Answer:
mR²/2
Explanation:
Here is the complete question
An object of radius′
R′ and mass ′
M′ is rolling horizontally without slipping with speed ′
V′
. It then rolls up the hill to a maximum height h = 3v²/4g. The moment of inertia of the object is (g= acceleration due to gravity)
Solution
Since it rolls without slipping, there is no friction. So, its initial mechanical energy at the horizontal surface equals its final mechanical energy at the top of the hill.
Since the object is rolling initially, and on horizontal ground, it initial energy is kinetic and made up of rotational and translational kinetic energy.
So, E = K + K'
E = 1/2mv² + 1/2Iω² where m = mass of object, v = speed of object, I = moment of inertia of object and ω = angular speed of object = v/r where v = speed of object and R = radius of object.
Also, the final mechanical energy of the object, E' is its potential energy at the top of the hill. So, E' = mgh.
Since E = E',
1/2mv² + 1/2Iω² = mgh
substituting the values of ω and h into the equation, we have
1/2mv² + 1/2Iω² = mgh
1/2mv² + 1/2I(v/R)²= mg(3v²/4g)
Expanding the brackets, we have
1/2mv² + 1/2Iv²/R²= 3mv²/4
Dividing through by v², we have
1/2m + I/2R²= 3m/4
Subtracting m/2 from both sides, we have
I/2R² = 3m/4 - m/2
Simplifying, we have
I/2R² = m/4
Multiplying through by 2R², we have
I = m/4 × 2R²
I = mR²/2
