Calculating Rod's Speed After Free Fall

In summary, a thin uniform rod of mass M and length L, positioned vertically above an anchored frictionless pivot point and allowed to fall to the ground, strikes the ground with a tangential velocity of v_t = sqrt(3gL/4). The potential energy and moment of inertia of the rod should be calculated using the distributed mass along its length, rather than assuming it is concentrated at its center of mass. This yields the correct value for the tangential velocity at the free end of the rod just before it hits the ground.
  • #1
jdstokes
523
1

Homework Statement



From the GRE 0177 practice exam:

A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point, as shown above, and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?

The Attempt at a Solution



[itex]\frac{MgL}{2} = \frac{1}{2}I \omega^2 \implies[/itex]
[itex]MgL = \frac{1}{12} M L^2 \omega^2 \implies[/itex]
[itex]\omega = \sqrt{\frac{12g}{L}} \implies[/itex]
[itex]v_t = L\omega = \sqrt{12gL}[/itex]

Answer
[itex]v_t=\sqrt{3gL}[/itex]

Any help would be appreciated.
 
Physics news on Phys.org
  • #2
0.5*m*g*L + 0 + 0 = 0 + 0.5*m*vc^2 + 0.5*Io*omega^2, where vc = tangential velocity of rod centroid, and Io = rod centroidal mass moment of inertia.
 
Last edited:
  • #3
Hi nvn,

Your equation is incorrect. Why have you put a point mass term on the RHS?
 
  • #4
Oh I understand now, the moment of inertia requires a modification because the object is not rotating about the center of mass. Thanks.
 
  • #5
The moment of inertia of a rod is only (1/12)*M*L^2 if it's rotating around the center. If it's rotating around an end it's (1/3)*M*L^2. Ach, I see you have already figured it out. Good job!
 
  • #6
jdstokes, nvn, Dick,

I was searching the analysis for this particular falling rod problem as I'm currently trying to figure out the speed of the free end of the rod just before it hits the ground.

If we are assuming all the mass of the rod as located at its center of mass when we calculate the potential energy (Ep=M.g.L/2), why are we not considering the same thing (mass located at its center of mass) when we calculate Moment of Inertia? In this case I=M(L/2)^2?

The next question is about the kinetic energy before the rod hits the ground.
Shouldn't we consider two components of the kinetic energy: one due to angular velocity and another due to tangential velocity?

Taking into considerations these two assumptions I got that Vend=SQRT(2gL)

Jvicens
 
  • #7
jvicens: The mass moment of inertia of the rod is centroidal as stated in post 2. The rod mass is distributed along its length, not concentrated at its center. Try looking up Io again; your formula appears incorrect.

The answer to your second question is, yes, we should, and we are. See post 2. Your current answer appears to be incorrect.
 
  • #8
nvn,

I understand what you said for my second question but I was initially still confused about the moment of inertia that should be used in these calculations
I thought that the expression for potential energy 0.5xMxgxL was derived assuming the entire mass of the rod located at its center of mass and this was the wrong assumption. With calculus is easy to see that the above mentioned expression is the result of integrating dEp along the length of the rod from zero to L. As we have not considered the total mass concentrated in the center of mass of the rod then we should not consider either the moment of inertia as coming from a concentrated mass located in the center of mass. I should have taken Io as the one from a rod rotating by one of its ends (Io= 1/3xMxL^2).
The new calculation yields Vend = SQRT(3xgxL/4). Is this in agreement with your calculations?
Jvicens
 
  • #9
jvicens: You can instead work the problem that way, which is a good method, but it gives the same answer (which is stated in the last line of post 1), not the answer you listed.
 

Related to Calculating Rod's Speed After Free Fall

1. How do you calculate the speed of an object after free fall?

The speed of an object after free fall can be calculated using the equation: speed = square root of (2 x acceleration due to gravity x distance fallen). This equation is derived from the equation for average velocity, which is velocity = distance/time.

2. What is the acceleration due to gravity?

The acceleration due to gravity is the rate at which an object accelerates towards the Earth's surface due to the force of gravity. It is approximately 9.8 meters per second squared (m/s^2) near the Earth's surface.

3. How do you measure the distance fallen in a free fall scenario?

The distance fallen can be measured using a stopwatch and measuring the time it takes for the object to fall. The distance can then be calculated using the equation: distance = 1/2 x acceleration due to gravity x (time)^2.

4. What factors can affect the speed of an object after free fall?

The speed of an object after free fall can be affected by the object's mass, the acceleration due to gravity, and air resistance. Objects with greater mass will have a higher terminal velocity and therefore a higher speed after free fall. The acceleration due to gravity can also vary depending on the location on Earth. Air resistance can also play a role in slowing down the object and affecting its final speed.

5. Can the speed of an object after free fall ever exceed the speed of light?

No, according to the theory of relativity, the speed of light is the maximum speed that can be reached by any object. Therefore, the speed of an object after free fall can never exceed the speed of light.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
835
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
530
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
34
Views
802
  • Introductory Physics Homework Help
Replies
7
Views
398
Replies
7
Views
1K
Replies
39
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Back
Top