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 heighth-3.85R. A R (a) What is its speed at point ? (b) How large is the normal force on the bead at point if its mass is 4.85 g?
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 heighth-3.85R. A R (a) What is its speed at point ? (b) How large is the normal force on the bead at point if its mass is 4.85 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.
A
(a) What is its speed at point ?
(b) How large is the normal force on the bead at point if its mass is 4.85 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.
Part 2 of 4 - Categorize
The speed at the top of the loop can be found from the conservation of energy for the bead-track-Earth system and the normal force can be found from Newton's second law.
Part 3 of 4 - Analyze
(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, we have
U₁ + K₁ = Uf + Kf.
Now we know that U; = mgh and K; = 0 because the bead starts from rest at height h. At point, the bead is at height [
Your response differs from the correct answer by more than 100%.R with speed v, so we have
mgh + 0 = mg
= mg(C
X
Your response differs from the correct answer by more than 100%.R) + mv².
Substituting the given expression for h in terms of R, gives
mal
X
Your response differs from the correct answer by more than 10%. Double check your calculations.R) = mg( [
Your response differs from the correct answer by more than 100%.R) + mv².
Now we can solve for v in terms of R and g.
X
v²
= Your response differs from the correct answer by more than 100%. R
x
Your response differs from the correct answer by more than 100%.gR
x
X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F984d2a30-d316-417a-99a5-942cbb14579b%2Fff51fe62-8d59-4944-85d9-95768c3f0094%2Fph5t27j_processed.png&w=3840&q=75)
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.
A
(a) What is its speed at point ?
(b) How large is the normal force on the bead at point if its mass is 4.85 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.
Part 2 of 4 - Categorize
The speed at the top of the loop can be found from the conservation of energy for the bead-track-Earth system and the normal force can be found from Newton's second law.
Part 3 of 4 - Analyze
(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, we have
U₁ + K₁ = Uf + Kf.
Now we know that U; = mgh and K; = 0 because the bead starts from rest at height h. At point, the bead is at height [
Your response differs from the correct answer by more than 100%.R with speed v, so we have
mgh + 0 = mg
= mg(C
X
Your response differs from the correct answer by more than 100%.R) + mv².
Substituting the given expression for h in terms of R, gives
mal
X
Your response differs from the correct answer by more than 10%. Double check your calculations.R) = mg( [
Your response differs from the correct answer by more than 100%.R) + mv².
Now we can solve for v in terms of R and g.
X
v²
= Your response differs from the correct answer by more than 100%. R
x
Your response differs from the correct answer by more than 100%.gR
x
X
Expert Solution
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Step 1
Given data:
Height, h = 3.85R
mass , m = 4.85g
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