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Q7.27:
Prove the result that the velocity $v$ of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height $h$ is given by $v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$ using dynamical consideration (i.e. by consideration of forces and torques). Note $k$ is the radius of gyration of the body about its symmetry axis, and $R$ is the radius of the body. The body starts from rest at the top of the plane.

Solution :

Step 1: Establishing the Mechanical System

Consider a rigid body of mass $M$, radius $R$, and radius of gyration $k$ about its central axis, rolling down an inclined plane of inclination angle $\theta$. The body is released from rest at a height $h$. The forces acting on the body are:

  • Gravity ($Mg$) acting vertically downward.
  • Normal reaction ($N$) perpendicular to the inclined plane.
  • Friction ($f$) acting upward along the inclined plane.

Step 2: Equations of Motion (Translational and Rotational)

Applying Newton's Second Law for translational motion along the incline [Per Newton's Second Law, $F_{net} = Ma$]:

$Mg \sin\theta - f = Ma$ --- (Equation 1)

Applying the rotational analog of Newton's Second Law about the center of mass [Per $\tau = I\alpha$]:

$fR = I\alpha$ --- (Equation 2)

Since the body rolls without slipping, the linear acceleration $a$ and angular acceleration $\alpha$ are related by the condition $a = R\alpha$. Substituting $I = Mk^2$ and $\alpha = a/R$ into Equation 2:

$fR = (Mk^2)\left(\frac{a}{R}\right) \implies f = \frac{Mk^2a}{R^2}$ --- (Equation 3)

R f Inclined Plane (Angle θ)

Step 3: Deriving Linear Acceleration

Substitute the expression for $f$ from Equation 3 into Equation 1:

$Mg \sin\theta - \frac{Mk^2a}{R^2} = Ma$

$g \sin\theta = a\left(1 + \frac{k^2}{R^2}\right)$

$a = \frac{g \sin\theta}{1 + k^2/R^2}$ --- (Equation 4)

Step 4: Kinematic Solution for Velocity

The distance $d$ traveled along the incline is related to the height $h$ by $d = \frac{h}{\sin\theta}$. Using the kinematic equation of motion starting from rest ($u=0$):

$v^2 = u^2 + 2ad \implies v^2 = 2 \left( \frac{g \sin\theta}{1 + k^2/R^2} \right) \left( \frac{h}{\sin\theta} \right)$

$v^2 = \frac{2gh}{1 + k^2/R^2}$

Step 5: Final Derivation

Taking the square root of both sides:

$v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$

Body Type $k^2/R^2$ Velocity $v$
Ring 1 $\sqrt{gh}$
Disc 1/2 $\sqrt{4gh/3}$
Solid Sphere 2/5 $\sqrt{10gh/7}$

Final Solution: The translational velocity of a rolling body at the bottom of an incline of height $h$ is mathematically proven to be $v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$.


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