
A solid cylinder and a hollow cylinder which are both of the same mass and same external diameter are released from a similar height at the same time on an inclined plane. Both roll down without slipping. Which one is reaching the bottom first?
A. solid cylinder
B. both together
C. one with higher density
D. hollow cylinder
Answer
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Hint: Acceleration of a mass is having relation with moment of inertia. As the moment of inertia increases, the acceleration for the mass or the object which is rolling will get decreased. It is due to the inverse relation between them.
Complete step-by-step answer:
In an inclined plane when a body is getting rolled down, the acceleration is also there which is on the basis of the moment of inertia of the object.
We know that,
As the moment of inertia of a body is higher, then the acceleration of that object will be less. This can be given by the formula,
$a=\dfrac{g\sin \theta }{1+\dfrac{I}{M{{R}^{2}}}}$
Where $a$ the acceleration of a body is, $g$ is the acceleration due to gravity, $\theta $ is the angle of inclination, $I$ is the moment of inertia of the object, $M$ is the mass of the object and $R$ is the radius of the body.
Now for an equal mass and equal radius,
As we know that,
${{I}_{solidcylinder}}<{{I}_{hollowcylinder}}$
Therefore we can write that,
${{a}_{solidcylinder}}>{{a}_{hollowcylinder}}$
Therefore the solid cylinder will be having higher acceleration. And hence the correct answer will be option A.
Note: As the angle is getting increased, the acceleration of the object is also increased. According to the increase of the angle, the component of force which is parallel to the incline will increase and the component of force perpendicular to the incline will decrease. It is the parallel component of the weight vector that is resulting in the acceleration.
Complete step-by-step answer:
In an inclined plane when a body is getting rolled down, the acceleration is also there which is on the basis of the moment of inertia of the object.
We know that,
As the moment of inertia of a body is higher, then the acceleration of that object will be less. This can be given by the formula,
$a=\dfrac{g\sin \theta }{1+\dfrac{I}{M{{R}^{2}}}}$
Where $a$ the acceleration of a body is, $g$ is the acceleration due to gravity, $\theta $ is the angle of inclination, $I$ is the moment of inertia of the object, $M$ is the mass of the object and $R$ is the radius of the body.
Now for an equal mass and equal radius,
As we know that,
${{I}_{solidcylinder}}<{{I}_{hollowcylinder}}$
Therefore we can write that,
${{a}_{solidcylinder}}>{{a}_{hollowcylinder}}$
Therefore the solid cylinder will be having higher acceleration. And hence the correct answer will be option A.
Note: As the angle is getting increased, the acceleration of the object is also increased. According to the increase of the angle, the component of force which is parallel to the incline will increase and the component of force perpendicular to the incline will decrease. It is the parallel component of the weight vector that is resulting in the acceleration.
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