Answer:
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
Answer:
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
