A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity, of 2 √ 2 rads -1 . The radius of cylinder must be (Take g = 10 ms -2 )

Question

A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity, of 2 √ 2 rads -1 . The radius of cylinder must be (Take g = 10 ms -2 ) (A) 5 cm (B) 0.5 cm (C)  √10 cm (D)  √ 5 m

Tardigrade Admin · Accepted Answer

beginarrayl v=√(2 g h/1+k2 / R2)=√(2 g h/1+(1 / 2))=√(4 g h/3)=√(4 × 10 × 3/3) 2 √10 ω=(v/R) ⇒ R=(v/ω)=(2 √10/2 √2)=√5 m endarray

Q. A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity, of 2 $2 r a d s^{- 1}$ . The radius of cylinder must be
(Take g = 10 $m s^{- 2}$ )

5 cm

0.5 cm

$10$ cm

$5$ m

10 cm

$v = \frac{2 g h}{1 + k ^{2} / R ^{2}} = \frac{2 g h}{1 + ( 1/2 )} = \frac{4 g h}{3} = \frac{4 \times 10 \times 3}{3} 210 ω = \frac{v}{R} \Rightarrow R = \frac{v}{ω} = \frac{2 10}{2 2} = 5 m$

Q. A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity, of 2 2​rads−1 . The radius of cylinder must be (Take g = 10 ms−2 )