Question

Problem 10.32

Answer 1

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 2

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 $I$ 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, $M$ is the mass of the beam and $L$ is the length of the object.

The expression for the torque is:

$\tau=F\times r$                                          …… (3)

Here, $F$ is the force and $r$ 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 $F$, $1\text{ m}$ for $r$, $2\text{ m}$ for $L$ and $5.00\text{ kg}$ for $M$ 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 $F$, $-1\text{ m}$ for $r$, $2\text{ m}$ for $L$ and $5.00\text{ kg}$ for $M$ 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 brainly.com/question/3923773

2. A 700kg car driving at 29m/s brainly.com/question/9484203

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.