[amp_mcq option1=”50 J” option2=”150 J” option3=”200 J” option4=”400 J” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{B) 150 J}}$.<\/p>\n
The kinetic energy of a rolling object is given by the equation:<\/p>\n
$$KE = \\frac{1}{2}mv^2 + \\frac{1}{2}I\\omega^2$$<\/p>\n
where $m$ is the mass of the object, $v$ is the velocity of the center of mass, $I$ is the moment of inertia, and $\\omega$ is the angular velocity.<\/p>\n
In this case, we are given that $m = 4\\text{ kg}$, $v = 5\\text{ m\/s}$, and $I = 3\\text{ kg m}^2$. We can calculate the angular velocity from the equation:<\/p>\n
$$\\omega = \\frac{v}{r}$$<\/p>\n
where $r$ is the radius of the object. In this case, $r = 0.5\\text{ m}$, so $\\omega = 10\\text{ rad\/s}$.<\/p>\n
Substituting these values into the equation for kinetic energy, we get:<\/p>\n
$$KE = \\frac{1}{2}(4\\text{ kg})(5\\text{ m\/s})^2 + \\frac{1}{2}(3\\text{ kg m}^2)(10\\text{ rad\/s})^2 = 150\\text{ J}$$<\/p>\n
Therefore, the kinetic energy of the disc is $\\boxed{\\text{150 J}}$.<\/p>\n
Option A is incorrect because it is the kinetic energy of a point mass with the same mass and velocity as the disc. Option C is incorrect because it is the kinetic energy of a solid cylinder with the same mass and radius as the disc. Option D is incorrect because it is the kinetic energy of a hollow cylinder with the same mass and radius as the disc.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”50 J” option2=”150 J” option3=”200 J” option4=”400 J” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[645],"tags":[],"class_list":["post-6776","post","type-post","status-publish","format-standard","hentry","category-applied-mechanics-and-graphic-statics","no-featured-image-padding"],"yoast_head":"\n