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Question: A hoop rolls down an inclined plane. The fraction of its total kinetic energy that is associated rot...

A hoop rolls down an inclined plane. The fraction of its total kinetic energy that is associated rotational motion is:
A .1:2
B .1:3
C .1:4
D. 2:3

Explanation

Solution

The total kinetic energy is the sum of translational kinetic energy and rotational energy. The translational kinetic energy is the energy required to displace the object and rotational kinetic energy is the energy required to rotate the object.

Formula used:
The total kinetic energy of the object is given as:
$$$$K.E(total)= (1/2)Iω2+(1/2)MV2K.E\left( total \right)=\text{ }\left( 1/2 \right)I\omega {}^\text{2}+\left( 1/2 \right)MV{}^\text{2}
In the above formula, the first term corresponds to the rotational kinetic energy and other is translational kinetic energy.

Complete step-by-step answer:
We know that, the Translational motion is the motion when an object is displaced from one place to another, and Rotational motion is the motion in which objects spin(or rotate) about the certain axis or revolve around the fixed axis.
Therefore, the total kinetic energy is the sum of both energies: K.E(total)= (1/2)Iω2+(1/2)MV2K.E\left( total \right)=\text{ }\left( 1/2 \right)I\omega {}^\text{2}+\left( 1/2 \right)MV{}^\text{2}
Also, the moment of inertia about the fixes axis through its center of rotation is given as,
I=(1/2)MR2I=(1/2)MR{}^\text{2}
Hence, the total kinetic energy equation gets changed and is given by:K.E(total)= (1/4)MR2ω2+(1/2)MV2K.E\left( total \right)=\text{ }\left( 1/4 \right)MR{}^\text{2}\omega {}^\text{2}+\left( 1/2 \right)MV{}^\text{2}
From the above formula, the rotational kinetic energy is given by:
K.E(rot.)=(1/4)MR2ω2K.E(rot.)=(1/4)M{{R}^{2}}{{\omega }^{2}}
Also, the translational kinetic energy is given as:
K.E(trans.)=(1/2)MV2K.E(trans.)=\left( 1/2 \right)MV{}^\text{2}
We have to find the ratio between the above two kinetic energies,
K.E.(rot)/K.E=(1/4)MR2ω2/(1/2)MV2 K.E.(rot)/K.E=(1/4)MR2ω2/(1/2)M(Rω)2 K.E.(rot)/K.E=1/2 \begin{aligned} & K.E.(rot)/K.E=(1/4)M{{R}^{2}}{{\omega }^{2}}/(1/2)M{{V}^{2}} \\\ & K.E.(rot)/K.E=(1/4)M{{R}^{2}}{{\omega }^{2}}/(1/2)M{{(R\omega )}^{2}} \\\ & K.E.(rot)/K.E=1/2 \\\ \end{aligned}

Therefore, the fraction of total kinetic energy associated with rotational motion is in the ratio 1:2.Hence, option A is the right answer.

So, the correct answer is “Option A”.

Note: The total kinetic energy consists of both translational and rotational kinetic energies. The moment of inertia used is the moment of inertia about the center of rotation. Also, the kinetic energy of the object depends on the mass and speed of the body.