Physics, 31.03.2020 04:39, ramseynikki87

A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence of gravity. What is the linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

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answered: 999supergirl

The linear velocity is v=4.08m/s

Explanation:

According to the law of conservation of energy

The potential energy possessed by the hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

The kinetic energy can be mathematically represented as

$KE = \frac{mv^2}{2} + \frac{Iw}{2}$

Where is the moment of inertia possessed by the hoop which is mathematically represented as

I = mr^2

Here R is the radius of the hoop

is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

$w = \frac{v}{r}$

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

Now expressing the above statement mathematically

$potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}$

$mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}$

=> $mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}$

=> $mgh = \frac{mv^2}{2} + \frac{mv^2}{2}$

=> mgh = mv^2

=> $v = \sqrt{gh}$

Substituting values

$v = \sqrt{9.81 * 1.7}$

v=4.08m/s

answered: dayaan009

vcm = 4.08 m/s

Explanation:

Assuming the inclined plane is frictionless, and that the hoop is rolling without slippling, the total mechanical energy of the hoop must be conserved:

$\Delta K + \Delta U = 0 (1)$

As the hoop starts from rest, the change in kinetic energy will be equal to the final kinetic energy of the hoop.This kinetic energy will have a part of rotational kinetic energy (due to the rotation around an axis passing through the center of the hoop), and a translational part, due to the translation of the center of mass.This kinetic energy can be written as follows:

$K_{tot} = \frac{1}{2} * I_{hoop} *\omega^{2} + \frac{1}{2} * m *v_{cm}^{2} (2)$

where Ihoop, is the moment of inertia of the hoop = m*r².If the hoop rolls without slipping, there exists a fixed relationship between the angular velocity ω and the velocitity of the center of mass vcm, as follows:

$\omega = \frac{v_{cm} }{r}$

Replacing the values of I and ω in (2), we have:

$K_{tot} = \frac{1}{2} * m*r^{2} * (\frac{v_{cm} }{r} )^{2} + \frac{1}{2} * m *v_{cm}^{2} = \\ \\ K_{tot} = m*v_{cm} ^{2}$

This change in kinetic energy must be equal to the change in gravitational potential energy.If we choose as our zero reference level the bottom of the incline, the change in gravitational potential energy is as follows:

$\Delta U = U_{f} - U_{o} = 0 - m*g*h = - m*g*h$