A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-the loop whi

Question

A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-the loop which has a radius 5 m. The acceleration of gravity is 9.8 m/s^2.

Required:
How large is the normal force on it if its mass is 5g?

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Neala 3 years 2021-08-22T20:52:25+00:00 1 Answers 26 views 0

Answers ( )

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    2021-08-22T20:53:41+00:00

    Answer:

     N₁ = 393.96 N   and  N = 197.96 N

    Explanation:

    In This exercise we must use Newton’s second law to find the normal force. Let’s use two points the lowest and the highest of the loop

    Lowest point, we write Newton’s second law n for the y-axis

              N -W = m a

    where the acceleration is ccentripeta

              a = v² / r

               

              N = W + m v² / r

              N = mg + mv² / r

             

    we can use energy to find the speed at the bottom of the circle

    starting point. Highest point where the ball is released

               Em₀ = U = m g h

    lowest point. Stop curl down

               Em_{f} = K = ½ m v²

               Emo = Em_{f}

               m g h = ½ m v²

               v² = 2 gh

    we substitute

                 N = m (g + 2gh / r)

                N = mg (1 + 2h / r)

    let’s calculate

              N₁ = 5 9.8 (1 + 2 17.6 / 5)

              N₁ = 393.96 N

    headed up

    we repeat the calculation in the longest part of the loop

              -N -W = – m v₂² / r

                N = m v₂² / r – W

                 N = m (v₂²/r  – g)

    we seek speed with the conservation of energy

               Em₀ = U = m g h

    final point. Top of circle with height 2r

                 Em_{f} = K + U = ½ m v₂² + mg (2r)

                  Em₀ =   Em_{f}

                mgh = ½ m v₂² + 2mgr

                 v₂² = 2 g (h-2r)

    we substitute

                N = m (2g (h-2r) / r – g)

                N = mg (2 (h-r) / r 1) = mg (2h/r  -2 -1)

                 N = mg (2h/r  – 3)

                N = 5 9.8 (2 17.6 / 5 -3)

                N = 197.96 N

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