Question 2: Centre of Mass and Rotational Motion

Question

A sphere pure rolls on a rough inclined
plane with initial velocity 2.8 m/s. Find the
maximum distance on the inclined plane.

Accepted Answer

Given, initial velocity of sphere, $$u=2.8 \mathrm{~m} / \mathrm{s}$$
Acceleration on the inclined plane,
$$a=\frac{g \sin 	heta}{1+\frac{k^2}{R^2}}$$ .... (i)
where, $$k$$ and $$R$$ are the radius of gyration and radius of sphere, respectively.
Moment of inertia of sphere,
$$\begin{aligned}
I & =m k^2=\frac{2}{5} m R^2 \
\frac{k^2}{R^2} & =\frac{2}{5} \quad 	ext{.... (ii)}
\end{aligned}$$
From Eqs. (i) and (ii), we have,
$$\begin{aligned}
a & =\frac{g \sin 30^{\circ}}{1+\frac{2}{5}} \quad\left[\because 	heta=30^{\circ}, 	ext { given }ight] \
& =\frac{5 g}{14}=\frac{25}{7} \mathrm{~m} / \mathrm{s}^2 \quad\left[	herefore g=10 \mathrm{~m} / \mathrm{s}^2ight]
\end{aligned}$$
If $$s$$ be the maximum distance travelled by the sphere, then at maximum distance, $$v=0$$.
$$	herefore$$ From equation of motion,
$$\begin{aligned}
v^2 & =u^2-2 a s \
0 & =u^2-2 a s \
\Rightarrow \quad s & =\frac{u^2}{2 a}=\frac{(2.8)^2 	imes 7}{2 	imes 25}=1.09 \mathrm{~m} \simeq 1.38 \mathrm{~m}
\end{aligned}$$

Answer

2.74 m

Answer

5.48 m

Answer

1.38 m

Answer

3.2 m

Centre of Mass and Rotational Motion

A sphere pure rolls on a rough inclined plane with initial velocity 2.8 m/s. Find the maximum distance on the inclined plane.

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