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A uniform circular disc placed on a horizontal rough surface has initially a velocity `v_(0)` and an angular velocity `omega_(0)` as shown in the figure. The disc comes to rest after moving some distance in the direction of motion. Then `v_(0)//omega_(0)` is `:` A. `r//2`B. `r`C. `3r//2`D. 2 |
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Answer» Correct Answer - A `mV_(0)R-(mR^(2))/(2).omega_(0)=0` `(V_(0))/(omega_(0))=(R)/(2)` |
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