Circular Loop Question?
In the figure here, a solid brass ball of mass 0.113 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 0.19 m, and the ball has radius r << R.
(a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop?
(b) If the ball is released at height h = 5R, what is the magnitude of the horizontal force component acting on the ball at point Q?
A) .513 meters
B) ????? i just need B. its not 6.328
In the figure here, a solid brass ball of mass 0.113 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 0.19 m, and the ball has radius r << R.
(a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop?
(b) If the ball is released at height h = 5R, what is the magnitude of the horizontal force component acting on the ball at point Q?
A) .513 meters
B) ????? i just need B. its not 6.328
1 Answer
Assuming figure below is the correct figure.
(a) Done for sake of completeness.
This is what I get.
Explanation:
(a) Using Law of Conservation of energy
#"Initial Total E"="Final PE"+"Translational KE"+"Rotational KE"#
#mgh=mg(2R)+1/2mv^2+1/2Iomega^2#
where#omega=v/r# , as for rolling without slipping, the linear velocity#v# of the center of mass is equal to the angular velocity#omega# times the radius#r# and moment of inertia of a solid spherical ball is given as#I=2/5mr^2# .
With these substitutions above equation becomes
#mgh=mg(2R)+1/2mv^2+1/2(2/5mr^2)(v/r)^2#
#=>gh=2gR+1/2v^2+1/5v^2#
#=>gh=2gR+7/10v^2# ......(1)
When the ball is on the verge of leaving the track as it reaches the top of the loop, its weight must be equal to modulus of centrifugal force.
#(mv_min^2)/R=mg#
#=>v_min^2=gR# ........(2)
Substituting in (1) we get
#gh=2gR+7/10gR#
#=>h=27/10R#
Inserting given value of
#h=2.7xx0.19=0.513\ m#
(b) Using similar approach of Law of Conservation of energy for point
#mg(5R)=mg(R)+1/2mv_Q^2+1/2(2/5mr^2)(v_Q/r)^2#
#=>4mgR=7/10mv_Q^2#
We know that horizontal force component
#F=4mgxx10/7#
#F=40/7xx0.113/1000xx9.81#
#=>F=0.006\ N#
