Physics: Asolid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25.0 degrees with the horizontal. what s the speed of the disk after is has rolled 3.00m, measured along the plane?

Asolid disk is released from rest and rolls without

Ответ:

donavery24

11.01.2022

$v=4.07\frac{m}{s^2}$

Explanation:

According to the law of conservation of energy, when an object falls, its gravitational potential energy is changed to kinetic energy. In this case we have linear as well as rotational motion:

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

A solid disk has a moment of inertia equal to $\frac{mR^2}{2}$ . The height of the disk is given by $h=dsin\theta$ . Recall that $\omega=\frac{v}{R}$ . Replacing and solving for v:

$mgdsin(25^\circ)=\frac{mv^2}{2}+\frac{1}{2}\frac{mR^2}{2}\frac{v^2}{R^2}\\mgdsin(25^\circ)=\frac{3mv^2}{4}\\v^2=\frac{4gdsin(25^\circ)}{3}\\v=\sqrt{\frac{4(9.8\frac{m}{s^2})(3m)sin(25^\circ)}{3}}\\v=4.07\frac{m}{s^2}$

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Asolid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25.0 degrees with the horizontal. what's the speed of the disk after is has rolled 3.00m, measured along the plane?

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