What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a final velocity of 6.00 m/s? Express the moment of inertia as a multiple of MR2, where M is the mass of the object and R is its radius.

PE=KE=KE(transl) + KE(rot).

mgh = mv²/2 + I ω²/2=
= mv²/2 + I v²/2R².

I= (2R²/v²)(mgh- mv²/2)=
=mR²(2gh/v² -1) =
= mR²(2•9.8•2/36 -1)=
=mR²(1.09-1)=0.09 mR²

To find the moment of inertia of the object, we will use the principle of conservation of energy.

The initial potential energy of the object when it is at the top of the incline is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Similarly, the final kinetic energy of the object when it reaches the bottom of the incline is given by (1/2)mv^2, where v is the final velocity.

Since the object rolls without slipping, the final velocity of the object can be related to its radius, R, and angular velocity, ω, through the equation v = ωR. We can also relate the angular velocity to the moment of inertia, I, of the object using I = MR^2.

Therefore, we can equate the initial potential energy to the final kinetic energy to solve for the moment of inertia:

mgh = (1/2)mv^2

Simplifying and canceling out the mass:

gh = (1/2)v^2

Using the relationship between linear and angular velocity, v = ωR, we can rewrite the equation:

gh = (1/2)(ωR)^2

Simplifying:

gh = (1/2)ω^2R^2

Rearranging the equation to solve for the moment of inertia, I:

I = 2(gh)/ω^2R^2

Since we are asked to express the moment of inertia as a multiple of MR^2, we can replace ω^2R^2 with v^2, which is (ωR)^2:

I = 2(gh)/v^2

Plugging in the given values:

I = 2(9.8 m/s^2)(2.00 m)/(6.00 m/s)^2

Simplifying:

I = 0.653 MR^2

Therefore, the moment of inertia of the object rolling without slipping down the incline is 0.653 times MR^2.

To find the moment of inertia of the object, we first need to determine its rotational kinetic energy. The rotational kinetic energy (KE_rot) can be related to the linear kinetic energy (KE) by the equation:

KE_rot = (1/2) * I * ω^2

Where I is the moment of inertia and ω is the angular velocity.

In this case, since the object rolls without slipping, the linear kinetic energy can be expressed as:

KE = (1/2) * m * v^2

Where m is the mass of the object and v is its linear velocity.

The linear velocity (v) of the object can be related to its angular velocity (ω) by the equation:

v = ω * R

Where R is the radius of the object.

Now, let's calculate the linear velocity (v) of the object. We can use the law of conservation of energy:

Initial potential energy (PE_i) = Final kinetic energy (KE_f)

The initial potential energy can be expressed as:

PE_i = m * g * h

Where h is the height of the incline and g is the acceleration due to gravity.

The final kinetic energy (KE_f) can be expressed as:

KE_f = (1/2) * m * v^2

Setting these two equations equal to each other, we get:

m * g * h = (1/2) * m * v^2

Canceling out the mass (m), we get:

g * h = (1/2) * v^2

Now, we can solve for the linear velocity (v):

v = sqrt(2 * g * h)

Substituting this value of v into the equation relating v and ω, we get:

ω = v / R = sqrt(2 * g * h) / R

Finally, substituting the value of ω into the equation for rotational kinetic energy, we get:

KE_rot = (1/2) * I * (sqrt(2 * g * h) / R)^2

Simplifying this equation, we get:

KE_rot = (1/2) * I * (2 * g * h) / R^2

Cancelling out the factor of 2 and rearranging the terms, we can express the moment of inertia (I) as:

I = (M * R^2 * KE_rot) / (M * g * h)

Where M is the mass of the object, R is its radius, KE_rot is the rotational kinetic energy, g is the acceleration due to gravity, and h is the height of the incline.

Given that the final velocity (v) of the object is 6.00 m/s, we can use the equation for linear kinetic energy to find KE:

KE = (1/2) * m * v^2

Rearranging this equation, we can find the mass (m):

m = 2 * KE / v^2

Substituting the values of KE = (1/2) * m * v^2 and v = 6.00 m/s into this equation, we get:

m = 2 * (1/2) * m * (6.00 m/s)^2

Simplifying this equation, we can solve for the mass (m):

m = (2 * (1/2) * m * (6.00 m/s)^2) / 6.00 m/s^2

Simplifying further, we find:

m = (2 * 36) / 6

m = 12 kg

Now, substituting the values of M = 12 kg, R = radius (given), g = 9.8 m/s^2, and h = 2.00 m into the equation for moment of inertia (I), we can find the answer:

I = (M * R^2 * KE_rot) / (M * g * h)

I = (12 kg * R^2 * KE_rot) / (12 kg * 9.8 m/s^2 * 2.00 m)

Simplifying this equation, we find:

I = (R^2 * KE_rot) / (19.6)

Therefore, the moment of inertia (I) can be expressed as a multiple of MR^2.

I = (1/19.6) * (R^2 * KE_rot)

The answer's actually 12.