Question

A solid cylinder rolls without slipping along a horizontal plane with velocity 6m/s and reaches a fixed inclined plane inclined at 60° with the horizontal. There is sufficient friction to prevent slipping then, after collision with the incline, the body starts rolling up the inclined plane without rebound with the velocity v. Find the value of v (in m/s). The axis of cylinder is perpendicular to the line of greatest slope of inclined plane.

Answer 1

18

Answer 2

The problem involves a solid cylinder rolling without slipping on a horizontal plane, colliding with an inclined plane, and then rolling up the incline without slipping. The key to solving this problem is the conservation of angular momentum about the edge where the horizontal and inclined planes meet.

1. Initial Angular Momentum:

   • Before the collision, the cylinder has both translational and rotational kinetic energy. The angular momentum about the edge (O) is given by:

     $L_{before} = -\frac{3}{2}mRv_0$

     where:

     • $m$ is the mass of the cylinder
     • $R$ is the radius of the cylinder
     • $v_0$ is the initial velocity (6 m/s)

2. Final Angular Momentum:

   • After the collision, the cylinder rolls up the inclined plane with velocity $v$. The angular momentum about the same edge (O) is given by:

     $L_{after} = -\frac{1}{2}mRv$

3. Conservation of Angular Momentum:

   • Since the collision is assumed to be instantaneous and without external torques about the edge O (the impulsive forces act at O), angular momentum is conserved:

     $L_{before} = L_{after}$

     $-\frac{3}{2}mRv_0 = -\frac{1}{2}mRv$

     Solving for $v$:

     $v = 3v_0$

4. Calculating the Final Velocity:

   • Given $v_0 = 6$ m/s, the final velocity is:

     $v = 3 \times 6 = 18 \text{ m/s}$

Thus, the value of v is 18 m/s.