Question

# An inclined plane makes an angle of $$30^{\circ}$$ with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to :

A
g3
B
2g3
C
5g7
D
5g14

Solution

## The correct option is D $$\dfrac {5g}{14}$$$$a=\cfrac { mg\sin\theta -F }{ m } \\ \alpha =\cfrac { FR }{ I } =\cfrac { 5FR }{ 2M{ R }^{ 2 } } =\cfrac { 5F }{ 2MR }$$For no slipping , $$R\alpha =a$$$$\cfrac { 5F }{ 2M } =g\sin\theta -\cfrac { F }{ M } \Rightarrow \cfrac { F }{ M } =\cfrac { 2 }{ 7 } g\sin\theta \\ a=g\sin\theta .\cfrac { F }{ M } =\cfrac { 5 }{ 7 } g\sin\theta =\cfrac { 5 }{ 7 } g\sin30\\ =\cfrac { 5 }{ 14 } g$$Physics

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