Rotational Momentum and linear acceleration

In summary: Hi blackheart,In summary, a rigid, massless rod with three equal masses attached to it is free to rotate about a frictionless axle and is released from rest in a horizontal position. The linear acceleration of the particle labeled 3 at t=0 is incorrect and the correct value depends on the distance between the masses and the radius of rotation for mass 3. The maximum angular speed attained by the rod can be found by checking if angular momentum is conserved and the maximum speed attained by the particle labeled 2 can be found using the equation for torque.
  • #1
blackheart
10
0
1. A rigid, massless rod has three particles with equal masses attached to it as shown below. The rod is free to rotate in a vertical plane about a frictionless axle perpendicular to the rod through the point P and is released from rest in the horizontal position at t = 0. Assume that m and d are known. (Use the following as necessary: m, d, and g.)Sorry this is my diagram (I couldn't paste it for some reason)...

distance from P to m3 is (2d/3)
m1-------------m2-------P---------m3 THIS IS THE ROD WITH MASSES
distance between the masses is d
m1, m2, m3 are the same mass(d) Find the linear acceleration of the particle labeled 3 at t = 0.
(torque of mass 3) = (I of mass 3)(alpha)

alpha = 2g/7d

a=r(alpha)

a= 4g/3 but this is incorrect

(f) Find the maximum angular speed attained by the rod. ωf =

(Ii)(wi) = (If)(wf)
It starts from rest so w initial is 0, so I could not solve using this way...

(h) Find the maximum speed attained by the particle labeled 2. vf =

Relevant equations:
L(initial) = L final
L=Iw=mvrsin(theta)
torque = r x F
 
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  • #2
Hi blackheart,

blackheart said:
1. A rigid, massless rod has three particles with equal masses attached to it as shown below. The rod is free to rotate in a vertical plane about a frictionless axle perpendicular to the rod through the point P and is released from rest in the horizontal position at t = 0. Assume that m and d are known. (Use the following as necessary: m, d, and g.)


Sorry this is my diagram (I couldn't paste it for some reason)...

distance from P to m3 is (2d/3)
m1-------------m2-------P---------m3 THIS IS THE ROD WITH MASSES
distance between the masses is d
m1, m2, m3 are the same mass


(d) Find the linear acceleration of the particle labeled 3 at t = 0.
(torque of mass 3) = (I of mass 3)(alpha)

alpha = 2g/7d

a=r(alpha)

a= 4g/3 but this is incorrect


Just from looking at these last three equations, it looks like you are setting r=(14/3) d (which is about r=4.67 d). But what should r be for mass 3?

How are you getting that value for the angular acceleration?

(f) Find the maximum angular speed attained by the rod. ωf =

(Ii)(wi) = (If)(wf)

This equation is stating the the angular momentum is conserved. But that is not automatically true for a system; you have to check that it is true. What is the test for whether angular momentum is conserved or not?

Are there any other quantities that are conserved here?


It starts from rest so w initial is 0, so I could not solve using this way...

(h) Find the maximum speed attained by the particle labeled 2. vf =

Relevant equations:
L(initial) = L final
L=Iw=mvrsin(theta)
torque = r x F
 

1. What is rotational momentum and how is it different from linear momentum?

Rotational momentum is the measure of a rotating object's resistance to changes in its rotational motion. It is different from linear momentum, which is the measure of an object's resistance to changes in its linear motion.

2. How is rotational momentum calculated?

Rotational momentum is calculated by multiplying an object's moment of inertia (a measure of its distribution of mass) by its angular velocity (the rate at which it rotates around an axis). The formula for rotational momentum is L = Iω.

3. What is the relationship between rotational momentum and angular acceleration?

Rotational momentum is directly proportional to angular acceleration. This means that an increase in rotational momentum will result in an increase in angular acceleration, and vice versa.

4. Can rotational momentum be conserved?

Yes, rotational momentum can be conserved in a closed system where there are no external torques acting upon the system. This means that the total amount of rotational momentum in the system will remain constant.

5. How does rotational momentum affect an object's stability?

Rotational momentum plays a crucial role in an object's stability. Objects with a higher rotational momentum are harder to rotate and are therefore more stable. This is why spinning tops and gyroscopes can maintain their balance even when tilted or disturbed.

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