A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle θ with the horizontal. The acceleration of the disc will be 2/bg sinθ where b is ____________. (Round off to the Nearest Integer) (g = acceleration due to gravity, θ = angle as shown in figure)

Solution

Physics sir · 3 steps

Rolling splits gravity's pull between sliding and spinning. For a disc, one third goes into the spin.

1. Rolling formula. For a body rolling without slipping down an incline, friction at the contact point slows the slide and drives the spin. Combining mg sin θ - f = ma with faᵣ = Iα and α = a_c/a gives the standard result.

   a_c = (g sin θ)/(1 + I/(m a²))

2. Disc inertia. A solid disc about its central axis has I = 1/2ma², so I/ma² = 1/2. Then a_c = (g sin θ)/(1 + 1/2) = 2/3g sin θ.
3. So. Matching 2/3g sin θ with 2/bg sin θ gives b = 3.

The common slip: 4. Using the ring's moment of inertia ma² instead of the disc's 1/2ma² gives a = 1/2g sin θ, which makes b = 4.

The idea to keep. For any rolling body, a = (g sin θ)/(1 + k²/R²). Just plug in k²/R²: ring 1, disc 1/2, solid sphere 2/5. The larger it is, the slower the body rolls down.

Answer from the NTA final answer key. Solution by PrepWiser.