Two identical disks, with rotational inertia I (= 1/2 MR2), roll without slipping across a horizontal floor and then up inclines. Disk A rol

Question

Two identical disks, with rotational inertia I (= 1/2 MR2), roll without slipping across a horizontal floor and then up inclines. Disk A rolls up its incline without sliding. On the other hand, disk B rolls up a frictionless incline. Otherwise the inclines are identical. Disk A reaches a height 12 cm above the floor before rolling down again. Disk B reaches a height above the floor of:

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Huyền Thanh 6 months 2021-07-17T23:56:51+00:00 1 Answers 50 views 0

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    2021-07-17T23:57:51+00:00

    Answer:

    8cm

    Explanation:

    Here, two disc are identical and rolling on the horizontal surface

    Also,while disc is in rolling motion its kinetic energy is sum of rotational kinetic energy and transnational kinetic energy.

    Therefore,

    KE = \frac{1}{2}mv²+ \frac{1}{2}[tex]Iw[/tex]²

    For pure rolling of disc we have: v=Rw

    I=\frac{1}{2} mR²

    By substituting in KE eq, we get

    KE = \frac{1}{2}mv²+ \frac{1}{2}(\frac{1}{2} mR²)(\frac{v^{2} }{R^{2} })

    KE= \frac{1}{2} mv^{2} + \frac{1}{4} mv^{2}

    The total kinetic energy will convert into gravitational potential energy when disc roll over the inclined surface.

    mgH=\frac{1}{2} mv^{2} + \frac{1}{4} mv^{2} =>\frac{3}{4} mv^{2}

    mv^{2}= 4/3mgH

    If another disc rolls up on frictionless inclined plane then it will lose all its translational kinetic energy but rotational kinetic energy will remain as it is as there is no torque on the disc

    Therefore, mgh=  1/2mv²

    1/2mv²= 4/3mgH

    mgh=2/3mgH

    h=2/3H

    Height = 12cm is given

    h= 8cm

    Thus,  Disk B reaches a height of 8cm above the floor.

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