Finding Angular Acceleration of Uniform Rod

AI Thread Summary
To find the angular acceleration of a uniform rod pivoted at one end, the correct approach involves using the relationship between torque, angular acceleration, and moment of inertia, rather than relying on time-dependent equations. The angular acceleration is calculated to be 19.62 rad/s². For the center of mass acceleration components, the x-component is -14.7 m/s² and the y-component is -7.36 m/s². The center of mass moves in a circular path, and understanding the components of acceleration in circular motion is essential for solving part (b). This discussion emphasizes the importance of using appropriate formulas in rotational dynamics.
roam
Messages
1,265
Reaction score
12

Homework Statement



A uniform rod of length 0.750 m and mass 1.20 kg is pivoted at one end by a smooth pin as shown below. The rod is released from the vertical position and given a slight nudge to release it from the vertical position of unstable equilibrium.

[PLAIN]http://img175.imageshack.us/img175/9820/angacc40pc.png

When the rod is horizontal:

(a) Calculate its angular acceleration.

(b) Calculate the x and the y components of the acceleration of the centre of mass.

The Attempt at a Solution



First I focus on part (a). I have already obtained the angular velocity:

The potential energy of the system relative to the reference configuration is MgL/2 because the center of mass of the rod is at a height L/2 away from its position in the reference configuration. Conservation of mechanical energy for the system is:

Ki+Ui = Kf + Uf

0+½MgL=½Iω²+0

\omega = \sqrt{\frac{MgL}{\frac{1}{3}ML^2}} = \sqrt{\frac{3g}{L}}

using the given values I get ω=6.26 rad/s, which is the correct velocity.

Now I want to use the equation \alpha = \frac{\Delta \omega}{t} to find the angular acceleration. Here's where I'm stuck. Could anyone please show me how to find the time for this equation?

Correct answer to part (a) is 19.62 rad/s² and for part (b) it is -14.7 m/s²(x-component) and -7.36 m/s² (y-component)
 
Last edited by a moderator:
Physics news on Phys.org
No, you can not use

<br /> \alpha = \frac{\Delta \omega}{t}<br />,
as it is valid for uniform acceleration only, and you do not know if the acceleration is uniform; neither you know the time.

There is a formula betwee torque, angular acceleration, and moment of inertia. Use that.

ehild
 
ehild said:
No, you can not use

<br /> \alpha = \frac{\Delta \omega}{t}<br />,
as it is valid for uniform acceleration only, and you do not know if the acceleration is uniform; neither you know the time.

There is a formula betwee torque, angular acceleration, and moment of inertia. Use that.

ehild

Thank you very much, I got it. Now could you explain how to deal with part (b)? How do we calculate the x (or y) component of the acceleration of the centre of mass? I'm not quite sure what that even means since I don't have any worked examples in my textbook.
 
The centre of mass moves along a circle, and the angular velocity is not constant. What components of acceleration has a body performing circular motion?

ehild
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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