Question 278: Centre of Mass and Rotational Motion

Question

A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. when the particle reaches point b, its angular momentum about O will be :
(Take g = 10 m/s
2
)

Accepted Answer

Work Energy Theorem from A to B
Mgh = $${1 \over g}$$ mv$$_B^2$$ $$-$$ $${1 \over g}$$ mv$$_A^2$$
2gh = $$v_B^2 - v_A^2$$
2 $$ 	imes $$ 10 $$ 	imes $$ 10 = v$$_B^2$$ $$-$$ 5
2
v
B
= 15m/s
Angular momentum about 0
L
0
= mvr
= 20 $$ 	imes $$ 10
$$-$$3
$$ 	imes $$ 20
L
0
= 6 kg.m
2
/s

Answer

6 kg-m
2
/s

Answer

8 kg-m
2
/s

Answer

2 kg-m
2
/s

Answer

3 kg-m
2
/s

Centre of Mass and Rotational Motion

Choose the correct answer: