JEE Challenger
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PhyMedAdv 2022 • Paper 1

Acceleration of Rolling Solid Sphere Down Incline with Applied Forces

A solid sphere of mass 1 kg1\text{ kg} and radius 1 m1\text{ m} rolls without slipping on a fixed inclined plane with an angle of inclination θ=30∘\theta = 30^\circ from the horizontal. Two forces of magnitude 1 N1\text{ N} each, parallel to the incline, act on the sphere, both at distance r=0.5 mr = 0.5\text{ m} from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ______ m s−2\text{m s}^{-2}. (Take g=10 m s−2g = 10\text{ m s}^{-2}.)

Question Diagram 1
Official Numerical Answer2.85 to 2.86

Topics & Concepts

System of Particles and Rotational MotionRolling MotionTorque

Step-by-Step Solution

To find the acceleration of the solid sphere down the inclined plane, we analyze both its translational motion along the incline and its rotational motion about its center of mass.

1. System Parameters and Forces

  • Mass of the solid sphere, m=1 kgm = 1\text{ kg}
  • Radius of the sphere, R=1 mR = 1\text{ m}
  • Distance of applied forces from the center, r=0.5 mr = 0.5\text{ m}
  • Angle of inclination, θ=30∘\theta = 30^\circ
  • Acceleration due to gravity, g=10 m s−2g = 10\text{ m s}^{-2}
  • Moment of inertia of the solid sphere about its center of mass, I=25mR2I = \frac{2}{5} m R^2

From the given figure:

  • The force F1=1 NF_1 = 1\text{ N} acts at distance r=0.5 mr = 0.5\text{ m} above the center, directed up the incline.
  • The force F2=1 NF_2 = 1\text{ N} acts at distance r=0.5 mr = 0.5\text{ m} below the center, directed down the incline.
  • The component of gravitational force down the incline is: Fg=mgsin⁡θ=(1 kg)(10 m s−2)sin⁡30∘=5 NF_g = m g \sin\theta = (1\text{ kg})(10\text{ m s}^{-2})\sin 30^\circ = 5\text{ N}
  • Let ff be the force of static friction acting at the contact point, directed up the incline.

2. Equations of Motion

Translational Motion (down the incline):

Taking the direction down the incline as positive: Fnet=mgsin⁡θ−F1+F2−f=maF_{\text{net}} = m g \sin\theta - F_1 + F_2 - f = m a

Substituting the given values: 5−1+1−f=(1)a5 - 1 + 1 - f = (1) a 5−f=a  ⟹  f=5−a— (Equation 1)5 - f = a \implies f = 5 - a \quad \text{--- (Equation 1)}

Rotational Motion (about the center of mass):

Taking the clockwise direction (corresponding to rolling down the incline) as positive:

  • Torque due to friction ff: τf=+fR\tau_f = +f R (clockwise)
  • Torque due to force F1F_1: τ1=−F1r\tau_1 = -F_1 r (counter-clockwise)
  • Torque due to force F2F_2: τ2=−F2r\tau_2 = -F_2 r (counter-clockwise)

The net torque equation is: τnet=fR−F1r−F2r=Iα\tau_{\text{net}} = f R - F_1 r - F_2 r = I \alpha

For pure rolling without slipping, the condition connecting angular acceleration α\alpha and linear acceleration aa is: a=Rα  ⟹  α=aRa = R \alpha \implies \alpha = \frac{a}{R}

Substitute I=25mR2I = \frac{2}{5} m R^2 and α=aR\alpha = \frac{a}{R} into the torque equation: fR−F1r−F2r=(25mR2)(aR)f R - F_1 r - F_2 r = \left(\frac{2}{5} m R^2\right) \left(\frac{a}{R}\right) fR−(F1+F2)r=25mRaf R - (F_1 + F_2) r = \frac{2}{5} m R a

Dividing by RR: f−(F1+F2)rR=25maf - (F_1 + F_2) \frac{r}{R} = \frac{2}{5} m a

Substituting m=1 kgm = 1\text{ kg}, R=1 mR = 1\text{ m}, r=0.5 mr = 0.5\text{ m}, F1=1 NF_1 = 1\text{ N}, and F2=1 NF_2 = 1\text{ N}: f−(1+1)(0.51)=25(1)af - (1 + 1) \left(\frac{0.5}{1}\right) = \frac{2}{5} (1) a f−1=25a  ⟹  f=25a+1— (Equation 2)f - 1 = \frac{2}{5} a \implies f = \frac{2}{5} a + 1 \quad \text{--- (Equation 2)}

3. Solving for Acceleration

Equating the expressions for ff from Equation 1 and Equation 2: 5−a=25a+15 - a = \frac{2}{5} a + 1 4=a+25a4 = a + \frac{2}{5} a 4=75a4 = \frac{7}{5} a a=207 m s−2≈2.86 m s−2a = \frac{20}{7}\text{ m s}^{-2} \approx 2.86\text{ m s}^{-2}

Thus, the acceleration of the sphere down the plane is 207 m s−2\frac{20}{7}\text{ m s}^{-2} (or approximately 2.86 m s−22.86\text{ m s}^{-2}).

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Acceleration of Rolling Solid Sphere Down Incline with Applied Forces | Physics PYQ Solution - JEE Challenger