A particle moves from rest at A on the surface of a smooth circular cylinder of radius r as shown. At B it leaves the cylinder. The equation relating α and β

Question

A particle moves from rest at A on the surface of a smooth circular cylinder of radius r as shown. At B it leaves the cylinder. The equation relating α and β <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/dcbdef98d482cb0c5c143b8d628f9061-.png" /> (A) 3 sin α=2 sin β (B) 2 sin α=3 cosβ (C) 3 sin β=2 cos α (D) 2 sin β=3 cos α

Tardigrade Admin · Accepted Answer

By law of conservation of mechanical energy

(K + U)_{A} = (K + U)_{B}

0 + m g R cos α = \frac{1}{2} m v_{B}^{2} + m g R sin β

v_{B}^{2} = 2 g R (cos α - sin β)

At

B

since particle leaves contact with the surface hence normal reaction equals zero.

m g sin β = \frac{m v _{B}^{2}}{R}

m g sin β = \frac{m}{R} [2 g R (cos α - sin β)]

Q. A particle moves from rest at A on the surface of a smooth circular cylinder of radius $r$ as shown. At $B$ it leaves the cylinder. The equation relating $α$ and $β$

$3 sin α = 2 sin β$

$2 sin α = 3 cos β$

$3 sin β = 2 cos α$

$2 sin β = 3 cos α$

Q. A particle moves from rest at A on the surface of a smooth circular cylinder of radius r as shown. At B it leaves the cylinder. The equation relating α and β

3sinα=2sinβ

2sinα=3cosβ

3sinβ=2cosα

2sinβ=3cosα

By law of conservation of mechanical energy (K+U)A​=(K+U)B​ 0+mgRcosα=21​mvB2​+mgRsinβ vB2​=2gR(cosα−sinβ) At B since particle leaves contact with the surface hence normal reaction equals zero. mgsinβ=RmvB2​​ mgsinβ=Rm​[2gR(cosα−sinβ)]

Q. A particle moves from rest at A on the surface of a smooth circular cylinder of radius $r$ as shown. At $B$ it leaves the cylinder. The equation relating $α$ and $β$

$3 sin α = 2 sin β$

$2 sin α = 3 cos β$

$3 sin β = 2 cos α$

$2 sin β = 3 cos α$