# Plane kinetics of rigid bodies

• IsaacCB
In summary, to find the angular velocity of the 3.6m pole as it reaches the horizontal position, the formula ƩMo=Iθ''+Ʃma(vector)d is used. The moment of inertia (I) is 4.8m and the distance (d) is 1.8m. Substituting these values into the formula, we get 0=4.8m*θ''+1.62m^2+1.8mg*sin(θ)-1.62mg*cos(θ). Solving for θ'', we can then use the formula ω=θ' to find the angular velocity (ω) of the pole at the horizontal position.
IsaacCB

## Homework Statement

The uniform 3.6m pole is hinged to the truck bed and is released from the vertical position as the truck starts from rest with an acceleration of 0.9m/s^2. If the acceleration remains constant during the motion of the pole as it reaches the horizontal position what is the angular velocity of the pole as it reaches the horizontal position.

## Homework Equations

a(tangential)=mrθ''
a(normal) = mrω^2
ƩMo=Iθ''+Ʃma(vector)d
I=k^2m

## The Attempt at a Solution

Using the given formula ƩMo=Iθ''+Ʃma(vector)d to find θ''.

so 0= Iθ''+mrθ''+mg1.8sin(θ)-ma*1.8cos(θ)

haven't been able to get past this point

as I am not sure what values to use for the moment of inertia (I) and the distance (d).

Hello,

Thank you for your question. It seems that you are on the right track with using the formula ƩMo=Iθ''+Ʃma(vector)d to find the angular acceleration (θ''). In order to continue, we will need to determine the moment of inertia (I) and the distance (d).

The moment of inertia for a uniform rod rotating about one end is given by I=1/3*ml^2, where m is the mass of the rod and l is the length of the rod. In this case, the mass of the rod is not given, so we will use a variable m to represent it. The length of the rod is given as 3.6m. Therefore, the moment of inertia for the rod is I=1/3*m*(3.6)^2=4.8m.

Next, we need to determine the distance (d) from the pivot point (hinge) to the center of mass of the rod. This distance can be found by using the formula d=l/2, where l is the length of the rod. In this case, the distance from the pivot point to the center of mass is d=3.6/2=1.8m.

Now, we can substitute these values into the formula ƩMo=Iθ''+Ʃma(vector)d to find θ''.

So, 0=4.8m*θ''+m*1.8m*0.9m/s^2+mg*1.8m*sin(θ)-m*0.9m/s^2*1.8m*cos(θ)

Simplifying, we get: 0=4.8m*θ''+1.62m^2+1.8mg*sin(θ)-1.62mg*cos(θ)

From here, we can solve for θ'' by rearranging the terms and solving for it. Once we have the value for θ'', we can use the formula ω=θ' to find the angular velocity (ω) of the pole as it reaches the horizontal position.

I hope this helps. Let me know if you have any further questions. Good luck with your problem!

## 1. What is plane kinetics of rigid bodies?

Plane kinetics of rigid bodies is the study of the motion and forces acting on a rigid body in a two-dimensional space. It involves the analysis of translation, rotation, and the resulting forces and moments on the rigid body.

## 2. What is the difference between plane kinetics and general kinetics?

The main difference between plane kinetics and general kinetics is the dimensionality of the motion. Plane kinetics deals with motion in a two-dimensional space, while general kinetics deals with motion in a three-dimensional space.

## 3. How is the motion of a rigid body described in plane kinetics?

The motion of a rigid body in plane kinetics is described using translational and rotational motion. Translational motion refers to the body's linear movement in a straight line, while rotational motion refers to the body's movement around a fixed axis.

## 4. What are the equations used in plane kinetics of rigid bodies?

The equations used in plane kinetics of rigid bodies include Newton's laws of motion, the equations of motion for translation and rotation, and the equations for calculating forces and moments.

## 5. What are some real-life applications of plane kinetics of rigid bodies?

Plane kinetics of rigid bodies has many practical applications, including the analysis of car crashes, the design of aircraft and spacecraft, and the study of the motion of objects in sports such as gymnastics and figure skating.

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