1please somebody help
9 Answers
11Q2 A student throws a stick of length L up in yhe air. At the moment the stick leaves his hand, the speed of the stick's end is 0. The stick completes N turns and is caught by the student in the same way as is thrown by him i.e, the initial and final positions are same.find the height to which the centre of mass of the stick rose.
a) 3.14NL/4
b) 2.14NL/4
c) 5.14NL/4
d) 7.14NL/4
Q3 Two spherical balls of mass M & 5M and radii R & 2R respectively are released in free space with initial seperation b/w their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is
a) 2.5R
b) 4.5R
c) 7.5R
d) 1.5
106Q2. let velocity of COM be v and angular velocity be w
Now since the point of contact is at rest v = wl / 2 i.e. w = 2v / l
now once the rod is in air there is no external torque about the COM so w will not change
and at the highest point the rod will have PE and rotational KE
(datum line is the line from where the rod is released)
Also the time taken by COM to reach its initial position will be 2v/g (in this time the rod completes n revolutions)
so v = (ng) / (2v)
w = 2Î v = nÎ g/v
v2 = nÎ gl/2
since w = 2v / l
Now apply energy conservation
Iw2/2+ mv2/2 = mgh + Iw2/2
i.e. h = v2/2g
therefore h=Î nl/4
Ans. is (a)
106find the centre of mass from body of mass M.... taking its position as origin... Position of centre of mass = (M.0 + 5M.12R)/(M+5M) = 10R
When they collide..... takke position of body of mass M to be zero again... now, COM = 5M.3R/6M = 5R/2
Hence bosy of smaller mass wil move distance of 10R-2.5R=7.5R
Hence answer is (c)