Concept explainers
A single bead can slide with negligible friction on a stiff wire that has been bent into a circular loop of radius 15.0 cm as shown in Figure P6.48. The circle is always in a vertical plane and rotates steadily about its vertical diameter with a period of 0.450 s. The position of the bead is described by the angle θ that the radial line, from the center of the loop to the bead, makes with the vertical. (a) At what angle up from the bottom of the circle can the bead slay motionless relative to the turning circle? (b) What If? Repeat the problem, this time taking the period of the circle’s rotation as 0.850 s. (c) Describe how the solution to part (b) is different from the solution to part (a). (d) For any period or loop size, is there always an angle at which the bead can stand still relative to the loop? (e) Are there ever more than two angles? Arnold Arons suggested the idea for this problem.
Figure P6.48
(a)
The angle from the bottom of the circle for which the bead can stay motionless.
Answer to Problem 68CP
The angles from the bottom of the circle for which the bead can stay motionless are
Explanation of Solution
Given info: The radius of the circular loop is
The acceleration due to gravity is
The rough sketch of the force body diagram of the situation is shown below,
Figure (1)
The bead moves in a circle,
Here,
The speed of the bead is,
Here,
From the figure (1) the net force in
Here,
The net force in
Substitute
Substitute
The two possible solutions are,
and,
Substitute
Conclusion:
Therefore, the angles from the bottom of the circle for which the bead can stay motionless are
(b)
The angles from the bottom of the circle for which the bead can stay motionless for the time period as
Answer to Problem 68CP
The only one possible angle from the bottom of the circle for which the bead can stay motionless for time period
Explanation of Solution
Given info: The radius of the circular loop is
The acceleration due to gravity is
Form the part (a) equation (1).
The possible solutions are,
and,
Substitute
The above value is not possible.
Conclusion:
Therefore, the only one possible angle from the bottom of the circle for which the bead can stay motionless at time period
(c)
The difference between the solution of the part (a) and part (b).
Answer to Problem 68CP
The part (b) has only one solution as the time period is very large.
Explanation of Solution
Given info: The radius of the circular loop is
Form equation (1)
The possible solutions are,
and,
As the second solution depends directly on the square of the amplitude of the time period so, as the time period increases the value of
Conclusion:
Therefore, the part (b) has only one solution as the time period is very large.
(d)
The angle at which the bead can stand still relative to the loop.
Answer to Problem 68CP
The angle for which condition
Explanation of Solution
Given info: The radius of the circular loop is
Form equation (1)
The possible solutions are,
and,
As the range of the cosine function is
So, the value of the cosine of the angle is always less than
Conclusion:
Therefore, for the condition
(e)
Whether there are more than two angles.
Answer to Problem 68CP
The number of possible angles are
Explanation of Solution
Given info: The radius of the circular loop is
Form equation (1)
The possible solutions are,
and,
Form the above expression the second solution depends on time period, acceleration due to gravity and the radius of the hoop and for a given case the parameters has only one value so the second solution has only one answer.
Conclusion:
Therefore, the number of possible angles are
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