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A long, uniform rod of length L and mass M is pivoted about a frictionless, horizontal pin through one end. The rod is nudged from rest in a vertical position as shown in figure. At the instant the rod is horizontal, find (a) its angular speed

Since angular speed is the same throughout a rigid object, can't I just use the conservation of energy equation and solve for w? There is an example similar to this question in my text-book, but they found the angular speed at the center of mass. Why did they do that? Why is the center of mass being used so often in rotational motion?

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Yes,you can use conservation of energy. The key is to make sure you use the correct moment of inertia in the kinetic energy, and the correct change in potential energy.

I don't understand why you asked your second question. You state correctly that the angular speed is the same throughout a rigid object, so why not find the angular speed at the center of mass.

The center of mass is used for several reasons:

  • the moment of inertia (MoI) about the center of mass (NOT what you want in this case) is the smallest MoI for that object. If you pick a different point for rotation in a given plane, the MoI about that point will be larger.
  • if gravity is acting and producing a torque, the gravitational force is considered to act through the center of mass when calculating the torque.

EDIT: The reason you shouldn't use other points is because the center of mass is the effective average location of the mass. If you choose a different location to calculate the weight force application point on the object, you are misrepresenting the mass distribution of the object. In rotational problems, mass distribution is a primary consideration.

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  • $\begingroup$ In regards to my second question, I was wondering why I couldn't choose the right endpoint on the rod since the angular speed is the same for all points. If I were to choose that point, then the rod will fall a distance L, so the potential energy becomes MgL, correct? Regardless, I keep getting the wrong answer if I choose the endpoint instead of the center of mass. $\endgroup$ – Kevin May 13 at 2:32
  • $\begingroup$ @kevin, just saying you're getting a wrong answer is unlikely to generate a lot of assistance. You probably should add your attempt into the the question and show how it differs from the accepted solution. $\endgroup$ – BowlOfRed May 13 at 5:14
  • $\begingroup$ @Kevin See my edit. $\endgroup$ – Bill N May 13 at 15:48
  • $\begingroup$ Yes I did! Thank you so much! $\endgroup$ – Kevin May 14 at 4:27
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The center of mass is important because for the rod when it is not at a horizontal position each part of the rod is at a different height and hence has a different potential energy. However, we can treat the rod as if it is a point mass located at its center of mass rotating about the pivot.

Consider an arbitrary object of total mass $M$ with a mass distribution $\rho(\mathbf r)$. The potential energy of an infinitesimal mass element of the object located at position $\mathbf r$ is given by $$\text dU=g(y-y_0)\text dm=g(y-y_0)\rho(\mathbf r) \text dV$$ where $y$ is the height of the infinitesimal mass element relative to $y_0$ where $U=0$. Therefore, the total potential energy is given by $$U=\iiint_Vg(y-y_0)\rho(\mathbf r) \text dV=Mg(y_{com}-y_0)$$

So as you can see, we only need to consider the height of the center of mass to determine the potential energy of the entire rod.

This is a specific case of a more general idea that near the center of the Earth we can treat an extended body in the uniform gravitational field as if it was just being acted upon by gravity at its center of mass.

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You were correct when you stated that angular velocity about any point on the rod can be found by work energy theorem. As the body is performing pure rotational motion about the hinged point, I would suggest you to directly equate work done on centre of mass by gravity and the rotational energy of the rod about the hinge.

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Yes conservation of energy can be used to solve the question. The initial gravitational potential energy will be equal to the sum of translational kinetic energy and rotational kinetic energy.

Why Velocity of Centre of Mass?

The velocity of centre of mass is generally used in rotational mechanics because it is the weighted average of the velocities of all points. Velocity of centre of mass on average is the velocity of the whole system.

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