A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be

Question

A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be (A) (2/R)√g h (B) (2/R)√(g h/2) (C) (2/R)√(g h/3)  (D) (1/2 R)√g h

Tardigrade Admin · Accepted Answer

Loss in PE = Gain in KE mgh=(1/2)mv2+(1/2)T(ω )2 (v = R ω ) mgh=(1/2)m(R ω )2+(1/2)((m R2/2))(ω )2 gh=(3/4)R2ω 2 ⇒ ω 2=( text4 g h/3 R2) ⇒ ω =(2/R)√(g h/3)

Q. A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be

$\frac{2}{R} g h$

$\frac{2}{R} \frac{g h}{2}$

$\frac{2}{R} \frac{g h}{3}$

$\frac{1}{2 R} g h$

Loss in PE = Gain in KE
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} T (ω)^{2} (v = R ω)$
$m g h = \frac{1}{2} m (R ω)^{2} + \frac{1}{2} (\frac{m R ^{2}}{2}) (ω)^{2}$
$g h = \frac{3}{4} R^{2} ω^{2}$
$\Rightarrow ω^{2} = \frac{4 g h}{3 R ^{2}} \Rightarrow ω = \frac{2}{R} \frac{g h}{3}$

Q. A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be

R2​gh​

R2​2gh​​

R2​3gh​​

2R1​gh​

Loss in PE = Gain in KE mgh=21​mv2+21​T(ω)2(v=Rω) mgh=21​m(Rω)2+21​(2mR2​)(ω)2 gh=43​R2ω2 ⇒ω2=3R24gh​⇒ω=R2​3gh​​

$\frac{2}{R} g h$

$\frac{2}{R} \frac{g h}{2}$

$\frac{2}{R} \frac{g h}{3}$

$\frac{1}{2 R} g h$

Loss in PE = Gain in KE
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} T (ω)^{2} (v = R ω)$
$m g h = \frac{1}{2} m (R ω)^{2} + \frac{1}{2} (\frac{m R ^{2}}{2}) (ω)^{2}$
$g h = \frac{3}{4} R^{2} ω^{2}$
$\Rightarrow ω^{2} = \frac{4 g h}{3 R ^{2}} \Rightarrow ω = \frac{2}{R} \frac{g h}{3}$