00:01
Hi, in this question we are asked to find the distance s of the plunger that must be pulled back and released so that the ball will begin to leave the track when the value of the angle teta is equal to 135 degree.
00:16
So here w is the weight which acts downwards through the center of the ball and the normal force exerted as shown in the figure by the track is perpendicular to the surface.
00:28
The frictional force between the ball and the track has no components in the n direction.
00:33
So applying the equation of motion in the end direction since the ball leaves back the track when theta is 135 degree and set n is equal to 0.
00:45
So we have mg cos theta is equal to mv square by r for the ball to move in the track.
00:53
So upon substituting all the values according to the question, we have the value.
00:58
Of m as 0 .5 into g as 9 .8 1 into cost theta value is 45 degree since the angle will be 180 minus 135 degree which is equal to 45 degree so which is equal to 0 .5 into v square by the value of our given in the question is 1 .5 meter upon solving this we get the value of v as 3 .2257 meter per second.
01:30
And now applying the work energy theorem between the position 1, with theta is equal to 0 and position 2, when theta is equal to 135 degree, we have the normal force which does not do any work since the displacement is perpendicular...