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4 May, 04:30

A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m. The acceleration of gravity is 9.8 m/s 2. 1.1 m 0.2 m B D C A 3.7 g Find the normal force pressing on the track at A, where A is at the same level as the center of the loop. Answer in units of N.

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  1. 4 May, 06:45
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    Normal force = 0.326N

    Explanation:

    Given that:

    mass released from rest at C = 3.7 g = 3.7 * 10⁻³ kg

    height of the mass = 1.1 m

    radius = 0.2 m

    acceleration due to gravity = 9.8 m/s²

    We are to determine the normal force pressing on the track at A.

    To to that;

    Let consider the conservation of energy relation; which says:

    mgh = mgr + 1/2 mv²

    gh = gr + 1/2 v²

    gh - gr = 1/2v²

    g (h-r) = 1/2v²

    v² = 2g (h-r)

    However; the normal force will result to a centripetal force; as such, using the relation

    N = mv²/r

    replacing the value for v² = 2g (h-r) in the above relation; we have:

    Normal force = 2mg (h-r) / r

    Normal force = 2 * 3.7 * 10⁻³ * 9.8 (1.1 - 0.2) / 0.2

    Normal force = 0.065268/0.2

    Normal force = 0.32634 N

    Normal force = 0.326N
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