[amp_mcq option1=”1 rad\/sec2″ option2=”4 rad\/sec2″ option3=”6 rad\/sec2″ option4=”12 rad\/sec2″ correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{B}}$.<\/p>\n
The angular acceleration is the rate of change of angular velocity. It is calculated by taking the derivative of the angular velocity with respect to time. In this case, we are given that the angular distance is $0 = 2t^3 – 3t^2$. We can differentiate this equation to find the angular velocity:<\/p>\n
$$\\omega = \\frac{d}{dt} 0 = 6t^2 – 6t$$<\/p>\n
We can then differentiate the angular velocity equation to find the angular acceleration:<\/p>\n
$$\\alpha = \\frac{d}{dt} \\omega = 12t – 6$$<\/p>\n
We are asked to find the angular acceleration at $t = 1$ sec. Substituting this value into the angular acceleration equation, we get:<\/p>\n
$$\\alpha = 12(1) – 6 = 6 \\text{ rad\/sec}^2$$<\/p>\n
Therefore, the angular acceleration at $t = 1$ sec. is $\\boxed{\\text{B}}$, 6 rad\/sec$^2$.<\/p>\n
The other options are incorrect because they do not represent the correct value of the angular acceleration at $t = 1$ sec.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”1 rad\/sec2″ option2=”4 rad\/sec2″ option3=”6 rad\/sec2″ option4=”12 rad\/sec2″ 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-6672","post","type-post","status-publish","format-standard","hentry","category-applied-mechanics-and-graphic-statics","no-featured-image-padding"],"yoast_head":"\n