A solid cylinder of mass M and radius R rolls without slipping down an inclined plane making an angle 6 with the horizontal. Then its acceleration is

Question

A solid cylinder of mass M and radius R rolls without slipping down an inclined plane making an angle 6 with the horizontal. Then its acceleration is (A) (1/3) g sin θ (B) (2/3) g sin θ (C) (2/5) g sin θ (D) (2/7) g sin θ

Tardigrade Admin · Accepted Answer

Acceleration of a rolling body down an inclined plane is a=(g sin θ/1+(k2)R2) For solid cylinder, k2=(R2/2) ∴ a=(g sin θ/1+(1)2)=(2/5) g sin θ

Q. A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane making an angle 6 with the horizontal. Then its acceleration is

$\frac{1}{3} g sin θ$

$\frac{2}{3} g sin θ$

$\frac{2}{5} g sin θ$

$\frac{2}{7} g sin θ$

Acceleration of a rolling body down an inclined plane is
$a = \frac{g s i n θ}{1 + \frac{k ^{2}}{R ^{2}}}$
For solid cylinder, $k^{2} = \frac{R ^{2}}{2}$
$∴ a = \frac{g s i n θ}{1 + \frac{1}{2}} = \frac{2}{5} g sin θ$

Q. A solid cylinder of mass M and radius R rolls without slipping down an inclined plane making an angle 6 with the horizontal. Then its acceleration is

31​gsinθ

32​gsinθ

52​gsinθ

72​gsinθ