A) Solid sphere
B) Hollow sphere
C) Disc
D) All will reach in same time
Correct Answer: A
Solution :
Key Idea: In rolling without slipping acceleration of all the objects down the plane is different. Acceleration of rolling body down an inclined plane \[a=\frac{g\sin \theta }{1+\frac{I}{m{{R}^{2}}}}\] ?(i) For solid sphere; \[I=\frac{2}{5}M{{R}^{2}}\]\[\Rightarrow \]\[\frac{I}{M{{R}^{2}}}=\frac{2}{5}\] \[\therefore \] \[a=\frac{g\sin \theta }{1+\frac{2}{5}}=\frac{5}{7}g\sin \theta \] For hollow sphere;\[I=\frac{2}{5}M{{R}^{2}}\Rightarrow \frac{I}{M{{R}^{2}}}=\frac{2}{3}\] \[\therefore \] \[a=\frac{g\sin \theta }{1+\frac{2}{3}}=\frac{3}{5}g\sin \theta \] For disc;\[I=\frac{1}{2}M{{R}^{2}}\Rightarrow \frac{I}{M{{R}^{2}}}=\frac{1}{2}\] \[\therefore \]\[a=\frac{g\sin \theta }{1+\frac{1}{2}}=\frac{2}{3}g\sin \theta \] It is obvious that acceleration of solid sphere is maximum, so solid sphere will reach the ground in least time.You need to login to perform this action.
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