JEE Advanced 2017Paper 2
- Exam
- JEE Advanced
- Year
- 2017
- Paper
- Paper 2
- Subject
- Physics
- Chapter
- Rotational Motion
JEE Advanced 2017 · Paper 2 · Physics · Rotational Motion
Paragraph: One twirls a circular ring (of mass M and radius R )...
Paragraph:
One twirls a circular ring (of mass and radius ) near the tip of one's finger as shown in Figure 1 . In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is . The finger rotates with an angular velocity . The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is and the acceleration due to gravity is .
Question: The total kinetic energy of the ring is
Question: The total kinetic energy of the ring is
Options
- A
- BM{\omega}_{0}^{2}{\left(R - r\right)}^{2}
- C\dfrac{1}{2}M{\omega}_{0}^{2}{\left(R - r\right)}^{2}
- D\dfrac{3}{2}M{\omega}_{0}^{2}{\left(R - r\right)}^{2}
Correct answer
B. M{\omega}_{0}^{2}{\left(R - r\right)}^{2}
Step-by-step solution
{V}_{CM}= ( R - r ) {\omega}_{0}R{\omega}^{1} - \left(R - r\right){w}_{0} = r{\omega}_{0}R{\omega}^{1} - R{\omega}_{0} = 0= \dfrac{1}{2}M{\left(R - r\right)}^{2}{\omega}_{0}^{2} + \dfrac{1}{2}m{R}^{2}{\omega}_{0}^{2}r ≪ RR{\omega}_{0} \approx \left(R - r\right){\omega}_{0}KE \approx \dfrac{2}{2}M{\left(R - r\right)}^{2}{\omega}_{0}^{2}M{\omega}_{0}^{2}{\left(R - r\right)}^{2} is nearest possible option