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Chapter

# PHYS 101 Chapter Notes -Net Force, Angular Velocity, Angular Acceleration

Department
Physics
Course Code
PHYS 101
Professor
Andrew Debenedictis

Page:
of 2
First note that the system is in static equilibrium.
Therefore, net Torque and net force are 0.
Draw the FBD and list the different forces.
You should have the:
Torque for the weight of the beam + the Torque for the person's weight=Torque of the
Tension.
Using this torque formula, find the torque for each and equate them. Pay attention to the r
for each one from the info in the question.
Also note that the angle for the weight torques is the one between the beam closest to the
pivot point (origin) and the weight force. Notice the F geometry? It should indicate
something..Similarly, the angle for the tension is the angle between the tension and the line
of action extending from the radius. Hint the angle is theta+90 degree as indicated in the
drawing.
Now you can solve for x.
Part B:
You are asked to find the Horizontal and Vertical pin forces or similarly, nx and ny.
n=normal force.
In order to do this, you need to break the force Tension into its x and y components in your
FBD the same way you should have indicated the nx and ny components. You will find that
there is a relationship between the normal force components and the tension components.
The tricky part is knowing that there is a third angle you need to find. Let's call this angle
alpha. In the diagram, angle alpha is the angle between the wall and the cord. We need this
angle to find the angle between Tmax (Tension force) and its y component using the z
geometry. Using your trig knowledge and given the two other angles in the diagram, you
can solve for alpha.
Next, you want to solve for the y and x components for tension, using alpha and the
Tension force which was given.
Newton's Law:
Equate all the vertical forces in the upward direction to those in the downwards direction
and solve for ny.
Find what nx equals using the same instructions.
Part c:
At this point, the cord has broken. Therefore there will be an angular velocity and
centripetal acceleration b/c the beam is now free to rotate at the pivot point.
Find out what the formula is for a(centripetal) and also solve for a(tangential). Note for both
cases, r=x.
In the end, you want these values to solve for a(total) =
sqrt(a(tangential)^2+a(centripetal)^2).
The first step is to solve for a(tangential)= angular acceleration*r. But we need to know
angular acceleration. So, write the formula of the net Torque which equals to I*angular
acceleration. Remember that we are dealing with the total inertia which is the I(of the beam)
+ I(of the man). Pay attention to the inertia of the beam.
So the net Torque is equal to the Torque of the beam and the Torque of the man. Find those
equations and solve to find net torque. Make that equal to I(total)*angular acceleration and
solve for angular acceleration.
So now that we have calculated angular acceleration, we can solve for a(tangential). And
since we also know a(centripetal), we can calculate for a(total) which is
sqrt( a(tangential)^2+a(centripetal)^2).