Rod w/ Pivot and Attached Mass Pendulum

  • #1
8
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.
 

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,987
3,141
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##.
 
  • Like
Likes srekai
  • #3
8
0
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
260
37
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?
 
  • Like
Likes Delta2 and TSny
  • #5
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,546
6,438
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 Threads on Rod w/ Pivot and Attached Mass Pendulum

Replies
8
Views
166
Replies
0
Views
2K
Replies
2
Views
22K
Replies
14
Views
4K
  • Last Post
Replies
5
Views
964
Replies
5
Views
4K
  • Last Post
Replies
2
Views
4K
Replies
11
Views
3K
Replies
48
Views
9K
Replies
3
Views
2K
Top