A circular disc is rolling on a horizontal surface. Its total kinetic energy is $600\,J$. Its rotational kinetic energy is:
A. 200 J
B. 400 J
C. 300 J
D. 100 J
Answer
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Hint:Kinetic energy is an energy which a body possesses by virtue of being in motion. Moving object or particle depends not only on motion but also on its mass. Kinetic energy is a property of a moving object or a particle depends not only on its motion but also on its mass. Kinetic energy defined as the work is needed to be accelerating a body of a given mass.
Complete step by step answer:
Assuming the kinetic energy of rotational and kinetic energy of translation in total kinetic energy. That means, $K.{E_{total}} = K.{E_{rotati}} + K.{E_{trans}}$. Here kinetic energy in rotational is the center of mass into moment of inertia and angular velocity.
That is $\dfrac{1}{2}{I_c}{\omega ^2}$
The kinetic energy in translation is mass of object and center of velocity,
That is $\dfrac{1}{2}m{V_c}^2$
The total kinetic energy is,
$K.{E_{total}} = \dfrac{1}{2}I{\omega ^2} + \dfrac{1}{2}m{V^2}$
Here the total kinetic energy is 600 joules, and moment of inertia is $\dfrac{{m{R^2}}}{4}$and angular velocity $\omega = \dfrac{V}{R}$ by substituting all the values in kinetic total energy we get,
$600 = \dfrac{1}{2}m{V^2} + \dfrac{1}{2}\dfrac{{m{R^2}}}{2} \times \dfrac{{{V^2}}}{{{R^2}}} \\
\Rightarrow 600 = \dfrac{1}{2}m{V^2} + \dfrac{{m{V^2}}}{4} \\
\Rightarrow 600 = \dfrac{{3m{V^2}}}{4} \\
\Rightarrow m{V^2} = 600 \times \dfrac{4}{3} \times \dfrac{1}{2} \\
\Rightarrow m{V^2} = 400J \\ $
The total kinetic energy is the sum of rotational kinetic energy and translational kinetic energy. Thus the translational kinetic energy is 400 Joules. Then rotational kinetic energy is the difference if total kinetic and translational kinetic energy,
That is,
$K.{E_{rotati}} = K.{E_{total}} - K.{E_{trans}} \\
\Rightarrow K.{E_{rotati}} = 600 - 400 \\
\therefore K.{E_{rotati}} = 200\,J \\ $
Hence, the rotational kinetic energy is 200 joules and the correct option is A.
Note:An object is the energy that it possesses due to its motion; it is defined as the work needed to accelerate the body of a mass. The Circular disc on a horizontal surface and its rotational kinetic energy is about 200 joules. Kinetic energy S.I unit is in joules.
Complete step by step answer:
Assuming the kinetic energy of rotational and kinetic energy of translation in total kinetic energy. That means, $K.{E_{total}} = K.{E_{rotati}} + K.{E_{trans}}$. Here kinetic energy in rotational is the center of mass into moment of inertia and angular velocity.
That is $\dfrac{1}{2}{I_c}{\omega ^2}$
The kinetic energy in translation is mass of object and center of velocity,
That is $\dfrac{1}{2}m{V_c}^2$
The total kinetic energy is,
$K.{E_{total}} = \dfrac{1}{2}I{\omega ^2} + \dfrac{1}{2}m{V^2}$
Here the total kinetic energy is 600 joules, and moment of inertia is $\dfrac{{m{R^2}}}{4}$and angular velocity $\omega = \dfrac{V}{R}$ by substituting all the values in kinetic total energy we get,
$600 = \dfrac{1}{2}m{V^2} + \dfrac{1}{2}\dfrac{{m{R^2}}}{2} \times \dfrac{{{V^2}}}{{{R^2}}} \\
\Rightarrow 600 = \dfrac{1}{2}m{V^2} + \dfrac{{m{V^2}}}{4} \\
\Rightarrow 600 = \dfrac{{3m{V^2}}}{4} \\
\Rightarrow m{V^2} = 600 \times \dfrac{4}{3} \times \dfrac{1}{2} \\
\Rightarrow m{V^2} = 400J \\ $
The total kinetic energy is the sum of rotational kinetic energy and translational kinetic energy. Thus the translational kinetic energy is 400 Joules. Then rotational kinetic energy is the difference if total kinetic and translational kinetic energy,
That is,
$K.{E_{rotati}} = K.{E_{total}} - K.{E_{trans}} \\
\Rightarrow K.{E_{rotati}} = 600 - 400 \\
\therefore K.{E_{rotati}} = 200\,J \\ $
Hence, the rotational kinetic energy is 200 joules and the correct option is A.
Note:An object is the energy that it possesses due to its motion; it is defined as the work needed to accelerate the body of a mass. The Circular disc on a horizontal surface and its rotational kinetic energy is about 200 joules. Kinetic energy S.I unit is in joules.
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