A rigid H falls rotating about one of its legs. What's its angular velocity?

[DavidAp]

A rigid "H" falls rotating about one of its legs. What's its angular velocity?


A rigid body is made of three identical thin rods, each with length L = 0.340 m, fastened together in the form of a letter H, as suggested by the figure here. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of the H is vertical

Answer: 8.05 rad/s


I knew k = 1/2 Iw^2 so I went on to find k and I about each rod and adding them together to get their total kinetic energy(potential energy)/Inertia. My work is shown below.

When I plug this in though I get 9.60 "rad"/s. Why, what did I do wrong? After working on this problem for over an hour I (although my work doesn't show it, it took my a long time to get to this stage) I decided to us I = Icm + Mh^2, knowing that h=l, giving me a new inertia of (17M(L^2))/(12) but when plugging in I got 5.208 "rad"/s.

What am I doing wrong? How do you solve this problem?
Thank you for taking the time to review my question.

 

[ehild]

The moment of the cross rod is 1/3 mL^2, and that of the leg is 1/12 mL^2+mL^2, according to the Parallel Axis Theorem-17/12 mL^2 altogether.

The potential energy is wrong, it is not 2/3 mLg. Correct.

ehild

 

[DavidAp]

The moment of the cross rod is 1/3 mL^2, and that of the leg is 1/12 mL^2+mL^2, according to the Parallel Axis Theorem-17/12 mL^2 altogether.

The potential energy is wrong, it is not 2/3 mLg. Correct.

ehild

Hi there!

I know it's been a while since I've posted this but I've been scratching my head and I can't figure out what PE is. I was thinking that, after taking another look at the problem, the PE for the center rod is mg(l/2) because that's where it's center of mass is, but I'm not sure what the PE for the other rod is. I thought it would just be mgl making the total PE 3/2 mgl (ironically the reciprocal of what I originally thought PE was) but that didn't give me the right answer either.

Any hints on what I'm doing wrong or where I should look for the right answer?

 

[ehild]

We both fell into the same trap when calculating the moment of inertia of the leg. The axis of rotation is parallel with the rod, so the moment of inertia with respect to the CM of the leg is zero instead of 1/12 mL^2 (that would be true for the axis perpendicular to the rod).

So the total moment of inertia is 1/3 mL^2+mL^2.

ehild

 

[asteriatic]

It seems like you are on the right track with your approach. However, there may be some errors in your calculations. Let's break down the problem and see how we can solve it.

Firstly, we need to determine the moment of inertia of the H-shaped body about its axis of rotation. Since the body is symmetric, we can use the parallel axis theorem to calculate the moment of inertia about the axis passing through the center of mass of the body. The moment of inertia for each rod is given by I = (1/12)ML^2, where M is the mass of each rod and L is the length. Since there are three identical rods, the total moment of inertia of the body is (3/12)ML^2 = (1/4)ML^2.

Next, we need to consider the potential energy of the body at the starting position (when the plane of the H is horizontal). The potential energy for a rotating body is given by U = mgh, where m is the mass of the body, g is the acceleration due to gravity and h is the height of the center of mass from the axis of rotation. In this case, the center of mass is at a height of L/2 from the axis of rotation. Therefore, the potential energy at the starting position is U = (3/4)MgL.

Now, we can use the conservation of energy principle to solve for the angular velocity of the body at the final position (when the plane of the H is vertical). At this position, the potential energy is zero and the kinetic energy is at its maximum. Therefore, we can equate the initial potential energy to the final kinetic energy and solve for the angular velocity. This gives us:

(3/4)MgL = (1/2)Iw^2

Substituting the moment of inertia we calculated earlier, we get:

(3/4)MgL = (1/2)(1/4)ML^2w^2

Solving for w, we get:

w = √(3g/L)

Plugging in the given values, we get:

w = √(3(9.8)/0.340) = 8.05 rad/s

Therefore, the angular velocity of the body at the final position is 8.05 rad/s.

I hope this helps to clarify the problem and guide you in the right direction. Keep in