A small block of mass m = 0.100 kg can slide along the frictionless loop-the-loo
ID: 2261896 • Letter: A
Question
A small block of mass m = 0.100 kg can slide along the frictionless loop-the-loop, with loop radius R = 0.150 m, as shown in the figure. The block is given an initial velocity along the track of vi = 2.00 m/s at the point P, at height h = 2.00R above the bottom of the loop.
(a) Determine the normal force that the track exerts on the block when it is at the point Q. Include a FBD in your answer, and clearly identify the laws of physics that you are using to solve the problem.
(b) Determine the normal force that the track exerts on the block when it is at the point T, at the top of the loop. Include a FBD in your answer, and clearly identify the laws of physics that you are using to solve the problem.
(c) S is a point higher than P. The ball is released from rest at S, and still makes it around the track without falling off the loop at the top. Determine the minimum height h, above the base of the loop, of the point S. Explain your reasoning carefully. Hint: The smallest possible normal force at the top of the loop is zero. If the ball is any slower at the top, it falls off the track.
A small block of mass m = 0.100 kg can slide along the frictionless loop-the-loop, with loop radius R = 0.150 m, as shown in the figure. The block is given an initial velocity along the track of vi = 2.00 m/s at the point P, at height h = 2.00R above the bottom of the loop. Determine the normal force that the track exerts on the block when it is at the point Q. Include a FBD in your answer, and clearly identify the laws of physics that you are using to solve the problem. Determine the normal force that the track exerts on the block when it is at the point T, at the top of the loop. Include a FBD in your answer, and clearly identify the laws of physics that you are using to solve the problem. S is a point higher than P. The ball is released from rest at S, and still makes it around the track without falling off the loop at the top. Determine the minimum height h, above the base of the loop, of the point S. Explain your reasoning carefully. Hint: The smallest possible normal force at the top of the loop is zero. If the ball is any slower at the top, it falls off the track.Explanation / Answer
a) energy conservation:
mgh = 0.5mv^2
=> v^2 = 2g(h-R)
Normal force = mv^2/R = 0.1*2g(h-R)/0.15 = 0.1*2*9.8*2*0.15/0.15 =3.92 N
b) energy conservation:
mgh = 0.5mv^2
=> v^2 = 2g(h-2R)
Normal force = mv^2/R = 0.1*2g(h-2R)/0.15 = 0 N
c) minimum height is h=2R any point below this will cause object to drop off before it reaches the top. in (b) we saw that at the point h the velocity is 0. any point of release lower than h will cause it to fall off
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