Rotational Motion and moments of inertia

• FluffCorgi
In summary: Right. Out of curiosity I looked up the moment of inertia about the midpoint of a cylinder of length ##L## and radius ##r##; it is $$I=\frac{1}{12}m(3r^2+L^2)=\frac{1}{12}mL^2\left(1+3\frac{r^2}{L^2}\right).$$Using the given value L = 170 cm and assuming a reasonable arm radius of 2-3 cm, the added term ##3\frac{r^2}{L^2}## is of order 10-4. It seems that the thin rod approximation for the extended arms is good.

FluffCorgi

Homework Statement
A 65kg student is ice skating as shown. She is initially rotating at with an angular speed of 2.5rad/s with her arms straight out. At this time she can be modeled as a vertical and horizontal cylinder as shown.

She then brings her arms in, at which point she can be modeled as only a vertical cylinder with the properties shown.

The moment of inertia for the vertical cylinders spinning around their axes is I = 1/2mr2.
The moment of inertia for the horizontal cylinder spinning around it's center, perpendicular to it's axis is I = (1/12)md^2.

Determine all of the following moments of inertia, angular speeds and angular momenta relative to the axis of rotation.
Relevant Equations
arms
m = 6kg
d = 1.7m
core
m = 59kg
r = 0.18m
full body
m = 65kg
r = 0.18m
I have the moment of inertia for the core(initial) and full body(final) but my answer for the moment of inertia for the arms(initial) was incorrect.
Arms(initial) moment of inertia:(1/12)(6)(1.7^2)=1.445 this is incorrect for some reason
Core(initial) moment of inertia: .9558
Full Body(final): 1.053
I also have to find the angular velocity and I have done this correctly for the arms and core but I could not figure out how to find the angular velocity for the full body(final)
Full Body(final): I don't think I can solve this without the arms initial moment of inertia
Angular Momentum
Arms: my moment of inertia was incorrect so this was as well
Core: 2.3895
Full Body(final): My moment of inertia for the arms was incorrect so this was as well but this was my work to attempt to find it.
(1.445+.9558)2.5 = (1.053)w
w = 5.699 this was wrong but i tried to solve it anyway
I need help with the initial moment of inertia for the arms.

Attachments

• Annotation 2019-10-20 172630.png
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The moment of inertia of the outstretched arms is proportional ##r^2##. You have put in 1.7m for the radius. Do you think that is a reasonable length for a skater's arm?

Why is d defined as 1.7 if they don't want you to use it in the formula?

tnich said:
The moment of inertia of the outstretched arms is proportional ##r^2##. You have put in 1.7m for the radius. Do you think that is a reasonable length for a skater's arm?
Not quite. The moment of inertia of a rod of length ##L## about its midpoint is ##I=\frac{1}{12}mL^2##. Here ##L=1.7 ~\mathrm{m}## and I get the same value as OP for the moment of inertia of just the arms. I also think that 1.7 m is a reasonable length for the skater's wingspan which should be approximately equal to her height at under 6 ft. See here for details.

I think the website is incorrect.

FluffCorgi said:
I think the website is incorrect.
This wouldn't be the first (or last) time that a website is incorrect. Talk to the person who assigned this question. He/she needs to know.

FluffCorgi said:
Full Body(final): 1.053
I would take the retracted arms as point masses at the full radius of the torso. Similarly, the extended arms should not be taken as a uniform rod through the body. Take them as starting at the periphery of the torso.
tnich said:
You have put in 1.7m for the radius.
No, as the fingertip to fingertip span, as @kuruman notes.
kuruman said:
Not quite. The moment of inertia of a rod of length L about its midpoint is
@tnich only said it was proportional to r2 (but maybe didn't notice the 1/12).

haruspex said:
Similarly, the extended arms should not be taken as a uniform rod through the body. Take them as starting at the periphery of the torso.
Yes, but the figure accompanying the post says "At this time she can be modeled as a vertical and horizontal cylinder as shown." And further down , "The moment of inertia for the horizontal cylinder spinning about it's (sic) center, perpendicular to it's axis is ##\frac{1}{12}md^2##." Parameter ##d## is the length of the horizontal cylinder in the figure.

kuruman said:
Yes, but the figure accompanying the post says "At this time she can be modeled as a vertical and horizontal cylinder as shown." And further down , "The moment of inertia for the horizontal cylinder spinning about it's (sic) center, perpendicular to it's axis is ##\frac{1}{12}md^2##." Parameter ##d## is the length of the horizontal cylinder in the figure.
Ok, I hadn't noticed the attachment.
In effect it treats the shoulders and the space between as part of the arms.
I note also it says to treat the arms as a cylinder but provides the formula for a thin rod and no thickness for the arms.

haruspex said:
Ok, I hadn't noticed the attachment.
In effect it treats the shoulders and the space between as part of the arms.
I note also it says to treat the arms as a cylinder but provides the formula for a thin rod and no thickness for the arms.
Right. Out of curiosity I looked up the moment of inertia about the midpoint of a cylinder of length ##L## and radius ##r##; it is $$I=\frac{1}{12}m(3r^2+L^2)=\frac{1}{12}mL^2\left(1+3\frac{r^2}{L^2}\right).$$Using the given value L = 170 cm and assuming a reasonable arm radius of 2-3 cm, the added term ##3\frac{r^2}{L^2}## is of order 10-4. It seems that the thin rod approximation for the extended arms is good.

1. What is rotational motion?

Rotational motion is the movement of an object around an axis or point, rather than in a straight line. It can occur in both natural and man-made systems, such as planets orbiting around the sun or a wheel spinning on an axle.

2. How is rotational motion different from linear motion?

Rotational motion involves movement around an axis, whereas linear motion involves movement in a straight line. In rotational motion, the distance of the object from the axis and the angle of rotation are important factors, whereas in linear motion, only the distance traveled is relevant.

3. What is moment of inertia and how is it calculated?

Moment of inertia is a measure of an object's resistance to rotational motion. It is calculated by multiplying the mass of the object by the square of its distance from the axis of rotation.

4. How does the distribution of mass affect an object's moment of inertia?

The distribution of mass affects an object's moment of inertia because the farther the mass is from the axis of rotation, the larger the moment of inertia will be. This means that objects with more mass concentrated near the axis will have a smaller moment of inertia, while objects with mass spread out farther from the axis will have a larger moment of inertia.

5. What are some real-world applications of moments of inertia?

Moments of inertia are important in many everyday situations, such as when calculating the stability of a building or bridge, designing rotating machinery, or understanding the motion of a spinning top or gyroscope. They are also used in sports, such as figure skating and diving, to control rotational movements and maintain balance.

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