Rod w/ Pivot and Attached Mass Pendulum

In summary: In this case, the center of mass of the system is the pivot, the mass at the end of the rod, and the mass that is attached.
  • #1
srekai
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0

Homework Statement


A solid rod of length L and mass M has a pivot through its center and is originally horizontal. Another mass 2M is then attached firmly to one end of the rod, and released. What is the maximum speed of the mass 2M attained thereafter? (Cornell 2009)

Homework Equations


Not 100% sure, see solution below
Approach 1:
PE = mgh
KE = ##\frac{1}{2}mv^2##

Approach 2:
Torque: ##\tau = rFsin \theta##
Moment of Inertia: ##I = mr^2##
Torque: ##\tau = I \alpha##

The Attempt at a Solution


I approached this two ways, first as a potential energy problem via a pendulum.
The max velocity of this object will be achieved when the pivot is 90 degrees.
So I set the half of the arm with the mass to be a pendulum with mass ##\frac{5M}{2}##
The starting height of the pendulum will be half of the rod's length ##\frac{L}{2}##

Setting KE = PE
$$\frac{1}{2} \cdot \frac{5M}{2} \cdot v^2 = \frac{5M}{2} \cdot g \cdot \frac{L}{2}$$
$$v = \sqrt{gL}$$

This doesn't seem quite right to me of course, so I tried to approach this as a problem with torque
Setting torque equal to ##\tau = \frac{L}{2} \frac{5M}{2} \cdot g \cdot sin(\theta)##
Then moment of Inertia is ##I = \frac{5M}{2} * \frac{L}{2}^2##

Not sure how to clear the rest of this approach.
 
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  • #2
Energy conservation is the way to go because the acceleration is not constant so you cannot use the standard kinematic equations. For the kinetic energy you should use the rotational form which means you need to find the moment of inertia about the pivot. You also need to find where the center of mass of this thing is and find how far it drops. That should be the ##y## in ##mgy##.
 
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  • #3
So, am I correct in stating that the moment of inertia is ##I = mr^2 = \frac{5M}{2} \cdot \frac{L}{2}^2##?

Then, I would solve for KE = PE, where KE = ## \frac{1}{2} I \omega^2 = mgy = ## PE

Thereby, I solve for ##\omega## as ##\omega = \sqrt{\frac{2mgy}{I}}##, and then v = ##r \omega = \sqrt{2gy}##

My main question now is, how would I calculate y? I tried using the center of mass equation ##m_1 r_1 = m_2 r_2##, and this means that ##r_1 = 5r_2##. Not sure how to go beyond this.

Supposing I use the CoM equation as this ##x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1+m_2}##, with ##x_1## as 0, and ##x_2## as L, I can set ##x_{cm} = \frac{5}{6} L##, is this my value?
 
Last edited:
  • #4
I actually don't think you need to find the center of mass of the entire system. Does the center of mass of the rod (without the attached mass) change?
 
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  • #5
person123 said:
I actually don't think you need to find the center of mass of the entire system
Yes, it is almost always easier to deal with moments of inertia, momenta, energy... separately for the simple components rather than as a compound body.
 

Related to Rod w/ Pivot and Attached Mass Pendulum

1. What is a "Rod w/ Pivot and Attached Mass Pendulum"?

A "Rod w/ Pivot and Attached Mass Pendulum" is a type of physical system used in experiments and demonstrations to study the principles of simple harmonic motion. It consists of a rigid rod attached to a fixed pivot point, with a mass attached to the end of the rod. When the rod is set in motion, the mass will swing back and forth in a predictable pattern.

2. How does a "Rod w/ Pivot and Attached Mass Pendulum" work?

The motion of the "Rod w/ Pivot and Attached Mass Pendulum" is governed by the laws of physics, specifically the principles of simple harmonic motion. As the pendulum swings back and forth, it experiences a restoring force that is proportional to its displacement from its equilibrium position. This causes the pendulum to oscillate in a regular pattern.

3. What factors affect the motion of a "Rod w/ Pivot and Attached Mass Pendulum"?

The motion of a "Rod w/ Pivot and Attached Mass Pendulum" is affected by several factors, including the length of the rod, the mass of the pendulum, and the strength of the gravitational field. Other factors such as air resistance and friction may also have an impact on the motion of the pendulum.

4. What can we learn from studying a "Rod w/ Pivot and Attached Mass Pendulum"?

Studying a "Rod w/ Pivot and Attached Mass Pendulum" can help us understand the principles of simple harmonic motion, which are important in many areas of science and engineering. It can also help us explore concepts such as energy conservation, damping, and resonance.

5. How is a "Rod w/ Pivot and Attached Mass Pendulum" relevant to real-world applications?

"Rod w/ Pivot and Attached Mass Pendulum" systems are used in various real-world applications, such as clock mechanisms, seismometers for measuring earthquakes, and even amusement park rides. By studying how these systems behave, we can better understand and design technologies that rely on similar principles.

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