Question

A sphere starts rolling down an incline of inclination theta find the speed of its centre and has covered distance l

Answer 1

Explanation:

Given A sphere starts rolling down an incline of inclination theta find the speed of its centre and has covered distance  l

Now sphere is rolling down so potential energy is being reduced.

So reduced potential energy = Kinetic energy (linear) + Rotational energy.

This will be rolling and point of contact has not slipped and velocity at any instant = R ω c

So decrease in potential energy = Increase in linear energy + Rotational kinetic energy.

       So mgl sin θ = 1/2 m v^2 + 1/2 I ω^2

                          = 1/2 mv^2 + 1/2 x 2/5 mr^2 x v^2 / R^2

                             = 1/2 mv^2 + mv^2 / 5

Now mgl sin θ = 7 v^2 / 10

Or v = √10 gl sin θ / 7

Answer 2

The speed of the centre when the sphere has covered a distance l is $v=\left(\frac{10}{7} g l \sin \theta\right)^{1 / 2}.$

Explanation:

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

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

Where m is the sphere’s mass, and r is the sphere’s radius.

Therefore,$\frac{K 2}{r 2}=\frac{2}{5}$.

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

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

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

$\text { And, } v^{2}=u^{2}+2 a s$

$=0+2 a l$

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

Therefore, $\boldsymbol{v}=\left(\frac{10}{7} g l \sin \theta\right)^{1 / 2}$.