| A disc of radius \[0.1\,m\] rolls without sliding on a horizontal surface with a velocity of \[6\,m/s\]. It then ascends a smooth continuous track as shown in figure. The height upto which it will ascend is :\[\left( g=10\,m/{{s}^{2}} \right)\] |
 |
A) \[2.4\,m\]
B) \[0.9\,m\]
C) \[2.7\,m\]
D) \[1.8\,m\]
Correct Answer: D
Solution :
| Let m be the mass of the disc. Then translational kinetic energy of the disc is: |
| \[{{k}_{r}}=\frac{1}{2}m{{v}^{2}}\] ...(i) |
| When it ascends on a smooth track its rotational kinetic energy will remain same while translational kinetic energy will go on decreasing. |
| At highest point. |
| \[{{k}_{r}}=mgh\] or \[\,\frac{1}{2}m{{v}^{2}}=mgh\] or \[h=\frac{{{v}^{2}}}{2g}=\frac{{{(6)}^{2}}}{2\times 10}=1.8m\] |
You need to login to perform this action.
You will be redirected in
3 sec
