8.2 A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and thebead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a heighth = 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 - ConceptualizeSince the bead is released above the top of the loop, it starts with enough potential energy to reach point andstill have excess kinetic energy. The energy of the bead at point is proportional to h and g. If it is movingrelatively slowly, the track will exert an upward force on the bead, but if it is whipping around quickly, the normalforce 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 energyof the system at point 4, we haveU¡ + K¡ = U₁ + Kf.Now we know that U₁height 3.85Your 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 atmgh + 0 = mg 5.55mg(XYour response differs from the correct answer by more than 100%.R) +=Substituting the given expression for h in terms of R, givesmg 3.85R) = mg (XYour 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(RgR2) + 1/2mv².5.55%.R) + 1/2mv².

Name: 8.2 A bead with a hole through it slides on a wire track. The wire is
Uploaded: 2024-03-20T06:57:16.000Z
Channel: Bartleby
Description: 8.2 A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and thebead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a heighth = 3.85R.R(a) What