
A disc at rest at the top of an inclined plane of height "h' rolls down without slipping and acquires a velocity " on reaching the bottom. If the same disc slides down a smooth incline and acquires the h/3 same velocity on reaching the bottom the height of smooth incline is:
A. \[\dfrac{h}{3}\]
B. \[\dfrac{h}{2}\]
C. \[\dfrac{{2h}}{3}\]
D. \[h\]
Answer
555.3k+ views
Hint: First memorize the law of energy conservation and then use the formula of Potential energy to make an equation and then compare Potential Energy and Rotational K.E , then substitute the equation value to get the height.
Formula used:
Conservation of energy:
Potential Energy = Rotational K.E + Translation K.E
\[mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}(\dfrac{{m{r^2}}}{2})\dfrac{{{v^2}}}{{{r^2}}}\]
Where, $m$ is the mass of the disc, $h$ is the height of the inclined, $v$ is the velocity of the disc, $r$ is the radius of the disc.
Complete step by step answer:
According to the energy Conservation, we have
Potential Energy = Rotational K.E + Translation K.E
\[mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}(\dfrac{{m{r^2}}}{2})\dfrac{{{v^2}}}{{{r^2}}}\]
\[\Rightarrow mgh = \dfrac{1}{2}m{v^2} + \dfrac{{m{v^2}}}{4}\]
\[\Rightarrow mgh = \dfrac{{3m{v^2}}}{4}\]
\[\Rightarrow {v^2} = \dfrac{{4gh}}{3}\] - 1st Equation
Here we have the value of , Substitute this value as shown below to get the height.Now, when the disk moves with smooth inclined. In Smooth inclined Rotational K.E is zero.
\[mgh' = \dfrac{1}{2}m{v^2}\]
\[\Rightarrow gh' = \dfrac{1}{2} \times \dfrac{{4gh}}{3}\] [ by 1st Equation]
\[\therefore h' = \dfrac{{2h}}{3}\]
Thus the height is \[\dfrac{{2h}}{3}\].
Hence,option C is the correct answer.
Additional Information:
Rotational Energy - Rotational energy or angular kinetic energy is kinetic energy due to the rotational motion of an object and you can say it is a part of its total kinetic energy.
Translational Kinetic Energy - Translational kinetic energy of a body is equal to one-half the product of its mass, m, and the square of its velocity.
Note: Potential Energy is defined as the energy stored in an object. It can be further divided into various types named Elastic Potential energy, Gravitational potential energy, Electric Potential Energy, etc. The potential energy of an object at height h is transformed into kinetic energy when it falls down. That is why the magnitude of kinetic energy just before reaching the ground will be equal to the potential energy at height h i.e. mgh.
Formula used:
Conservation of energy:
Potential Energy = Rotational K.E + Translation K.E
\[mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}(\dfrac{{m{r^2}}}{2})\dfrac{{{v^2}}}{{{r^2}}}\]
Where, $m$ is the mass of the disc, $h$ is the height of the inclined, $v$ is the velocity of the disc, $r$ is the radius of the disc.
Complete step by step answer:
According to the energy Conservation, we have
Potential Energy = Rotational K.E + Translation K.E
\[mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}(\dfrac{{m{r^2}}}{2})\dfrac{{{v^2}}}{{{r^2}}}\]
\[\Rightarrow mgh = \dfrac{1}{2}m{v^2} + \dfrac{{m{v^2}}}{4}\]
\[\Rightarrow mgh = \dfrac{{3m{v^2}}}{4}\]
\[\Rightarrow {v^2} = \dfrac{{4gh}}{3}\] - 1st Equation
Here we have the value of , Substitute this value as shown below to get the height.Now, when the disk moves with smooth inclined. In Smooth inclined Rotational K.E is zero.
\[mgh' = \dfrac{1}{2}m{v^2}\]
\[\Rightarrow gh' = \dfrac{1}{2} \times \dfrac{{4gh}}{3}\] [ by 1st Equation]
\[\therefore h' = \dfrac{{2h}}{3}\]
Thus the height is \[\dfrac{{2h}}{3}\].
Hence,option C is the correct answer.
Additional Information:
Rotational Energy - Rotational energy or angular kinetic energy is kinetic energy due to the rotational motion of an object and you can say it is a part of its total kinetic energy.
Translational Kinetic Energy - Translational kinetic energy of a body is equal to one-half the product of its mass, m, and the square of its velocity.
Note: Potential Energy is defined as the energy stored in an object. It can be further divided into various types named Elastic Potential energy, Gravitational potential energy, Electric Potential Energy, etc. The potential energy of an object at height h is transformed into kinetic energy when it falls down. That is why the magnitude of kinetic energy just before reaching the ground will be equal to the potential energy at height h i.e. mgh.
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