A solid sphere of mass 1   kg and radius 1   m rolls without slipping on a fixed inclined plane with an angle of inclination θ = 30 ° from the horizontal. Two forces of magnitude 1   N each, parallel to the incline, act on the sphere, both at distance r = 0 . 5   m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is m   s - 2 . (Ta

Step-by-step solution

The forces acting on the solid sphere is shown below. Here, N is the normal force acting on sphere and weight m g is acting downwards. Taking torque about contact point. τ → = m g R sin 30 ⊗ + 1 × 1 ⊙ = 10 × 1 × 1 2 - 1 (Taking ⊗ as positive) Then, we have ⇒ 5 - 1 = I sphere about tangent α Using parallel axis theorem, τ = 2 5 m R 2 + m R 2 α = 7 5 m R 2 α ⇒ α = 20 7 rad   s - 2 So, acceleration of sphere down the plane