Saved Bookmarks
| 1. |
A circular ring of mass 10 kg rolls along a horizontal floor. The center of mass of the ring has a speed 1.5 m/s. The work required to stop the ring is |
|
Answer» Solution :Given mass of a circular ring, m = 10kg speed of CENTRE of mass of the ring i.e., linear speed of ring,u = 15 m/s `K_i = K_("rotational") + K_("linear") = 1/2 I omega^2 + 1/2 mv^2` Here , `I = mR^2 " and " omega= v/R` ` = 1/2 mR^2 (v/R)^2 + 1/2 mv^2 = 1/2 mv^2 + 1/2 mv^2 = mv^2` `= 10 xx (1.5)^2 = 22.5 J` According to work-energy theorem, work required to stop the ring ` omega`change in kinetic energy ` = K_f - K_i = 0 - 22.5 = -22.5J` |
|