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:

Answer 1

Answer:

Disc B will reach to height of 8 cm height above the floor

Explanation:

As we know that two disc are identical and rolling on the horizontal surface

Now we know that while disc is in rolling motion its kinetic energy is sum of rotational kinetic energy and transnational kinetic energy.

So we have

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now we know that for pure rolling of disc

$v = R\omega$

$I = \frac{1}{2}mR^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v^2}{R^2})$

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

now when disc roll over the inclined surface then its total kinetic energy will convert into gravitational potential energy

so we will have

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

$mgH = \frac{3}{4}mv^2$

$mv^2 = \frac{4}{3}mgH$

here we know that H = 12 cm

now when 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

So we have

$mgh = \frac{1}{2}mv^2$

so we have

$mgh = \frac{2mgH}{3}$

so we have

$h = \frac{2}{3}H$

put H = 12 cm

$h = 8 cm$

#Learn

Topic : Rolling on inclined plane

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