KCET · Physics · Rotational Motion
Three bodies a ring \( (R) \), a solid cylinder \( (C) \) and a solid sphere \( (S) \) having same mass and same radius roll down the inclined plane without slipping. They start from rest, if \( v_{R}, v_{c} \) and \( v_{s} \) are velocities of respective bodies on reaching the bottom of the plane, then
- A \( v_{R}=v_{C}=v_{S} \)
- B \( v_{R}>v_{C}>v_{S} \)
- C \( v_{R} < v_{C} < v_{S} \)
- D \( v_{R}=v_{C}>v_{S} \)
Answer & Solution
Correct Answer
(C) \( v_{R} < v_{C} < v_{S} \)
Step-by-step Solution
Detailed explanation
The velocity of a body rolling without slipping down oninclined plane is given as
\(v=\sqrt{\frac{2 g h}{1+K^{2} / R^{2}}}\)
Here, \( K \) is the radius of gyration.
For ring, \( K=R \)
\(\Rightarrow v_{R}=\sqrt{g h}\)
For solid sphere, \( K=\sqrt{\frac{2}{5}} R \)
\(\Rightarrow v_{s}=\sqrt{\frac{10 g h}{7}}\)
For solid cylinder, \( K=\frac{R}{\sqrt{2}} \)
\(\Rightarrow v_{c}=\sqrt{\frac{4 g h}{3}}\)
\(v=\sqrt{\frac{2 g h}{1+K^{2} / R^{2}}}\)
Here, \( K \) is the radius of gyration.
For ring, \( K=R \)
\(\Rightarrow v_{R}=\sqrt{g h}\)
For solid sphere, \( K=\sqrt{\frac{2}{5}} R \)
\(\Rightarrow v_{s}=\sqrt{\frac{10 g h}{7}}\)
For solid cylinder, \( K=\frac{R}{\sqrt{2}} \)
\(\Rightarrow v_{c}=\sqrt{\frac{4 g h}{3}}\)
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