A solid sphere of mass ' m ' and radius ' r ' is allowed to roll without slipping from the highest point of an inclined plane of length ' L ' and makes an angle 30 with the horizontal. The speed of the particle at the bottom of the plane is v1. If the angle of inclination is increased to 45 while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2. The ratio v12:v22 is
- 1:3
- 1:2
- 1: 2
- 1: 3
Answer
(3) 1: 2
Solution
Rolling without slipping: friction does no work, so mechanical energy is conserved (h=L ).
[figure]
mgL = 12mv2+ 12 ( 25mr2 ) v2r2= 710mv2 v2= 107gL
v12v22= 30 45 = 1 2
Answer: (3)