Question

A uniform solid sphere of radius (R = 1m) is rolling without sliding
one of radius (R = 1m) is rolling without sliding on a horizontal surface with a
Velocity vo = 14 m/sec. It collides with an obstacle of height (h = R/2) inelastically, the angular speed
of the sphere just after the collision is
(1) 4.5 rad/sec (2) 9 rad/sec
(3) 3 rad/sec (4) 18 rad/sec​

Answer 1

Answer:

9 rad/s I guess

Explanation:

Answer 2

Given that,

Radius R = 1m

Velocity = 4 m/s

Height of obstacle $h=\dfrac{R}{2}$

Angular velocity remain conserved about contact point.

We need to calculate the angular velocity of the sphere just after the collision

Using conservation of momentum

$L_{i}=L_{f}$

$mvR\sin\theta+\dfrac{2}{5}mR^2\omega=mv'R +\dfrac{2}{5}mR^2\omega'$

Put the value into the formula

$R^2\omega\times\sin60+\dfrac{2}{5}R^2\omega=R^2\omega'+\dfrac{2}{5}R^2\omega$

$R^2\omega\times\dfrac{1}{2}+\dfrac{2}{5}R^2\omega=R^2\omega'+\dfrac{2}{5}R^2\omega$

$\dfrac{5R^2\omega+4R^2\omega}{10}=\dfrac{5R^2\omega'+2R^2\omega'}{5}$

$\dfrac{9R^2\omega}{10}=\dfrac{7R^2\omega'}{5}$

$\omega'=\dfrac{9}{14}\omega$

$\omega'=\dfrac{9}{14}\times\dfrac{v}{R}$

Put the value into the formula  

$\omega'=\dfrac{9}{14}\times\dfrac{14}{1}$

$\omega'=9\ rad/s$

Hence, the angular speed  of the sphere just after the collision is 9 rad/sec.

(2) correct option