Question

A sphere starts rolling down an incline of inclination O. Find the speed of its center when it has covered a distance L. Please it's urgent. The best answer will be marked Brainliest.

Answer 1

Answer:-

The speed of the centre when the sphere has covered a distance L is

$\large\bf{v}=(\sqrt\frac{10}{7}g l \sin \theta)$

• Concept:-

It is understood that the sphere is rolling with an inclination of $\theta$ in an inclined plane. So, as per the principle,

$\large{M \cdot i, I=M k^{2}=\frac{2}{5} M r^{2}}$

Where,

$\bf{m}$ is the sphere’s $\bf{mass}$

and

$\bf{r}$ is the sphere’s $\bf{radius}$

Therefore,

$\large\frac{K 2}{r 2}=\frac{2}{5}$

As we know that acceleration is $\\ \large{a=\frac{g \sin \theta}{1+k^{2} / r^{2}}}$

This is equal to $\\ \large\frac{g \sin \theta}{1}+\frac{2}{5}$

$=\large\frac{5}{7} \: g \sin \theta$

And,

$v^{2}=u^{2}+2 a s$

$=0+2 a l$

$V 2=\frac{2.5}{7} g \sin \theta \times l$

Therefore,

$\large\bf{v}=\left(\frac{10}{7} g l \sin \theta\right)^{1 / 2}$

or

$\large\bf{v}=(\sqrt\frac{10}{7}g l \sin \theta)$

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