Calculating CW & CCW Torques for Point Masses on a Rod

AI Thread Summary
The discussion focuses on calculating clockwise (cw) and counterclockwise (ccw) torques for point masses on a rod. A horizontal massless rod, 6 meters long, has a 2kg mass at each end and is pivoted at its center, leading to equilibrium where the sum of torques equals zero. Participants clarify that the torque is calculated using the formula torque = Fd, and that both sides of the rod exert equal force due to symmetry. Another problem involves a 4-meter rod with a mass of 5kg, where the net torque and angular acceleration are calculated, prompting questions about the distribution of mass and forces. The conversation emphasizes understanding torque calculations and the effects of mass distribution on angular motion.
steveo0
Messages
3
Reaction score
0

Homework Statement


Find all cw and ccw torques. I don't know how to post the picture, but I was absent from my physics class today for AMC so I missed out. First problem: A horizontal massless rod of length 6 meters is pivoted about its center. There is a 2kg point mass on each end of the rod.


Homework Equations


torque = Fd


The Attempt at a Solution


So I looked at the textbook, but they only provide examples with cylinders. So I just need one example for this and I think I can pretty much do the rest. Do I calculate the Force for each one? or do I use the moment of Inertia?
 
Physics news on Phys.org
Welcome to PF.

You mean it's a see-saw.

If it is in equilibrium then the sum of the Torques are 0.

The F*2m on one side is equal to the F*2m on the other.
 
oh. thanks. lol that's that only example where they're both equal. so uh.

1. Homework Statement
Find all cw and ccw torques. Find the net torque. Find the angular acceleration. A differential horizontal rod has a length of 4 meters and a mass of 5kg. It is pivoted about its center. 2. Homework Equations
torque = Fd
torque = (I)(alpha)3. The Attempt at a Solution
I found Fg = -49 N
torque = Fxd = -98 N/m
angular acceleration = torque / moment of inertia = -98 / (20/3) = -14.7 m/s^2?
is this right?
 
Does the rod have mass?
 
yeah 5kg
 
If it is pivoted about the center where is the net force?

Is there a weight on one end as well?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top