A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h = 3.85R. (a) What is its speed at point ? (b) How large is the normal force on the bead at point if its mass is 5.55 g?

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A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the
bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height
h = 3.85R.
R
(a) What is its speed at point ?
(b) How large is the normal force on the bead at point if its mass is 5.55 g?
Part 1 of 4 - Conceptualize
Since the bead is released above the top of the loop, it starts with enough potential energy to reach point and
still have excess kinetic energy. The energy of the bead at point is proportional to h and g. If it is moving
relatively slowly, the track will exert an upward force on the bead, but if it is whipping around quickly, the normal
force will push it toward the center of the loop.
Transcribed Image Text:A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h = 3.85R. R (a) What is its speed at point ? (b) How large is the normal force on the bead at point if its mass is 5.55 g? Part 1 of 4 - Conceptualize Since the bead is released above the top of the loop, it starts with enough potential energy to reach point and still have excess kinetic energy. The energy of the bead at point is proportional to h and g. If it is moving relatively slowly, the track will exert an upward force on the bead, but if it is whipping around quickly, the normal force will push it toward the center of the loop.
(a) We define the bottom of the loop as the zero level for the gravitational potential energy. For the total energy
of the system at point 4, we have
U¡ + K¡ = U₁ + Kf.
Now we know that U₁
height 3.85
Your response differs from the correct answer by more than 10%. Double check your calculations.R with speed v,
so we have
=
mgh and Ki = 0 because the bead starts from rest at height h. At point, the bead is at
mgh + 0 = mg 5.55
mg(
X
Your response differs from the correct answer by more than 100%.R) +
=
Substituting the given expression for h in terms of R, gives
mg 3.85
R) = mg (
X
Your response differs from the correct answer by more than 100%.R) +
Now we can solve for v in terms of R and g.
v² = g(
R
gR
2) + 1/2mv².
5.55
%.R) + 1/2mv².
Transcribed Image Text:(a) We define the bottom of the loop as the zero level for the gravitational potential energy. For the total energy of the system at point 4, we have U¡ + K¡ = U₁ + Kf. Now we know that U₁ height 3.85 Your response differs from the correct answer by more than 10%. Double check your calculations.R with speed v, so we have = mgh and Ki = 0 because the bead starts from rest at height h. At point, the bead is at mgh + 0 = mg 5.55 mg( X Your response differs from the correct answer by more than 100%.R) + = Substituting the given expression for h in terms of R, gives mg 3.85 R) = mg ( X Your response differs from the correct answer by more than 100%.R) + Now we can solve for v in terms of R and g. v² = g( R gR 2) + 1/2mv². 5.55 %.R) + 1/2mv².
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