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

Show answers

Answers

answered: 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.

answered: 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

answered: Guest

Ithink true dkfkfkfkdkkdkdkdkddkdk

answered: Guest

it would be kevlar ^^

Other questions on the subject: Physics

Physics, 22.06.2019 00:20, thedocgalloway

You have been consulted by the mobile phone company hookup about a new augmented reality product called “brows”. they want to develop an alternative to google glasses, which, you recall, is a proposed new product which plans to superimpose a whole range of wireless google services over the user’s ordinary vision of the world. these appear in little circles and ovals, and let you phone people, take photographs, navigate across cities and inside buildings, find shops and update your social media. brows would do this too, but not be limited to special google services. the user would wear a small cylinder-shaped device designed to fit on top of their existing eyewear (glasses or sunglasses). it would project a reversed image onto the inside of the lenses, the reflection of which would be in focus for the user. instead of voice control, the user would control the device by eye and neck movements, which the device would track. the device would connect via bluetooth to the user’s phone, which would supply most of the computing power and wireless connectivity. hookup also wants to know what existing services would be most in demand, and which would not work well with brows. using your knowledge of augmented reality, wearable technology and interface design, is the project feasible? if not, why not? if feasible, how could the product need to be modified for a better experience?

Answers: 2

continue

Physics, 22.06.2019 13:50, Geo777

Two students, each riding bicycles, start from the same apartment building and ride to the same building on campus, but each takes a different route. the first student rides 1100 m due east and then turns due north and travels another 1430 m before arriving at the destination. the second student heads due north for 1930 m and then turns and heads directly toward the destination. at the turning point, how far is the second student from the destination?

Answers: 3

continue

Physics, 22.06.2019 14:40, allenpaietonp9v8sv

What is the orbital period of a spacecraft in a low orbit near the surface of mars? the radius of mars is 3.4×106m.

Answers: 2

continue

Physics, 22.06.2019 22:20, queenbee2994

One hazard of space travel is debris left by previous missions. there are several thousand masses large enough to detect by radar orbiting the earth, but there are far greater numbers of very small masses such as flakes of paint. calculate the force exerted by a 0.110 mg chip of paint that strikes a space shuttle window at a relative speed of 5.00 ✕ 103 m/s and sticks, given the collision lasts 6.00 ✕ 10-8 s. such a collision chipped the window of the ill-fated challenger in june 1983, causing $50,000 of damage.

Answers: 2

continue

Do you know the correct answer?

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