Moment of Inertia for Combination of Rod and Ball

[TMO]

Homework Statement

[PLAIN]http://www.smartphysics.com/images/content/mechanics/ch15/momentofinertia2new.png

What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)

m_rod (hence denoted as m_r) = 7.11 kg
m_sphere (hence denoted as m_s) = 35.55 kg
r_rod (this is the length of the rod, and it will hence be denoted as r_r) = 5.8 m
r_sphere (this is the radius of the sphere, hence denoted as r_s) = 1.45 m

Homework Equations

$$I = I_{center of mass} + md^2$$ (parallel axis theorem)
$$I_{sphere} = \frac{2}{5}mr^2$$
$$I_{rod at end} = \frac{1}{3}ml^2$$

The Attempt at a Solution

I figured that the moment of inertia must include both the moment of a rod spinning at its edge and the moment resulting from the parallel axis theorem. This leads to the first part.

$$\frac{1}{3}m_r(r_r)^2 + m_r(r_s/2)^2$$

Doing something similar to the sphere gives a similar equation

$$\frac{2}{5}m_s(r_s)^2 + m_s(r_s/2)^2$$

Then I just combine these together, if the hyperphysics page is anything to go by (http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html). This yields the following.

$$\frac{1}{3}m_r(r_r)^2 + m_r(r_s/2)^2 + \frac{2}{5}m_s(r_s)^2 + m_s(r_s/2)^2$$

However, this isn't correct. Why is this, and how can I fix this?

 

[vkash]

Homework Statement

[PLAIN]http://www.smartphysics.com/images/content/mechanics/ch15/momentofinertia2new.png

What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)

m_rod (hence denoted as m_r) = 7.11 kg
m_sphere (hence denoted as m_s) = 35.55 kg
r_rod (this is the length of the rod, and it will hence be denoted as r_r) = 5.8 m
r_sphere (this is the radius of the sphere, hence denoted as r_s) = 1.45 m

Homework Equations

$$I = I_{center of mass} + md^2$$ (parallel axis theorem)
$$I_{sphere} = \frac{2}{5}mr^2$$
$$I_{rod at end} = \frac{1}{3}ml^2$$

The Attempt at a Solution

I figured that the moment of inertia must include both the moment of a rod spinning at its edge and the moment resulting from the parallel axis theorem. This leads to the first part.

$$\frac{1}{3}m_r(r_r)^2 + m_r(r_s/2)^2$$

Doing something similar to the sphere gives a similar equation

$$\frac{2}{5}m_s(r_s)^2 + m_s(r_s/2)^2$$

Then I just combine these together, if the hyperphysics page is anything to go by (http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html). This yields the following.

$$\frac{1}{3}m_r(r_r)^2 + m_r(r_s/2)^2 + \frac{2}{5}m_s(r_s)^2 + m_s(r_s/2)^2$$

However, this isn't correct. Why is this, and how can I fix this?

If the axis of rotation is arrow line
moment of inertia of the rod will Zerp. Do you; Because elementry formula for moment of inertia is I=mr². R=0 for all the particles of the rod so moment of inertia of the rod is0.
Now come to sphere; moment of inertia of sphere is 2/5mr². that's the answer.

if axis of rotation is doted line(COM)
apply theorem of perpendicular axis.
moment of inertia about an axis passing through COM(dotted line in fig) and perpendicular to your screen will same since mass configuration is same around both the axises. Let me say it be I
then I+I=2/5mr²
This is short cut.
It must correct. IF this is wrong please tell me.

 

[Doc Al]

I figured that the moment of inertia must include both the moment of a rod spinning at its edge and the moment resulting from the parallel axis theorem. This leads to the first part.

$$\frac{1}{3}m_r(r_r)^2 + m_r(r_s/2)^2$$

To use the parallel axis theorem, you must start with the rod's moment of inertia about its center of mass, not its end.