|
35. Problem 12.15P (HRW) / 12.29 (HR) We are given that a small solid marble of mass m and radius r will roll without slipping along the loop-the-loop track if it is released from rest somewhere on the straight section of track. We have to find (a) The minimum height h above the bottom of the track from where the marble must be released to ensure that it does not leave the track at the top of the loop (the radius of the loop-the-loop is R); (b) the horizontal component of the force acting on the marble at the point Q when it is released from height 6R above the bottom of the track. The following diagram illustrates the setup.
|
|
Solution: Click For PDF Version We will solve this problem using the principle of conservation of energy. As the marble ball has a finite size and it rolls without slipping, magnitude of its rotational velocity will change with change in speed of its centre of mass. The speed of the ball at any instant when it is moving in the loop-the-loop can be found by equating at that instant the total kinetic energy of the ball with the change in the potential energy of the ball. Let the speed of the centre of mass of the marble ball at some instant be
As the ball is released when it is at rest, its K.E. is equal to the
change in its potential energy. When the change in the vertical height of the
ball from its starting position is
where
The other key concept that we will use in answering the problem is that when
a body of mass
a) The minimum speed that the ball must have so that it does not leave the track when it is at the top of the loop-the-loop can be found by using the condition that at that location the centripetal force is provided by the weight of the ball. That is
Therefore, the minimum height
b) The horizontal component of the force acting on the ball when it is at the
point
|
. At that
instant its angular velocity
will be
, as the
radius of the marble ball is
and the
ball is rolling without slipping. Moment of inertia,
,
of a spherical ball of mass
and
radius
.
Therefore, the kinetic energy of the ball, when its speed is
, the
change in potential energy
is the
acceleration due to gravity. From conservation of energy, we get
moves in
a circular path of radius
with
speed
.
The direction of the centripetal force is along the line joining the body with
the centre of the circular path, which changes as the body moves along its
circular path.
above the
bottom of the track from which the ball has to be dropped so that it does not
leave the top of the loop-the-loop will be given by the equation
on the
track will be the centripetal force. As the ball is released from a height 6
above the base of the track and the point
and the
horizontal component of the force at
