Question

A block of mass 2 kghangs from the rim of a wheel of radius 0.5 m . On releasing from rest the block falls through height in . The moment of inertia of the wheel will be

• A. 1 kg-m²
• B. 3.2 kg-m²
• C. 2.5 kg-m²
• D. 1.5 kg-m²

Answer 1

Answer: (D)

1.5 kg-m²

Answer 2

On releasing from rest the block falls through 5m height in 2 sec.

$5 = 0 + \frac { 1 } { 2 } a ( 2 ) ^ { 2 }$ [As $S = u t + \frac { 1 } { 2 } a t ^ { 2 }$ ]

$\therefore$ $a = 2.5 \mathrm {~m} / \mathrm { s } ^ { 2 }$

Substituting the value of a in the formula $a = \frac { g } { 1 + \frac { I } { m R ^ { 2 } } }$

and by solving we get

⇒ $2.5 = \frac { 10 } { 1 + \frac { I } { 2 \times ( 0.5 ) ^ { 2 } } }$ ⇒