A solid sphere rolls without slipping down an inclined plane of angle $\theta$. What fraction of its total kinetic energy is rotational?

## CORE INFORMATION\n✓ **Answer**: [Option C]\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Kinetic Energy of Rolling Motion\n\t**Related Topics**: Rotational Kinetic Energy, Translational Kinetic Energy, Moment of Inertia\n\t**Prerequisites**: Understanding of rotational and translational kinetic energy, moment of inertia for a solid sphere.\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Solid sphere rolling without slipping\n\t* Inclined plane with angle $\theta$\n**Method & Steps**:\n\t1. Define total kinetic energy:\n\t\tReasoning: Total kinetic energy of a rolling object is the sum of its translational and rotational kinetic energies.\n\t\tFormula/Rule: $KE_{total} = KE_{trans} + KE_{rot}$\n\t2. Express translational kinetic energy:\n\t\tReasoning: Translational kinetic energy is given by $\frac{1}{2}mv^2$, where m is the mass and v is the linear velocity.\n\t\tFormula/Rule: $KE_{trans} = \frac{1}{2}mv^2$\n\t3. Express rotational kinetic energy:\n\t\tReasoning: Rotational kinetic energy is given by $\frac{1}{2}I\omega^2$, where I is the moment of inertia and $\omega$ is the angular velocity.\n\t\tFormula/Rule: $KE_{rot} = \frac{1}{2}I\omega^2$\n\t4. Relate linear and angular velocity:\n\t\tReasoning: For rolling without slipping, the linear velocity is related to the angular velocity by v = r$\omega$, where r is the radius of the sphere.\n\t\tFormula/Rule: $v = r\omega$\n\t5. Find the moment of inertia of a solid sphere:\n\t\tReasoning: The moment of inertia of a solid sphere about its center is $\frac{2}{5}mr^2$.\n\t\tFormula/Rule: $I = \frac{2}{5}mr^2$\n\t6. Substitute and calculate rotational kinetic energy:\n\t\tWorking: $KE_{rot} = \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{5}mv^2$\n\t\tKey Point: Substituting $\omega = v/r$ and $I = \frac{2}{5}mr^2$\n\t7. Calculate total kinetic energy:\n\t\tWorking: $KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$\n\t\tKey Point: Sum of the translational and rotational kinetic energy\n\t8. Find the fraction of rotational kinetic energy:\n\t\tWorking: $\frac{KE_{rot}}{KE_{total}} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$\n\t\tResult: Fraction of rotational kinetic energy is 2/7\n\t\tVerification: The result is unitless as it is a fraction.\n ## OPTION ANALYSIS\n[A] :x: Incorrect. This would be the fraction if the rotational and translational kinetic energies were equal\n[B] :x: Incorrect. This is the fraction of rotational KE divided by the translational KE.\n[C] ✓: Correct. The fraction of the total kinetic energy that is rotational is 2/7.\n[D] :x: Incorrect. This is not the correct fraction obtained from the calculation.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Forgetting to use the correct moment of inertia for a solid sphere. [How to avoid]: Remember the correct formula for the moment of inertia, $I = \frac{2}{5}mr^2$.\n\t\t2.[Error 2]: Not correctly relating linear and angular velocity. [How to avoid]: Always remember the relation $v = r\omega$ when dealing with rolling without slipping.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: The fraction of rotational kinetic energy for a rolling object depends only on the moment of inertia and not on the mass or velocity.\n\t\t2.[Memory aid]: The fraction can be generalized as $\frac{I}{I+mr^2}$ which will give $\frac{2/5}{2/5+1} = \frac{2}{7}$ for a solid sphere.

A solid sphere rolls without slipping down an inclined plane of inclination $\theta$. The acceleration of its center of mass is:

## CORE INFORMATION\n✓ **Answer**: Option B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Rotational and Translational Motion\n\t**Related Topics**: Rolling without slipping, Inclined plane, Newton's second law\n\t**Prerequisites**: Understanding of translational and rotational motion, moment of inertia, free body diagrams\n## SOLUTION PATHWAY\n**Given Information**:\n\t* A solid sphere rolling down an inclined plane\n\t* Inclination of the plane is $\theta$\n\t* Rolling without slipping\n**Method & Steps**:\n\t1. Identify the forces acting on the sphere:\n\t\tReasoning: The forces acting are gravity ($mg$) acting vertically downwards, normal force ($N$) perpendicular to the plane, and friction ($f$) acting along the plane opposite to the motion.\n\t\tFormula/Rule: None\n\t2. Resolve gravity into components along and perpendicular to the incline:\n\t\tWorking: Gravity component along the incline is $mg\sin\theta$ and perpendicular is $mg\cos\theta$.\n\t\tKey Point: Normal force balances the perpendicular component of gravity, i.e., $N = mg\cos\theta$.\n\t3. Apply Newton's second law for translational motion:\n\t\tReasoning: The net force along the incline causes linear acceleration ($a$) of the center of mass.\n\t\tFormula/Rule: $mg\sin\theta - f = ma$\n\t4. Apply Newton's second law for rotational motion:\n\t\tReasoning: The torque due to friction causes angular acceleration ($\alpha$).\n\t\tFormula/Rule: $\tau = I\alpha$, where $\tau = fR$ and $I = \frac{2}{5}mR^2$ for a solid sphere, and $\alpha = \frac{a}{R}$ for rolling without slipping.\n\t5. Substitute and solve for acceleration:\n\t\tWorking: $fR = \frac{2}{5}mR^2\frac{a}{R} \implies f = \frac{2}{5}ma$. Substitute $f$ into the translational equation: $mg\sin\theta - \frac{2}{5}ma = ma \implies mg\sin\theta = \frac{7}{5}ma \implies a = \frac{5}{7}g\sin\theta$\n\t\tKey Point: The friction is static friction and does not perform work here, it just changes the direction of the force.\n\t6. Final step:\n\t\tResult: The acceleration of the center of mass is $a = \frac{5}{7}g\sin\theta$.\n\t\tVerification: The derived acceleration is less than $g\sin\theta$, which is expected due to the rotational inertia.\n ## OPTION ANALYSIS\n[A] :x: This is the acceleration if the sphere were sliding down the incline without rolling. The moment of inertia reduces the linear acceleration.\n[B] ✓: This is the correct acceleration for a solid sphere rolling without slipping down the incline.\n[C] :x: This is the acceleration of a free-falling object along an incline (no rotation).\n[D] :x: This acceleration is greater than gravity's component along the incline and not physically possible.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1. [Error 1]: Forgetting to consider rotational motion: Solve it as a sliding block. This leads to an incorrect acceleration of $g\sin\theta$.\n\t\tHow to avoid: Remember to account for the rotational inertia of the rolling object.\n\t\t2. [Error 2]: Incorrect moment of inertia: Using the wrong moment of inertia for a solid sphere. Using the moment of inertia of a hoop, for example.\n\t\tHow to avoid: Double-check the moment of inertia formula based on the object's shape.\n\t**Quick Tips**:\n\t\t1. [Helpful shortcut]: The acceleration of a rolling object down an incline is always less than $g\sin\theta$.\n\t\t2. [Memory aid]: Remember the formula for a solid sphere's moment of inertia: $I = \frac{2}{5}mR^2$

