Question

A solid sphere is under pure rolling on a rough fixed incline plane of angle $\theta$. Choose the correct options if only contact forces and gravity are acting:

• A. Frictional force will be down the incline if sphere rolls up the incline and it will be up the incline if sphere rolls down the incline.
• B. Frictional force will be down the incline whether spheres rolls up the incline or down the incline
• C. Frictional force and acceleration of the body will increase with the increase in angle of incline plane
• D. Velocity and acceleration of the point of contact of sphere with incline will be zero during the motion.

Answer 1

Answer: (A), (C)

A, C

Answer 2

Analysis of the motion:

Let's denote the acceleration of the center of mass as $a_{cm}$ and the angular acceleration as $\alpha$. The radius of the sphere is $R$ and its mass is $M$. The moment of inertia of a solid sphere about its center of mass is $I = \frac{2}{5}MR^2$.

Forces:

1. Gravitational force: $Mg$ acting vertically downwards. Its component along the incline is $Mg \sin\theta$ (down the incline) and perpendicular to the incline is $Mg \cos\theta$.
2. Normal force: $N$ acting perpendicular to the incline, balancing $Mg \cos\theta$. So, $N = Mg \cos\theta$.
3. Frictional force: $f$ acting parallel to the incline. Its direction depends on the tendency of slipping.

Equations of motion:

1. Translational motion along the incline: $F_{net,x} = Ma_{cm}$
2. Rotational motion about the center of mass: $\tau_{net} = I\alpha$. The only force providing torque about the center of mass is friction, $fR$.
3. Pure rolling condition: For pure rolling (no slipping), the velocity of the point of contact with the incline is zero. This implies $v_{cm} = R\omega$ (where $v_{cm}$ is the speed of the center of mass and $\omega$ is the angular speed). Differentiating this, we get $a_{cm} = R\alpha$.

Case 1: Sphere rolls down the incline.

The sphere's center of mass moves down the incline ($a_{cm}$ is positive downwards). The angular velocity $\omega$ is clockwise. The tendency of the point of contact is to slip down the incline. To prevent this, the frictional force $f$ must act up the incline.

1. Translational: $Mg \sin\theta - f = Ma_{cm}$ (Equation 1)
2. Rotational: $fR = I\alpha = \frac{2}{5}MR^2 \alpha$ (Clockwise torque is positive)
3. Pure rolling: $a_{cm} = R\alpha \implies \alpha = a_{cm}/R$

Substitute $\alpha$ into the rotational equation: $fR = \frac{2}{5}MR^2 \frac{a_{cm}}{R} \implies f = \frac{2}{5}Ma_{cm}$

Substitute $f$ into the translational equation: $Mg \sin\theta - \frac{2}{5}Ma_{cm} = Ma_{cm}$ $Mg \sin\theta = \frac{7}{5}Ma_{cm}$ $a_{cm} = \frac{5}{7}g \sin\theta$ (down the incline)

Now find the frictional force: $f = \frac{2}{5}M \left(\frac{5}{7}g \sin\theta\right) = \frac{2}{7}Mg \sin\theta$ (up the incline)

Case 2: Sphere rolls up the incline.

The sphere's center of mass moves up the incline, but its acceleration $a_{cm}$ will be down the incline (decelerating). The angular velocity $\omega$ is counter-clockwise. The gravitational component $Mg \sin\theta$ acts down the incline. This force tries to make the sphere slip down. To prevent this, the frictional force $f$ must act down the incline.

Let's take the positive x-direction as up the incline.

1. Translational: $-Mg \sin\theta - f = Ma_{cm}$ (Equation 2) (Here $a_{cm}$ will be negative, indicating deceleration)
2. Rotational: $fR = I\alpha = \frac{2}{5}MR^2 \alpha$ (Counter-clockwise torque is positive)
3. Pure rolling: $a_{cm} = R\alpha \implies \alpha = a_{cm}/R$

Substitute $\alpha$ into the rotational equation: $fR = \frac{2}{5}MR^2 \frac{a_{cm}}{R} \implies f = \frac{2}{5}Ma_{cm}$

Substitute $f$ into the translational equation: $-Mg \sin\theta - \frac{2}{5}Ma_{cm} = Ma_{cm}$ $-Mg \sin\theta = \frac{7}{5}Ma_{cm}$ $a_{cm} = -\frac{5}{7}g \sin\theta$ (down the incline, magnitude $\frac{5}{7}g \sin\theta$)

Now find the frictional force: $f = \frac{2}{5}M \left(-\frac{5}{7}g \sin\theta\right) = -\frac{2}{7}Mg \sin\theta$. The negative sign means friction acts opposite to the assumed positive x-direction (up the incline), so friction acts down the incline.

Summary of friction directions:

• Rolling down: Friction is up the incline. ($f = \frac{2}{7}Mg \sin\theta$)
• Rolling up: Friction is down the incline. ($f = \frac{2}{7}Mg \sin\theta$)

Option (A) Frictional force will be down the incline if sphere rolls up the incline and it will be up the incline if sphere rolls down the incline.

• Rolling up: Friction is down the incline. (Correct)
• Rolling down: Friction is up the incline. (Correct) So, option (A) is correct.

Option (B) Frictional force will be down the incline whether spheres rolls up the incline or down the incline.

• Incorrect, as friction is up the incline when rolling down.

Option (C) Frictional force and acceleration of the body will increase with the increase in angle of incline plane.

• Rolling down: $a_{cm} = \frac{5}{7}g \sin\theta$, $f = \frac{2}{7}Mg \sin\theta$. As $\theta$ increases (from $0$ to $90^\circ$), $\sin\theta$ increases, so both $a_{cm}$ and $f$ increase.
• Rolling up: $|a_{cm}| = \frac{5}{3}g \sin\theta$, $|f| = \frac{2}{3}Mg \sin\theta$. As $\theta$ increases, both $|a_{cm}|$ and $|f|$ increase. So, option (C) is correct.

Option (D) Velocity and acceleration of the point of contact of sphere with incline will be zero during the motion.

• For pure rolling, the velocity of the point of contact is indeed zero.
• However, the acceleration of the point of contact is not zero.

Therefore, option (D) is incorrect.

Both (A) and (C) are correct options.