disc at rest at the top of a rough inclined plane. If rolls without slipping. At the bottom of the inclined plane there is a vertical groove of radius R. In order to loop the groove, the minimum height of inclined plane is

The problem involves the conservation of mechanical energy and the condition for circular motion.

Step 1: Identify the initial and final conditions The disc starts from rest at the top of the inclined plane and ends up at the bottom just before it enters the vertical loop.

Step 2: Apply the principle of conservation of mechanical energy The mechanical energy of the disc is conserved because there are no non-conservative forces doing work. The initial mechanical energy is all potential energy, and the final mechanical energy is all kinetic energy.

Step 3: Set up the energy conservation equation The initial potential energy is mgh, where m is the mass of the disc, g is the acceleration due to gravity, and h is the height of the inclined plane. The final kinetic energy is (1/2)Iω² + (1/2)mv², where I is the moment of inertia of the disc, ω is the angular velocity, and v is the linear velocity. Since the disc rolls without slipping, v = Rω, where R is the radius of the disc.

Step 4: Solve the energy conservation equation for h Substituting v = Rω into the kinetic energy expression gives (1/2)Iω² + (1/2)m(Rω)² = mgh. Simplifying this gives (1/2)(mR² + mR²)ω² = mgh, or ω² = 2gh/R².

Step 5: Apply the condition for circular motion The minimum condition for the disc to complete the loop is that the centripetal force at the top of the loop is equal to the weight of the disc, or mv²/R = mg. Substituting v = Rω gives m(Rω)²/R = mg, or ω² = g/R.

Step 6: Solve the two equations for ω² together to find h Setting 2gh/R² = g/R gives h = R/2.

So, the minimum height of the inclined plane for the disc to loop the groove is half the radius of the groove.