a solid sphere, a solid cylinder, a spherical shell, and a hoop all have the same mass and radius. each are rolling on a horizontal surface with the same center of mass speed, and then they roll up identical inclines. which one goes the least distance up its incline?

In physics, the term "moment of inertia" refers to the quantitative measurement of a body's inertia with respect to rotation, or the resistance that a body shows when a torque is applied to change its speed of rotation around an axis (turning force).
The axis may or may not be fixed, and it may or may not be internal or external.
The formula for moment of inertia is =

I = L/ω

I = inertia

L = angular momentim

ω = angular velocity

In the question it is given that a solid sphere, solid cylinder, a spherical shell and a hoop all having the same mass(m) , radius(r) and linear center of mass speed(v) roll up identical inclines

Since all the bodies have same mass, radius and linear speed as such according to conservation of energy for every object,

Rotational Energy + Kinetic Energy = Potential Energy

Iω^2/2 + mv^2/2 = mgh

where, the symbols represent-

I = moment of inertia

ω = angular speed

v= linear speed

m = mass

h = height

We know the

I for solid sphere = 2mr^2/5

I for solid cylinder = mr^2/2

I for spherical shell = 2mr^2/3

I for hoop = mr^2/2

As per the given informations kinetic energy of all bodies is same therefore, on the value of I or the moment of inertia depends which body will cover the highest or least distance up its incline

I for solid sphere is minimum among all the given bodies and thus it will attain the least distance.

Hence, the solid sphere goes the least distance up its incline.

To learn more about moment of inertia from the given link

https://brainly.com/question/3406242

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