| A wheel of radius R is rolling along a horizontal surface with a speed \[u.\] A pebble trapped on the wheel gets separated from the highest point of the wheel arrives at position P (figure). The horizontal range PQ of the pebble is: |
 |
A) \[u\,\sqrt{R/g}\]
B) \[2u\,\sqrt{R/g}\]
C) \[4u\,\sqrt{R/g})\]
D) \[{{u}^{2}}\,\sqrt{R/g}\]
Correct Answer: C
Solution :
| Considering the vertical motion of the projectile (pebble), the time taken to reach the ground after leaving the wheel is given by \[2R=0+(1/2)g{{t}^{2}},\] from which t=2√(R/g) [We have used the equation of the one dimensional motion, \[x={{x}_{0}}+{{v}_{0}}t+(1/2)a{{t}^{2}}]\]. The horizontal range of the pebble is PQ = horizontal velocity \[\times \]time of flight The center of mass of the wheel is moving with speed u. The pebble at the topmost point of the wheel is moving horizontally with speed u with respect to the wheel so that the horizontal velocity of the pebble with respect to the ground is \[u+u=2u.\] Therefore, horizontal range \[PQ=2u\times 2\surd (R/g)=4u\surd (R/g)\] |
You need to login to perform this action.
You will be redirected in
3 sec
