A spherically symmetric object, with radius R = 0.700 m and mass M = 1.90 kg, rolls without slipping across a horizontal floor, with velocity V = 1.200 m/s. It then rolls up an incline with an angle of inclination θ = 40.0 deg and comes to rest a distance d = 0.177 m up the incline, before reversing direction and rolling back down. Find the moment of inertia of this object about an axis through its center of mass.

I have no idea where to start

To find the moment of inertia of the object about an axis through its center of mass, you can use the principle of conservation of mechanical energy.

First, we need to calculate the initial kinetic energy when the object is rolling without slipping across the horizontal floor. The kinetic energy of a rolling object can be split into two components: translational KE (1/2 * M * V^2) and rotational KE (1/2 * I * w^2), where I is the moment of inertia about the axis through its center of mass and w is the angular velocity.

Since the object is rolling without slipping, the linear velocity V can be related to the angular velocity w through the equation V = R * w, where R is the radius of the object.

So, the initial kinetic energy can be written as:
KE_initial = Translational KE + Rotational KE
= 1/2 * M * V^2 + 1/2 * I * (V/R)^2
= 1/2 * M * V^2 + 1/2 * I * (V^2 / R^2)
= 1/2 * M * V^2 (1 + (V^2 / R^2))

Next, we need to calculate the potential energy gained by the object as it rolls up the incline. The potential energy gained is given by m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height gained.

Since the distance d traveled up the incline is perpendicular to the incline, the height gained h can be calculated as:
h = d * sin(θ)

Now, the final kinetic energy when the object comes to rest at the top of the incline is zero, as it is at rest. Therefore, the total mechanical energy is conserved throughout the motion, so we can equate the initial kinetic energy to the potential energy gained:

1/2 * M * V^2 (1 + (V^2 / R^2)) = m * g * h

Simplifying the equation and substituting the values given:
1/2 * 1.9 kg * (1.2 m/s)^2 (1 + ((1.2 m/s)^2 / (0.7 m)^2)) = 1.9 kg * 9.8 m/s^2 * 0.177 m * sin(40°)

Now, solve this equation for the moment of inertia I to find your answer.