","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Angular velocity Quiz 1","text":"A man in a circus is rotating a long stick of length 6 m by holding it exactly at its center. Mass of this stick is assumed to be 2 kg. What will be the angular velocity of the tip of the stick if it is rotating with an energy of 30 J? ","comment":{"@type":"Comment","text":" Rotational kinetic energy is given by the equation, $E=\\dfrac{1}{2}I{{\\omega }^{2}}$ . The moment of inertia of a rod about its center is given by $\\dfrac{m{{l}^{2}}}{12}$. Here, the length will be measured from the center. By rearranging the formula for energy and substituting all known quantities, we can find angular velocity. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{20}\\text{ rad/s}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{80}\\text{ rad/s}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{10 rad/s}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{40}\\text{ rad/s}$$ ","position":0,"answerExplanation":{"@type":"Comment","text":"$$ \\text{ Rotational kinetic energy is given by the equation,}\\\\ E=\\dfrac{1}{2}I{{\\omega }^{2}}\\\\ \\text{Here, the man is rotating a rod about its center. The moment of inertia of a rod about its center is given by,}\\\\ I=\\dfrac{m{{l}^{2}}}{12}\\\\ \\Rightarrow l=3\\text{ }m\\\\ \\text{As he is rotating it about the center. So,}\\\\ I=\\dfrac{m{{l}^{2}}}{12}\\\\ \\Rightarrow I =\\dfrac{2\\times 3\\times 3}{12}\\\\ \\Rightarrow I=\\dfrac{3}{2}\\text{ kg-}{{\\text{m}}^{\\text{2}}}\\\\ \\text{Now, angular velocity can be found using the formula,}\\\\ E=\\dfrac{1}{2}I{{\\omega }^{2}} \\\\ \\Rightarrow \\omega =\\sqrt{\\dfrac{2E}{I}} \\\\ \\Rightarrow \\omega =\\sqrt{\\dfrac{2\\times 30}{\\dfrac{3}{2}}}\\\\ \\Rightarrow \\omega =\\sqrt{\\dfrac{4\\times 30}{3}}\\\\ \\therefore \\omega=\\sqrt{40}\\,rad/s \\\\ \\text{Therefore, angular velocity of the road will be} \\sqrt{40}\\text{ rad/s}.$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Angular velocity Quiz 1","text":"A wheel is rotating about 20 rotations per minute initially. Suddenly a torque of 20 N-m is applied on the wheel for 2 seconds. Then what will be the final angular velocity of the wheel if it is having a moment of inertia of $$4\\text{ kg-}{{\\text{m}}^{2}}$$? ","comment":{"@type":"Comment","text":"Torque is defined as the product of moment of inertia of a body and its angular acceleration. That is, $\\tau =I\\times \\dfrac{{{\\omega }_{f}}-{{\\omega }_{i}}}{\\Delta t}$. Substituting quantities in this equation will give the final angular velocity. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 10 rpm","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 20 rpm","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 40 rpm","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 30 rpm","position":2,"answerExplanation":{"@type":"Comment","text":"$$ \\text{Torque is given by the relation,}\\\\ \\tau =I\\alpha \\\\ \\Rightarrow \\tau =I\\times \\dfrac{\\Delta \\omega }{\\Delta t} \\\\ \\tau =I\\times \\dfrac{{{\\omega }_{f}}-{{\\omega }_{i}}}{\\Delta t}\\\\ \\Rightarrow {{\\omega }_{f}}=\\dfrac{\\tau \\times \\Delta t}{I}+{{\\omega }_{i}} \\\\ \\Rightarrow {{\\omega }_{f}}=\\dfrac{20\\times 2}{4}+20\\\\ \\therefore {{\\omega }_{f}} =30\\text{ }rpm \\\\ \\text{Therefore, final angular velocity of the wheel will be}\\,30\\text{ }rpm.$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}