The correct Option is "$\frac{5}{7}g\sin\theta$"

A solid cylinder of mass $M$ and radius $R$ rolls without slipping down a rough inclined plane of inclination $\theta$. The coefficient of static friction is $\mu_s$. What is the minimum value of $\mu_s$ to prevent slipping?

## CORE INFORMATION\n✓ **Answer**: [C]\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Rotational Dynamics and Friction\n\t**Related Topics**: Inclined Plane, Rolling Motion, Torque, Newton's Second Law\n\t**Prerequisites**: Understanding of forces, torque, moment of inertia, and conditions for rolling without slipping.\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Mass of cylinder = M\n\t* Radius of cylinder = R\n\t* Inclination angle = $\theta$\n\t* Coefficient of static friction = $\mu_s$\n\t* Rolling without slipping\n**Method & Steps**:\n\t1. Analyze the forces acting on the cylinder:\n\t\tReasoning: The forces acting on the cylinder are gravity (Mg), normal force (N), and static friction (f_s).\n\t\tFormula/Rule: Newton's Second Law: $\sum F = ma$\n\t2. Write the equations of motion:\n\t\tWorking: Along the incline: $Mg\sin\theta - f_s = Ma$ (1)\n\t\t\tPerpendicular to the incline: $N = Mg\cos\theta$\n\t\t\tTorque about the center of mass: $f_sR = I\alpha$ (2)\n\t\t\tFor a solid cylinder, $I = \frac{1}{2}MR^2$\n\t\tKey Point: For rolling without slipping, $a = R\alpha$ (3)\n\t3. Substitute the moment of inertia and relation of a & alpha into equation (2)\n\t\tWorking: $f_sR = \frac{1}{2}MR^2\frac{a}{R}$ => $f_s = \frac{1}{2}Ma$ (4)\n\t4. Substitute (4) into (1) to find acceleration:\n\t\tWorking: $Mg\sin\theta - \frac{1}{2}Ma = Ma$ => $Mg\sin\theta = \frac{3}{2}Ma$ => $a = \frac{2}{3}g\sin\theta$\n\t5. Substitute acceleration into (4) to find the friction force:\n\t\tWorking: $f_s = \frac{1}{2}M(\frac{2}{3}g\sin\theta)$ => $f_s = \frac{1}{3}Mg\sin\theta$\n\t6. The condition to prevent slipping is $f_s \le \mu_sN$. Since we are looking for the minimum value, we use $f_s = \mu_sN$\n\t\tWorking: $\frac{1}{3}Mg\sin\theta = \mu_s Mg\cos\theta$ => $\mu_s = \frac{1}{3}\frac{\sin\theta}{\cos\theta}$\n\t7. Final Answer\n\t\tResult: $\mu_s = \frac{1}{3}\tan\theta$\n\t\tVerification: Dimensional analysis confirms that $\mu_s$ is dimensionless.\n ## OPTION ANALYSIS\n[A] :x: This is the coefficient of friction when the object slides down the plane without rotation\n[B] :x: This value is incorrect based on the calculation\n[C] ✓: This is the minimum coefficient of static friction derived from the above steps\n[D] :x: This value is incorrect based on the calculation\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Forgetting the relationship between linear and angular acceleration (a = R\alpha) when considering rolling motion.\n\t\t2.[Error 2]: Incorrectly applying the moment of inertia for a solid cylinder.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: When rolling without slipping, the friction force does not dissipate energy, it is static friction.\n\t\t2.[Memory aid]: Remember the direction of the static friction is up the incline to provide the necessary torque for rotation.

The correct Option is "$\frac{tan\theta}{3}$"

NEET Questions

📑 Chapter:Systems Of Particles And Rotational Motion

🎯Topic:Rolling Motion

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :2 mins

🔄Ncert Concepts :Rolling Motion, Acceleration, Newton's Laws

10.

A ring of mass $m$ and radius $r$ rolls down an inclined plane of angle $heta$ without slipping. A block of the same mass slides down the same incline. The ratio of the velocities of the ring and the block when they reach the bottom is: