin horizontal, 2.00-m long, 5.00-kg uniform beam that lies along the east-west direction is acted on by two forces. at the east end of the beam, a 200-n force pushes downward. at the west end of the beam, a 200-n force pushes upward. what is the angular acceleration of the beam?
a: 1.33 × 102 rad/s2 south
b: 240 rad/s2 south
c: zero
d: 240 rad/s2 north
e: 1.33 × 102 rad/s2 north

Answers: 2

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Answers

Answer from: NycLife

Answer : $\alpha = -240\ rad/s^{2}$ south

Explanation :

Given that,

Force = 200 N

Length = 2.00 m

Mass = 5.00 kg

We know that , the formula of the angular acceleration

$\alpha = \dfrac{\tau}{I}\ rad/s^{2}$

We know the moment of inertia of rod

$I = \dfrac{ML^{2}}{12}$

Now, the torque is

$\tau = F\times r$

The angular acceleration

$\alpha = \dfrac{F\times r\times12}{ML^{2}}$

$\alpha = \dfrac{((200\ N\times-1\ m) + (-200\ N)\times(1\ m))\times12}{5\ kg\times2\ m\times2\ m}$

$\alpha = \dfrac{-400Nm\times12}{5kg\times4m^{2}}$

$\alpha = -240\ rad/s^{2}$

Negative sign represent of south direction.

Hence, this is the required solution.

Answer from: gm2

The angular acceleration of the beam is $\boxed{240\text{ rad/s}^2}$ in south direction.

Further Explanation:

The angular acceleration is the acceleration of the object in the circular motion in other words it is the rate of change in the angular velocity of the object.

The moment of inertia is the tendency withstands the action or effect of the angular acceleration and directly proportional to the mass and square of the length of the object.

Given:

The force applied on the beam is $200\text{ N}$ .

The length of the beam is $2.00\text{ m}$ .

The mass of the uniform beam is $5.00\text{ kg}$ .

Concept:

The expression for the angular acceleration is:

$\alpha=\dfrac{\tau}{I}$ …… (1)

Here, $\alpha$ is the angular acceleration, $\tau$ is the torque and is the moment of inertia.

The expression for the moment of inertia of a thing beam about its center is:

$I=\dfrac{{M{L^2}}}{{12}}$ …… (2)

Here, is the mass of the beam and is the length of the object.

The expression for the torque is:

$\tau=F\times r$ …… (3)

Here, is the force and is the perpendicular distance from the center.

Compare equation (1), (2) and (3), we get:

$\alpha=\dfrac{{F\times r\times12}}{{M{L^2}}}$ …… (4)

Let ${\alpha_1}$ is the angular acceleration due to force applied on the east end of the beam and ${\alpha _2}$ is the angular acceleration due to force applied on the west end of the beam.

The force applied on the east end in downward direction therefore the force is taken negative and distance from the center to the end is positive but the force applied on the west end in upward direction, thereforethe force is taken positive and distance from the center to the end is negative

Substitute ${\alpha _1}$ for $\alpha$ , $-200\text{ N}$ for , $1\text{ m}$ for , $2\text{ m}$ for and $5.00\text{ kg}$ for in equation (4).

$\begin{aligned}{\alpha_1}&=\frac{{({-200}\text{ N})\times({1{\text{ m}}})\times 12}}{{({5{\text{ kg}}}){{( {2{\text{ m}}})}^2}}}\\&=-120{\text{ rad/s}^2}\\\end{aligned}$

Substitute ${\alpha _2}$ for $\alpha$ , $200\text{ N}$ for , $-1\text{ m}$ for , $2\text{ m}$ for and $5.00\text{ kg}$ for in equation (4).

$\begin{aligned}{\alpha_2}&=\frac{{( {200{\text{ }}N})\times({-1{\text{ m}}})\times 12}}{{({5{\text{ kg}}}){{({2{\text{ m}}})}^2}}}\\&=-120{\text{ rad/s}^2}\\\end{aligned}$

The expression for the resultant angular acceleration is:

$\begin{aligned}\alpha&=-120{\text{ rad/s}^2}+({-120{\text{ rad/s}^2}})\\&=-120{\text{ rad/s}^2}-120{\text{ rad/s}^2}\\&=-240{\text{ rad/s}^2}\\\end{aligned}$

The negative sign show the movement of the beam in anti-clock wise.

Thus, the angular acceleration of the beam is $\boxed{240\text{ rad/s}^2}$ in south direction.

Learn More:

1. Conservation of momentum

2. A 700kg car driving at 29m/s

Answer Details:

Grade: High School

Subject: Physics

Chapter: Rotational mechanics.

Keywords:

Thin, beam, horizontal east-wesr direction, forces, 200 N, pushed, 240 rad/s2, acceleration, rotation, two forces, east end, upwaard, downward,torque, moment of inertia.

Problem 10.32 a very thin horizontal, 2.00-m long, 5.00-kg uniform beam that lies along the east-wes

Answer from: Quest

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

Answer from: Quest

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000

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Problem 10.32
a very thin horizontal, 2.00-m long, 5.00-kg uniform beam that lies along the ea...

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