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. $\frac{g}{3}$
• B. $\frac{2g}{3}$
• C. $\frac{5g}{7}$
• D. $\frac{5g}{14}$

Answer 1

The correct option is D $\frac{5g}{14}$

$a=\frac{mg\sin \theta -F}{m}\alpha =\frac{FR}{I}=\frac{5FR}{2M{R}^2}=\frac{5F}{2MR}$

For no slipping , $R\alpha =a$

$\frac{5F}{2M}=g\sin \theta -\frac{F}{M}\Rightarrow \frac{F}{M}=\frac{2}{7}g\sin \theta a=g\sin \theta .\frac{F}{M}=\frac{5}{7}g\sin \theta =\frac{5}{7}g\sin 30=\frac{5}{14}g$