Simple Harmonic Motion of a Bent Rod

AI Thread Summary
The discussion revolves around calculating the angular frequency of a bent rod oscillating about a fulcrum. Participants emphasize the role of gravity as the restoring force and the need to apply Newton's second law for rotational motion to derive the equation of motion. The moment of inertia is identified as crucial, with a suggested value of ML^2/4 for the bent rod. There is uncertainty regarding the correct representation of torque and angles when the rod is displaced. Overall, the focus is on establishing the relationship between torque, moment of inertia, and angular acceleration to solve for the angular frequency.
HelpMeJebus
Messages
2
Reaction score
0

Homework Statement



A uniform rod of mass m and length 2L is bent into 2 straight segments about its
midpoint. The angle between the 2 segments is α. The bent rod is balanced on a sharp
fulcrum. Find the angular frequency for small oscillations if the rod is displaced from
equilibrium and released.

know total mass of rod: m
length: 2L
angle between rods is: alpha

Homework Equations



x(t)=A cos((omega)(t)+(phi))
ma=??



The Attempt at a Solution



I've tried to draw a diagram explaining the motion, and forces involved and I can tell that the "tension" or force that the rod applies at the center of mass of each length of the rod is responsible for the restoring force, however I am not sure how to treat this in order to write Newton's second law so that I can solve the 2nd order differential equation for the angular frequency. Help is greatly appreciated!
 
Physics news on Phys.org
HelpMeJebus said:
I've tried to draw a diagram explaining the motion, and forces involved and I can tell that the "tension" or force that the rod applies at the center of mass of each length of the rod is responsible for the restoring force,
Gravity supplies the restoring force.
however I am not sure how to treat this in order to write Newton's second law so that I can solve the 2nd order differential equation for the angular frequency.
What torque does gravity exert about the pivot? What's the rotational inertia about the pivot?
 
Aha, another one of these rotational problems. I thought I had escaped them.

Gravity does indeed supply the restoring force and I fell into the same old physics student trap.

Approaching this with Newton's second law for rotations, we get the sum of all the external torques= (Moment of Inertia)*(Angular Acceleration).

I am aware that torque is: R (vector) X Force (vector), but I am still unsure as to how to label the angles. Let's say we displace it a small angle theta, I am unsure how to treat this. Qualitatively it is easy to see that the torque will be greater for the "upper" end of the bent rod when it is displaced, because the angle between the weight and the "lower" part of the rod is approaching parallel, but I am not sure how to notate this in quantitative terms.

As for the moment of inertia, it is the sum of all of the individual mass particles and their radius from the axis of rotation squared. In this case I believe it would be ML^2/4. Now all I need is a way to represent those torques...

(I)(d^2 (theta/dt) +??(theta)=0, where the ??'s actually give me the square of the angular velocity.

Thanks for the quick, and very helpful response!
 
HelpMeJebus said:
I am aware that torque is: R (vector) X Force (vector), but I am still unsure as to how to label the angles. Let's say we displace it a small angle theta, I am unsure how to treat this. Qualitatively it is easy to see that the torque will be greater for the "upper" end of the bent rod when it is displaced, because the angle between the weight and the "lower" part of the rod is approaching parallel, but I am not sure how to notate this in quantitative terms.
For the purpose of calculating the torque, realize that gravity can be considered to act at the center of mass of the bent rod. Find the center of mass and its distance from the pivot.

As for the moment of inertia, it is the sum of all of the individual mass particles and their radius from the axis of rotation squared. In this case I believe it would be ML^2/4.
How did you arrive at this value? Hint: What's the moment of inertia of a single rod about one end?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top