Question

solve physics question 3, JEE Advanced 2025

Answer 1

$0 \le F \le Mg\left(\frac{3\sqrt{3}-2}{4}\right)$

Answer 2

The problem asks for the range of values of an applied force $F$ for which a uniform solid cylinder rolls without slipping on a rough inclined plane.

1. Define Coordinate System and Forces: Let the inclined plane make an angle $\theta$ with the horizontal. Let the positive x-axis be up the incline and the positive y-axis be perpendicular to the incline, pointing outwards. Let the positive angular direction be clockwise (as seen from the side, which corresponds to rolling up the incline).

Forces acting on the cylinder:

• Gravitational force: $Mg$ acting vertically downwards.
  • Component parallel to incline: $Mg \sin\theta$ (down the incline)
  • Component perpendicular to incline: $Mg \cos\theta$ (into the incline)
• Normal force: $N$ acting perpendicular to the incline, upwards.
• Applied force: $F$ acting parallel to the incline, upwards.
• Friction force: $f$ acting parallel to the incline. Its direction depends on the tendency of slipping.

2. Equations of Motion:

• Translational Motion (along the incline): Let $a_{CM}$ be the acceleration of the center of mass up the incline. $F_{net, x} = Ma_{CM}$ $F - Mg \sin\theta + f = Ma_{CM}$ (Equation 1) (Here we assume friction $f$ acts up the incline. We will verify this assumption later.)

• Translational Motion (perpendicular to the incline): $F_{net, y} = 0$ (no acceleration perpendicular to the incline) $N - Mg \cos\theta = 0 \implies N = Mg \cos\theta$ (Equation 2)

• Rotational Motion (about the center of mass): Let $\alpha$ be the angular acceleration. The moment of inertia of a solid cylinder about its axis is $I = \frac{1}{2}MR^2$. Torques about CM:

  • Force $F$ creates a clockwise torque: $\tau_F = F \times R$
  • Friction force $f$ creates a counter-clockwise torque: $\tau_f = f \times R$ $\tau_{net} = I\alpha$ $FR - fR = I\alpha$ (Equation 3)

3. Rolling Without Slipping Condition: For rolling without slipping, the point of contact has zero acceleration. This implies a relationship between linear and angular acceleration: $a_{CM} = R\alpha$ (Equation 4)

4. Solve for $a_{CM}$ and $f$: Substitute $I = \frac{1}{2}MR^2$ and $\alpha = a_{CM}/R$ into Equation 3: $FR - fR = \frac{1}{2}MR^2 \left(\frac{a_{CM}}{R}\right)$ $FR - fR = \frac{1}{2}MRa_{CM}$ Divide by $R$: $F - f = \frac{1}{2}Ma_{CM}$ (Equation 3')

Now we have a system of two equations (Equation 1 and Equation 3') with two unknowns ($a_{CM}$ and $f$):

1. $F - Mg \sin\theta + f = Ma_{CM}$ 3'. $F - f = \frac{1}{2}Ma_{CM}$

Add (1) and (3'): $(F - Mg \sin\theta + f) + (F - f) = Ma_{CM} + \frac{1}{2}Ma_{CM}$ $2F - Mg \sin\theta = \frac{3}{2}Ma_{CM}$ $a_{CM} = \frac{2}{3M}(2F - Mg \sin\theta)$ (Equation A)

Now, substitute $a_{CM}$ back into Equation 3' to find $f$: $f = F - \frac{1}{2}Ma_{CM}$ $f = F - \frac{1}{2}M \left(\frac{2}{3M}(2F - Mg \sin\theta)\right)$ $f = F - \frac{1}{3}(2F - Mg \sin\theta)$ $f = F - \frac{2}{3}F + \frac{1}{3}Mg \sin\theta$ $f = \frac{1}{3}F + \frac{1}{3}Mg \sin\theta$ (Equation B)

5. Check Friction Direction and Condition for No Slipping: From Equation B, since $F \ge 0$ (pulling force upwards) and $Mg\sin\theta > 0$, the friction force $f$ will always be positive. This confirms our initial assumption that friction acts up the incline. This also means that for rolling without slipping, the tendency of the cylinder is to slip downwards relative to the contact point.

For rolling without slipping, the static friction condition must be satisfied: $|f| \le \mu_s N$. Using $N = Mg \cos\theta$ from Equation 2: $\frac{1}{3}F + \frac{1}{3}Mg \sin\theta \le \mu_s Mg \cos\theta$ Multiply by 3: $F + Mg \sin\theta \le 3\mu_s Mg \cos\theta$ $F \le 3\mu_s Mg \cos\theta - Mg \sin\theta$ $F_{max} = Mg(3\mu_s \cos\theta - \sin\theta)$

Now, substitute the given values: $\theta = 30^\circ$, $\mu_s = 0.5$. $\sin 30^\circ = \frac{1}{2}$ $\cos 30^\circ = \frac{\sqrt{3}}{2}$

$F_{max} = Mg \left(3 \times 0.5 \times \frac{\sqrt{3}}{2} - \frac{1}{2}\right)$ $F_{max} = Mg \left(\frac{3\sqrt{3}}{4} - \frac{1}{2}\right)$ $F_{max} = Mg \left(\frac{3\sqrt{3} - 2}{4}\right)$

This is the upper limit for $F$. What about the lower limit? The friction force $f = \frac{1}{3}F + \frac{1}{3}Mg \sin\theta$ must be non-negative for it to act up the incline. As established, this is always true for $F \ge 0$. The problem does not specify that $F$ must be positive. If $F$ were negative (pulling down), the direction of friction might reverse. However, a "pulling upwards" force usually implies $F \ge 0$.

Let's check the condition for friction to be static: $f \ge 0$. This is already satisfied for $F \ge 0$. So the range of $F$ for rolling without slipping is $0 \le F \le F_{max}$. The question asks for the range of values for $F$ if the cylinder rolls without slipping. This implies that $F$ can be zero or positive.

Final Range: $0 \le F \le Mg \left(\frac{3\sqrt{3} - 2}{4}\right)$

Numerical approximation: $\sqrt{3} \approx 1.732$ $F_{max} = Mg \left(\frac{3 \times 1.732 - 2}{4}\right) = Mg \left(\frac{5.196 - 2}{4}\right) = Mg \left(\frac{3.196}{4}\right) = 0.799 Mg$

So the range is approximately $0 \le F \le 0.799 Mg$.