A uniform disc of mass m and radius R is pivoted at point P and is free to rotate in vertical plane. The centre C of disc is initially in horizontal position with P as shown in figure. If it is released from this position, then its angular acceleration when the line PC is inclined to the horizontal at an angle θ is

Question

A uniform disc of mass m and radius R is pivoted at point P and is free to rotate in vertical plane. The centre C of disc is initially in horizontal position with P as shown in figure. If it is released from this position, then its angular acceleration when the line PC is inclined to the horizontal at an angle θ is (A) (2 g cos θ/3 R) (B) (g sin θ/2 R) (C) (2 g sin θ/R) (D) (2 g sin θ/3 R)

Tardigrade Admin · Accepted Answer

τ=I α m g(R cos θ)=(3/2) m r2 α α=(2 g cos θ/3 r)

$\frac{2 g c o s θ}{3 R}$

$\frac{g s i n θ}{2 R}$

$\frac{2 g s i n θ}{R}$

$\frac{2 g s i n θ}{3 R}$

$τ = I α$
$m g (R cos θ) = \frac{3}{2} m r^{2} α$
$α = \frac{2 g c o s θ}{3 r}$

3R2gcosθ​

2Rgsinθ​

R2gsinθ​

3R2gsinθ​

τ=Iα mg(Rcosθ)=23​mr2α α=3r2gcosθ​

$\frac{2 g c o s θ}{3 R}$

$\frac{g s i n θ}{2 R}$

$\frac{2 g s i n θ}{R}$

$\frac{2 g s i n θ}{3 R}$

$τ = I α$
$m g (R cos θ) = \frac{3}{2} m r^{2} α$
$α = \frac{2 g c o s θ}{3 r}$