Rod + 2 Particle masses = angular acceleration

In summary, the problem involves a rotating rigid rod with two particles attached at the ends. The moment of inertia of the system is calculated using the formula Irod = 1/12ML^2 + the moment of inertia of the two particles. The net torque is found by subtracting the weights of the particles and the angle between the torque and the horizontal is calculated to be 140.1 degrees. However, the error in the calculation lies in finding the net force instead of the net torque and using the wrong formula for the moment of inertia of the particles.
  • #1
ninjagowoowoo
75
0
PROBLEM:
A rigid rod of mass 8.70 kg and length 2.30 m rotates in a vertical (x,y) plane about a frictionless pivot through its center. Particles m1 (mass=4.30 kg) and m2 (mass=2.60 kg) are attached at the ends of the rod. Determine the size of the angular acceleration of the system when the rod makes an angle of 50.1° with the horizontal.

I tried getting the moment of inertia using Irod = 1/12ML^2 + the Is of both the particles. Then I found the weight difference and used that to compute torque in torque = r x F. Finally I used those two answers to get angular acceleration. Of course this is wrong... can someone help me please? Or at least tell me what I did wrong... I think I may be confused about the angle between r and F. Anyway, any help would be apperciated! Thanks.
 
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  • #2
i have that problem for my physics hwk too :smile:
 
  • #3
haha, do you go to UCSB?
 
  • #4
ninjagowoowoo said:
I tried getting the moment of inertia using Irod = 1/12ML^2 + the Is of both the particles. Then I found the weight difference and used that to compute torque in torque = r x F. Finally I used those two answers to get angular acceleration. Of course this is wrong... can someone help me please? Or at least tell me what I did wrong... I think I may be confused about the angle between r and F.
This should work just fine. Show in detail what you did and then we can try to spot any errors.
 
  • #5
Irod = (1/12)ML^2 = (1/12)(8.7kg)(2.3^2)=3.83525
Iparticles = m1r1 + m2r2 = 4.3(1.15) + 2.60(1.15) = 7.935
Inet = 11.77025

Force due to gravity(net) = m1g - m2g = 42.14N - 25.48N = 16.66 N

Now this is the part where i think i may be wrong:

since the angle between r and the horizontal is 50.1 deg, then the angle between r and F is 140.1 deg.:confused:(I did try it with 50.1 deg and still no luck.)

Thus:
torque = rF(sin140.1)= 12.28953349

torque = I(alpha) = 12.28953349 = 11.77025(alpha)
so alpha = 1.04411 rad/s/s

which is wrong... is my angle wrong or is there more to it than that? Thanks for the help.
 
Last edited:
  • #6
ninjagowoowoo said:
Irod = (1/12)ML^2 = (1/12)(8.7kg)(2.3^2)=3.83525
Iparticles = m1r1 + m2r2 = 4.3(1.15) + 2.60(1.15) = 7.935
Here's your error: the moment of inertia of a particle is [itex]mr^2[/itex], not [itex]mr[/itex].

Force due to gravity(net) = m1g - m2g = 42.14N - 25.48N = 16.66 N
Subtracting the weights of the two particles to find the net force is a shortcut that is OK here, since they happen to be equidistant from the pivot point. But each particle should be thought of as exerting its own torque, not just its own weight. You need to find the net torque, not the net force.
...
Now this is the part where i think i may be wrong:

since the angle between r and the horizontal is 50.1 deg, then the angle between r and F is 140.1 deg.
That's perfectly OK. Since you don't know whether the heavier particle is above or below the horizontal, the angle could be 140.1 deg or 39.9 deg. But it doesn't matter, the torque is the same.
 

Related to Rod + 2 Particle masses = angular acceleration

What is angular acceleration?

Angular acceleration is a measure of how quickly an object's rotational velocity changes over time. It is usually denoted by the symbol α and is measured in radians per second squared (rad/s²).

How is angular acceleration calculated?

Angular acceleration is calculated by dividing the change in an object's angular velocity by the change in time. The formula for angular acceleration is α = (ω₂ - ω₁) / (t₂ - t₁), where ω is the angular velocity and t is the time.

What is the relationship between angular acceleration and linear acceleration?

Angular acceleration and linear acceleration are related by the radius of rotation. The formula for this relationship is a = αr, where a is the linear acceleration, α is the angular acceleration, and r is the distance from the center of rotation to the object.

How do particle masses affect the angular acceleration of a rod?

The masses of the particles on a rod will affect the rod's moment of inertia, which is a measure of an object's resistance to rotational motion. The larger the moment of inertia, the smaller the rod's angular acceleration will be. Therefore, the more massive the particles, the smaller the rod's angular acceleration will be.

Can angular acceleration be negative?

Yes, angular acceleration can be negative. This means that the object is slowing down its rotational motion. It can also be positive, which means the object is increasing its rotational motion.

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