Question
A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
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Answer
Consider a circularly symmetric rigid body, like a sphere or a wheel or a disc, rolling with friction down a plane inclined at an angle $9$ to the horizontal. If the frictional force on the body is large enough, the body rolls without slipping.
Let M and R be the mass and radius of the body. Let I be the moment of inertia of the body for rotation about an axis through its centre. Let the body start from rest at the top of the incline at a
height h. Let v be the translational speed of the centre of mass at the bottom of the incline. Then, its kinetic energy at the bottom of the incline is
$E=\frac{1}{2} M v^2\left[1+\frac{I}{M R^2}\right]=\frac{1}{2} M v^2(1+\beta)$
where $\beta=\frac{I}{M R^2}$.
If $k$ is the radius of gyration of the body, $I=M k^2$ and $\beta=\frac{I}{M R^2}=\frac{k^2}{R^2}$
From conservation of energy,
$ ( KE + PE )_{\text {initial }}=( KE + PE )_{\text {final }}$
$\therefore 0+M g h=\frac{1}{2} M v^2(1+\beta)+0$
$ \therefore M g h=\frac{1}{2} M v^2(1+\beta)$
$\therefore v^2=\frac{2 g h}{1+\beta}$
$\therefore v=\sqrt{\frac{2 g h}{1+\beta}}=\sqrt{\frac{2 g h}{1+\left(k^2 / R^2\right)}} $
Since $h=L \sin \theta$,
$v=\sqrt{\frac{2 g L \sin \theta}{1+\left(k^2 / R^2\right)}}$
Let a be the acceleration of the centre of mass of the body along the inclined plane. Since the body starts from rest,
$L=\frac{1}{2} a t^2$
$\therefore t=\sqrt{\frac{2 L}{a}}=\sqrt{\frac{2 L}{g \sin \theta} \cdot\left(1+\frac{k^2}{R^2}\right)}$
[Note : For rolling without slipping, the contact point of the rigid body is instantaneously at rest relative to the surface of the inclined plane. Hence, the force of friction is static rather than kinetic, and does no work on the body. Thus, the force of static friction causes no decrease in the mechanical energy of the body and we can use the principle of conservation of energy.]
Let M and R be the mass and radius of the body. Let I be the moment of inertia of the body for rotation about an axis through its centre. Let the body start from rest at the top of the incline at a
height h. Let v be the translational speed of the centre of mass at the bottom of the incline. Then, its kinetic energy at the bottom of the incline is
$E=\frac{1}{2} M v^2\left[1+\frac{I}{M R^2}\right]=\frac{1}{2} M v^2(1+\beta)$
where $\beta=\frac{I}{M R^2}$.
If $k$ is the radius of gyration of the body, $I=M k^2$ and $\beta=\frac{I}{M R^2}=\frac{k^2}{R^2}$
From conservation of energy,
$ ( KE + PE )_{\text {initial }}=( KE + PE )_{\text {final }}$
$\therefore 0+M g h=\frac{1}{2} M v^2(1+\beta)+0$
$ \therefore M g h=\frac{1}{2} M v^2(1+\beta)$
$\therefore v^2=\frac{2 g h}{1+\beta}$
$\therefore v=\sqrt{\frac{2 g h}{1+\beta}}=\sqrt{\frac{2 g h}{1+\left(k^2 / R^2\right)}} $
Since $h=L \sin \theta$,
$v=\sqrt{\frac{2 g L \sin \theta}{1+\left(k^2 / R^2\right)}}$
Let a be the acceleration of the centre of mass of the body along the inclined plane. Since the body starts from rest,
$L=\frac{1}{2} a t^2$
$\therefore t=\sqrt{\frac{2 L}{a}}=\sqrt{\frac{2 L}{g \sin \theta} \cdot\left(1+\frac{k^2}{R^2}\right)}$
[Note : For rolling without slipping, the contact point of the rigid body is instantaneously at rest relative to the surface of the inclined plane. Hence, the force of friction is static rather than kinetic, and does no work on the body. Thus, the force of static friction causes no decrease in the mechanical energy of the body and we can use the principle of conservation of energy.]
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