Question

A solid sphere of mass $1 \,kg$ and radius $1 \,m$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta=30^{\circ}$ from the horizontal Two forces of magnitude $1 \,N$ each, parallel to the incline, act on the sphere, both at distance $r=05 \,m$ from the center of the sphere, as shown in the figure The acceleration of the sphere down the plane is ____ $ms ^{-2}$ (Take $g=10 \,m s ^{-2}$)

Answer 1

Answer: 2.86

Answer 2

Given Data:
Mass of the sphere, \(m = 1 \, \text{kg}\)
The radius of the sphere, \(R = 1 \, \text{m}\)
The angle of inclination, \(\theta = 30^\circ\)
Two forces of 1N each acting at \(\frac{R}{2}\) and \(\frac{3R}{2}\) from the center
Gravitational acceleration, \(g = 10 \, \text{m/s}^2\)

Gravitational Force Component along the Incline:
\(mg \sin \theta = 1 \, \text{kg} \times 10 \, \text{m/s}^2 \times \sin 30^\circ = 5 \, \text{N}\)

Torque Calculation:
The torque due to gravitational force:
\(\tau_g = mg \sin \theta \times R = 5 \, \text{N} \times 1 \, \text{m} = 5 \, \text{Nm}\)
The torque due to the two applied forces:
Force at \(\frac{R}{2}\): \(\tau_1 = 1 \, \text{N} \times \frac{R}{2} = 1 \, \text{N} \times 0.5 \, \text{m} = 0.5 \, \text{Nm}\)

Force at \(\frac{3R}{2}\): \(\tau_2 = 1 \, \text{N} \times \frac{3R}{2} = 1 \, \text{N} \times 1.5 \, \text{m} = 1.5 \, \text{Nm}\)
Net torque:
\(\tau_{\text{net}} = \tau_g + \tau_1 - \tau_2 = 5 \, \text{Nm} + 0.5 \, \text{Nm} - 1.5 \, \text{Nm} = 4 \, \text{Nm}\)

Moment of Inertia for a Solid Sphere:
\(I_{\text{cm}} = \frac{2}{5} m R^2 = \frac{2}{5} \times 1 \, \text{kg} \times (1 \, \text{m})^2 = \frac{2}{5} \, \text{kg} \cdot \text{m}^2\)
For rolling without slipping:
\(I_{\text{total}} = I_{\text{cm}} + mR^2 = \frac{2}{5} m R^2 + m R^2 = \frac{2}{5} m R^2 + \frac{5}{5} m R^2 = \frac{7}{5} m R^2\)

Relating Torque to Angular Acceleration:
\(\tau_{\text{net}} = I_{\text{total}} \alpha\)
\(4 \, \text{Nm} = \frac{7}{5} \times 1 \, \text{kg} \times (1 \, \text{m})^2 \times \alpha\)
\(4 = \frac{7}{5} \alpha\)
\(\alpha = \frac{4 \times 5}{7} = \frac{20}{7} \, \text{rad/s}^2\)

Finding Linear Acceleration using the No-slip Condition:
\(a_{\text{cm}} = R \alpha\)
\(a_{\text{cm}} = 1 \, \text{m} \times \frac{20}{7} \, \text{rad/s}^2 = \frac{20}{7} \, \text{m/s}^2= 2.86 \, \text{m/s}^2\)

So, the answer is: \(2.86 \, \text{m/s}^2\)