A spherically symmetric object with radius of .7m and mass of 1.6kg rolls without slipping accross a horizontal surface with velocity of 1.7m/s. It then rolls up an invline with an angle of 28degrees and comes to rest a distance d of 2.3m up the incline, before rolling back down. Find the moment of inertia of this object about an axis through its center of mass.

Isn't I just= mr^2?

No, not for a solid sphere.

You can find it from the experiment.

Conservation of energy:

final energy= energy at top of incline
1/2mv^2 + 1/2 I w^2= mgh= 1.6*9.8*2.3sin28

but w=v*r, so
1/2 mv^2+1/2 I v^2 r^2= 1.6*9.8*2.3sin28
Solve for I.

No, the formula I = mr^2 only applies to a point mass rotating about an axis passing through its center of mass. In the case of a spherically symmetric object like the one described, we need to consider its moment of inertia about an axis that doesn't necessarily pass through its center of mass.

To find the moment of inertia of the object about an axis through its center of mass, we can use the parallel axis theorem. This theorem states that the moment of inertia about an axis parallel to an axis through the center of mass is given by the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance between the two axes.

First, let's find the moment of inertia of the object about an axis passing through its center of mass. We can use the formula for the moment of inertia of a solid sphere rotating about its diameter, which is given by (2/5) * (m * r^2).

Substituting the given values:
m = 1.6 kg (mass of the object)
r = 0.7 m (radius of the object)

So, the moment of inertia about an axis passing through the center of mass is:
I_cm = (2/5) * (1.6 kg) * (0.7 m)^2

Next, we need to find the distance between the center of mass and the axis of rotation, which is the radius of the object (0.7 m).

Using the parallel axis theorem, the moment of inertia about the axis through the center of mass is given by:
I_axis = I_cm + (m * d^2)

Substituting the given values:
d = 2.3 m (distance between the center of mass and the axis of rotation)
m = 1.6 kg (mass of the object)
I_cm = (2/5) * (1.6 kg) * (0.7 m)^2

So, the moment of inertia about an axis through the center of mass is:
I_axis = (2/5) * (1.6 kg) * (0.7 m)^2 + (1.6 kg) * (2.3 m)^2

Calculating this expression will give you the moment of inertia of the object about an axis through its center of mass.