Vertical acceleration of a rod rotating around a pivot

In summary, the center of mass of the rod experiences a vertical acceleration of 2g at the instant it is in a horizontal position.
  • #1
yankans
12
0

Homework Statement



A uniform rod of mass 1.6 kg is 6 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance of 6 m from the center of mass of the rod. Initially the rod makes an angle of 58 degrees with the horizontal. The rod is released from rest at an angle of 58 degrees with the horizontal.

What is the magnitude of the vertical acceleration of the center of mass of the rod at the instant the rod is in a horizontal position? The acceleration of gravity is 9.8 m/s^2 and the moment of inertia of the rod about its center of mass is 1/12 mL^2. Answer in units of m/s^2.

Homework Equations



torque = I (moment of inertia) * angular acceleration
tang. acceleration = R*angular acceleration

The Attempt at a Solution


torque = I * ang. acceleration
r mg = 1/12 m l^2 * ang. acc.
6mg = 1/12 m (6^2) * ang. acc.
6mg = 3m* ang. acc.
ang. acc = 2g
tang. acc. = r(ang. acc) = 2(6g)
tang. acc(vertical) = 12g

And it is wrong.
 
Physics news on Phys.org
  • #2
At the position indicated, what does the vertical component of the acceleration depend on?

Note that you'll need the rotational inertia about the pivot point.

(I assume that the "thin extension" is perpendicular to the rod.)
 
  • #3
The thin extension is attached to the rod in the same direction as the rod (i.e.) if I drew a picture it would sort of look like a corn dog:
------((((((((((((((
dist. from end of the thin extension to center of mass of rod = 6m
length of rod itself = 6m
so length of the thin extension = 3m
 
  • #4
yankans said:
The thin extension is attached to the rod in the same direction as the rod (i.e.) if I drew a picture it would sort of look like a corn dog:
------((((((((((((((
dist. from end of the thin extension to center of mass of rod = 6m
length of rod itself = 6m
so length of the thin extension = 3m
OK. My previous comments remain (except my guess as to how the extension was attached).

Was the initial angle of the rod 58 degrees above or below the horizontal?
 
  • #5
above the horizontal
 
  • #6
it's falling down
 
  • #7
OK, answer my question from post #2.
 
  • #8
hmm...I guess it would depend on gravity (I originally looked at this problem and thought that the answer was just g, but my physics teacher told me I was wrong), because the total vertical acceleration apparently also (I think) depends on the angular acceleration.
 
  • #9
yankans said:

The Attempt at a Solution


torque = I * ang. acceleration
r mg = 1/12 m l^2 * ang. acc.
6mg = 1/12 m (6^2) * ang. acc.
You are using the moment of inertia about the center of mass; you should be using the moment of inertia about the pivot. (Consider the parallel axis theorem.)
 
  • #10
Oh wow, I see, problem solved!

Thank you!
 

FAQ: Vertical acceleration of a rod rotating around a pivot

How does vertical acceleration affect the rotation of a rod around a pivot?

The vertical acceleration of a rod rotating around a pivot does not directly affect its rotation. However, it can affect the stability and balance of the rod, which can indirectly impact its rotation.

Can the vertical acceleration of a rod change its direction of rotation?

No, the vertical acceleration alone cannot change the direction of rotation of a rod around a pivot. The direction of rotation is determined by the initial velocity and the angular velocity of the rod.

How does the length of the rod affect its vertical acceleration?

The length of the rod does not have a direct impact on its vertical acceleration. However, a longer rod may experience more torque due to its increased moment arm, which can result in a higher vertical acceleration.

What factors can affect the vertical acceleration of a rod rotating around a pivot?

The vertical acceleration of a rod can be affected by factors such as the applied force, the mass distribution of the rod, the pivot point, and any external forces acting on the rod.

How can the vertical acceleration of a rod be calculated?

The vertical acceleration of a rod can be calculated using the equation a = α * r, where a is the vertical acceleration, α is the angular acceleration, and r is the distance between the pivot point and the center of mass of the rod.

Similar threads

Back
Top