Why Do Angular Accelerations Differ for Two Pivoted Rods of Varying Lengths?

AI Thread Summary
The discussion centers on the differences in angular accelerations between two pivoted rods of varying lengths when released from the same angle. Angular accelerations αA and αB depend on the initial angle θ due to the geometry of the rods. While both rods have the same initial vertical acceleration |ay| due to gravity, their horizontal accelerations |ax| are not equal, as centripetal forces affect them differently. Just before landing, the center of mass of the longer rod B has a greater speed than rod A, which can be explained through energy conservation principles involving gravitational potential energy and rotational kinetic energy. The conversation emphasizes the importance of understanding angular acceleration in relation to the center of mass and the effects of rod length on motion dynamics.
zhenyazh
Messages
55
Reaction score
0
hi,
could some on help me understand why the following is true?
for some of them i wrote exactly what i don't understand, for others
i really have no idea

Two uniform rods are connected to a table by pivots at one end. Rod B is longer than rod A. Both are released simultaneously from an initial angle θ. Neglect air friction. NOTATION: CM = center of mass; α = angular acceleration; |ay|= size of downward acceleration.

True: αA and αB are dependent on θ.
just don't understand why.

False: |ax| of the CM initially equals 0 for both rods.
this would be false because of the centr. force right?

True: |ay| is initially equal for the CM of A and B.
that is because it is g right?
True: Just before landing, the CM of B has a greater speed than the CM of A.
why so? i found an explanation on the internet but there they comapered mgh to 0.5Iw^2
and from there proved b's speed is larger therefore the answer.
but why can i use such an equation? why do i not take into consideration the kinetic energy
of center of mass and how do i define the height here? according to the center of mass?

False: Rods A and B hit the table at the same time.

False: αA and αB are the same initially.

True: αA and αB both increase with time.

thanks
 
Physics news on Phys.org
zhenyazh said:
True: αA and αB are dependent on θ.
just don't understand why.
How do you find the angular acceleration? What's Newton's 2nd law for rotation?

False: |ax| of the CM initially equals 0 for both rods.
this would be false because of the centr. force right?
Find the initial acceleration of the CM by first finding the initial angular acceleration.

True: |ay| is initially equal for the CM of A and B.
that is because it is g right?
The initial acceleration is not g. (It's not in free fall.)

True: Just before landing, the CM of B has a greater speed than the CM of A.
why so? i found an explanation on the internet but there they comapered mgh to 0.5Iw^2
and from there proved b's speed is larger therefore the answer.
but why can i use such an equation? why do i not take into consideration the kinetic energy
of center of mass and how do i define the height here? according to the center of mass?
If you measure I from the axis of rotation, you are automatically including the KE of the center of mass. The change in gravitational PE can be measured by the change in height of the center of mass.

False: Rods A and B hit the table at the same time.

False: αA and αB are the same initially.

True: αA and αB both increase with time.
Figure out α as a function of θ.
 
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