A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm / s. The kinetic energy of the ring is

Question

A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm / s. The kinetic energy of the ring is (A) 4 × 10-3 joules (B) 4 × 10-2 joules (C) 2 × 10-3 joules (D) 2 × 10-2 joules

Tardigrade Admin · Accepted Answer

Kinetic energy of ring, K=(1/2) m v2(1+(K2/R2)) =(1/2) 0.4 ×((10/100))2 ×(1+1) [∴ (K2/R2)=1] ⇒ K =4 × 10-3 Joule

Q. A uniform thin ring of mass $0.4 k g$ rolls without slipping on a horizontal surface with a linear velocity of $10 c m / s$ . The kinetic energy of the ring is

$4 \times 1 0^{- 3}$ joules

$4 \times 1 0^{- 2}$ joules

$2 \times 1 0^{- 3}$ joules

$2 \times 1 0^{- 2}$ joules

Kinetic energy of ring,
$K = \frac{1}{2} m v^{2} (1 + \frac{K ^{2}}{R ^{2}})$
$= \frac{1}{2} 0.4 \times (\frac{10}{100})^{2} \times (1 + 1)$
$[∴ \frac{K ^{2}}{R ^{2}} = 1]$
$\Rightarrow K = 4 \times 1 0^{- 3}$ Joule

Q. A uniform thin ring of mass 0.4kg rolls without slipping on a horizontal surface with a linear velocity of 10cm/s. The kinetic energy of the ring is

4×10−3 joules

4×10−2 joules

2×10−3 joules

2×10−2 joules