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Q7.18(a):
A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case?

Solution :

Given Variables & Initial Setup

Consider a solid sphere of mass $M$ and radius $R$ rolling down an inclined plane of height $h$ and angle of inclination $\theta$. The sphere is subject to the Principle of Conservation of Mechanical Energy. [Per the Law of Conservation of Energy in a closed system].

  • Initial Kinetic Energy ($K_i$) = $0$ (at the top, assuming rest).
  • Initial Potential Energy ($U_i$) = $Mgh$.
  • Final Potential Energy ($U_f$) = $0$ (at the bottom).
  • Final Kinetic Energy ($K_f$) = Translational Kinetic Energy + Rotational Kinetic Energy.

Step 1: Establishing the Energy Equation

The total kinetic energy at the bottom is given by:

$K_f = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$

Where $I$ is the Moment of Inertia of a solid sphere about its center of mass, $I = \frac{2}{5}MR^2$. [Per the Parallel Axis Theorem and standard geometric moments].

Step 2: Substitution and Simplification

Since the sphere is rolling without slipping, the condition $v = R\omega$ must hold, where $\omega = \frac{v}{R}$. Substituting $I$ and $\omega$ into the energy equation:

$K_f = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v}{R}\right)^2$
$K_f = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$

Step 3: Applying Conservation of Energy

By equating initial potential energy to final kinetic energy:

$Mgh = \frac{7}{10}Mv^2$

Solving for velocity ($v$):

$v^2 = \frac{10gh}{7} \implies v = \sqrt{\frac{10gh}{7}}$

Step 4: Vector and Geometric Representation

Base $h$ (Fixed Height) Angle $\theta$

Step 5: Analysis of Variables

The expression for the final velocity $v = \sqrt{\frac{10gh}{7}}$ demonstrates that the velocity is dependent solely on:

Variable Dependency Status
Acceleration due to gravity ($g$) Constant
Height ($h$) Fixed
Angle of inclination ($\theta$) Absent in final expression

Because the angle of inclination $\theta$ does not appear in the final energy-derived equation for velocity, the speed at the bottom of the plane is independent of the angle of inclination, provided the vertical height $h$ remains constant.

Final Solution: Yes, the sphere will reach the bottom with the same speed in each case, as the final velocity $v = \sqrt{\frac{10gh}{7}}$ is independent of the angle of inclination $\theta$.


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