Rotational motion and gravitational torque

[brethman86]

1. There is a figure showing two thin beams joined at right angles. The vertical beam is 15.0 kg and 1.00m long and the horizontal beam is 25.0 kg and 2.00m long.

Part a) askes to find the center of gravity to the two joined beams, taking the origin at the corner where the beams join.

Part b) calculating the gravitational torque.

2. Homework Equations : Xcg= (x1m1 + x2m2 + ...)/ (m1 + m2 +...)

Ycg= (y1m1 + y2m2 + ...)/(m1 + m2 + ...)

torque net= t1 +t2

t1=-x1m1g

t2=+2m2g

Tnet =T1 + T2

3. I have no idea where to start this problem. We do problems similar in class but with a mass at each end of a "massless" rod.

Starting off the right angle is oriented so the 25 kg 2.00 m bar would run along the + x-axis and the 15 kg 1.00m bar would run along the + y axis.

I set x1 m1 t1 as the 25 kg 2m bar.

To calculate the x coord for center of gravity I did Xcg= (25.0kg)(2.00m)+(0m)(15.0kg)/(25.0kg+15.0kg)

= 1.25m
I used zero as m2 because I did not think it would contribute to the x coord system.

Ycg= (1.0kg)(15.0kg)+(0m)(25.0kg)/(25.0kg + 15.0kg) = .375m

so for part A I got (1.25m+.375m)

Part B)

I set toque 1 for the 2m bar. T1= -x1m1g (neg becuase it want to rotate CW) so T1= -25kg(1.25m)(9.8m/s^2) = -306.3 Nm

torque 2 was for 1.0m bar. T2= +x2m2g T2 (I figured it would want to undergo CCW rotation motion so I thought it would be +) T2= +15kg(1.25m)(9.8m/s^2) =183.75 Nm

Toque net =T1+T2 -306.3 Nm + 183.8Nm = -122.5 Nm

I would really like some help in the equations and setting up. I feel like 122.5 Nm is to large of a torque.

 

[JoAuSc]

Your center of mass is wrong. You can calculate the center of mass of a bunch of beams by treating each beam as if it were a point particle located at its own center of mass, which should be half-way along the beam. In other words, use the half way points of the beams rather than the end points of the beams.

 

[baba89]

Hello,

I am happy to help you understand rotational motion and gravitational torque. Let's start by reviewing the concepts involved in this problem.

Rotational motion refers to the motion of an object around an axis or pivot point. In this problem, the beams are joined at a right angle, so the axis of rotation would be at the corner where they meet. The beams will rotate around this point when a torque is applied.

Gravitational torque, also known as the moment of force, is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied by the distance from the axis of rotation. In this problem, the gravitational torque is caused by the force of gravity acting on the beams.

Now, let's address the two parts of the problem.

Part a) Finding the center of gravity (CG) of the two joined beams is important because it is the point where the weight of the beams can be considered to act. This is also known as the balance point or the point where the beams would balance if they were placed on a fulcrum. To find the CG, we use the equation Xcg= (x1m1 + x2m2 + ...)/ (m1 + m2 +...) for the x coordinate and Ycg= (y1m1 + y2m2 + ...)/(m1 + m2 + ...) for the y coordinate. In this case, we can use the origin at the corner where the beams join, so the x coordinate would be the distance of the center of mass of the 25 kg beam (2.00m) and the y coordinate would be the distance of the center of mass of the 15 kg beam (1.00m). Plugging these values into the equations, we get Xcg= (25.0kg)(2.00m)+(0m)(15.0kg)/(25.0kg+15.0kg) = 1.25m and Ycg= (1.0kg)(15.0kg)+(0m)(25.0kg)/(25.0kg + 15.0kg) = 0.375m. Therefore, the CG of the two joined beams is located at (1.25m, 0.375m).

Part b) Calculating the gravitational torque involves finding the net torque acting on the beams. Torque is calculated by multiplying the force