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A wheel is released on a rough horizontal floor after imparting it an initial horizontal velocity `v_(0)` and angular velocity` omega_(0)` as shown in the figure below. Point `O` is the centre of mass of the wheel and point `P` is its instantaneous point of contact with the groundl. The radius of wheel is `r` and its radius of gyration about `O` is `k`. Coefficient of friction between wheel and ground is `mu`. A is a fixed point on the ground. If above question , distance travelled by centre of mass of the wheel before it stops is `-`A. `(v_(0)^(2))/(2mug)(1+(r^(2))/(k^(2)))`B. `(v_(0)^(2))/(2mug)`C. `(v_(0)^(2))/(2mug)(1+(k^(2))/(r^(2)))`D. None of these |
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Answer» Correct Answer - B Torque of friction about `A` is zero. `a_(cm)=- mu g` `0^(2)=v_(0)^(2)-2 mugsrArr S=(v_(0)^(2))/(2mug)` |
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