A uniform plank leans against a cylindrical body of radius R as shown in figure, the right end B of the plank slides at a constant speed V to the right whereas the cylindrical body moves at a constant speed of V₀ to the left side. The angular speed ω of the rod at this instant will be? Distance between centre of cylinder and the end B is x. Take V = 15 m/s, V₀ = 5 m/s, x = 5 m, R = 3 m.
Physics · Mixed Chapter
JEE Mains

Question

A uniform plank leans against a cylindrical body of radius R as shown in figure, the right end B of the plank slides at a constant speed V to the right whereas the cylindrical body moves at a constant speed of V₀ to the left side. The angular speed ω of the rod at this instant will be? Distance between centre of cylinder and the end B is x. Take V = 15 m/s, V₀ = 5 m/s, x = 5 m, R = 3 m.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The plank is tangent to the circular cylinder. In the right triangle formed by its centre C, tangent point A and end B, the contact radius CA=R is perpendicular to AB and CB=x. Therefore sinα=R/x, where α is the plank angle to horizontal. Step 2: Relative separation between B and the moving cylinder is increasing at rate dx/dt=V+V₀=15+5=20 m/s. Step 3: Differentiate sinα=R/x: cosα·dα/dt=−R(dx/dt)/x². With cosα=√(x²−R²)/x, |ω|=R(V+V₀)/[x√(x²−R²)]. Step 4: ω=3×20/[5×√(25−9)] =60/(5×4)=3 r…