A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ______ ms.
Show answer and solution
Answer: 40
Solution
For a solid sphere in pure rolling, total KE = (1/2)mv² + (1/2)Iω², where I = (2/5)mr² and ω = v/r.
Total KE = (1/2)mv² + (1/2)(2/5)mr²(v/r)² = (1/2)mv²(1 + 2/5) = (7/10)mv²
So: 2240 = (7/10)(2)v² = (7/5)v²
v² = 2240 × 5/7 = 1600
v = 40 m/s
Total KE = (1/2)mv² + (1/2)(2/5)mr²(v/r)² = (1/2)mv²(1 + 2/5) = (7/10)mv²
So: 2240 = (7/10)(2)v² = (7/5)v²
v² = 2240 × 5/7 = 1600
v = 40 m/s
[JEE Main 2023 · 29 Jan · Shift 1]