Question

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

Answer 1

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

Step-by-step 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

Answer 2

Step-by-step explanation:

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