The moment of inertia of a solid cylinder about its axis is given by 0.5MR2. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

Given that moment of inertia is

I=0.5MR²

K. E=?

We know that

w=v/R

Also

Translational K. Et is 1/2Mv²

And rotational K. Er is Iw²/2

Since w=v/R

Then K. Er=Iv²/2R²

Also I=0.5MR²

K. Er=0.5MR²v²/2R²

K. Er=Mv²/4

Then the ration of rotational Kinetic energy to transitional Kinetic energy is given as

K. Er/K. Et

Mv²/4 : Mv²/2

Mv²/4 * 2/Mv²

Then the ratio is 1/2

The ratio of the rotational kinetic energy to the translational kinetic energy is 1/2 or 0.5

Ratio of rotational kinetic energy to translational kinetic energy is 1/2 or 0.5

Explanation:

Rotational kinetic energy Er = (1/2) (Iω^ (2))

Where;

ω is the angular velocity and I is the moment of inertia around the axis of rotation

For a solid cylinder moment of Inertia (I) = 0.5 (mr^ (2)) or (mr^ (2)) / 2

Also, angular velocity (ω) = v/r

Where v is the linear velocity and r is the radius of curvature.

Thus plugging these into the equation for Er, we get;

Er = (1/2) [ (mr^ (2)) / 2][ (v/r) ^ (2) ]

So, Er = (1/4) [mr^ (2) ][ (v/r) ^ (2) ] = (1/4) mv^ (2)

Now, formula for translational kinetic energy is Et = (1/2) mv^ (2)

Thus, ratio or Er to Et is;

[ (1/4) mv^ (2) ]/[ (1/2) mv^ (2) ] = 1/2 or 0.5