Q129 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Physics
A solid sphere rolls down without slipping on a smooth inclined plane of inclination \(\sin ^{-1}(0.42)\). If the acceleration due to gravity is \(10 \mathrm{~ms}^{-2}\), the acceleration of the rolling sphere is
- A\(1 \mathrm{~ms}^{-2}\)
- B\(2 \mathrm{~ms}^{-2}\)
- Correct answerC\(3 \mathrm{~ms}^{-2}\)
- D\(4 \mathrm{~ms}^{-2}\)
Solution
Given, inclination of inclined plane,
\(\theta=\sin ^{-1}(0.42) \Rightarrow \sin \theta=0.42\)
Acceleration due to gravity,
\(g=10 \mathrm{~m} / \mathrm{s}^2\)
Acceleration of rolling solid sphere on the inclined plane without slipping is given by
\(a=\frac{g \sin \theta}{1+\frac{I}{M R^2}}\)
where, \(I=\) moment of inertia
\(M=\) mass of sphere
\(R=\) radius of sphere
But \(\quad I=\frac{2}{5} M R^2\)
\(\begin{aligned}
\therefore \quad a= & \frac{g \sin \theta}{\frac{2}{M R^2}}=\frac{5}{7} g \sin \theta \\
& 1+\frac{5}{M R^2} \\
& =\frac{5}{7} \times 10 \times 0.42=3 \mathrm{~m} / \mathrm{s}^2
\end{aligned}\)
Hence, the correct answer is option: 'C'.