Question

a solid sphere of radius R is gently placed on a rough horizontal ground with an initial angular speed Omega and no linear speed linear speed of sphere when it is start pure rolling is​

Answer 1

Question :

A SOLID SPHERE OF RADIUS R IS GENTLY PLACED ON A ROUGH HORIZONTAL GROUND WITH AN INITIAL ANGULAR SPEED w and no linear speed. LINEAR SPEED OF SPHERE WHEN IT STARTS PURE ROLLING IS WHAT?

Answer :

When the sphere starts rolling, the linear velocity, 'v' becomes equal to 'Rω' (condition for pure rolling)

                 v = at = $(\frac{f}{m})$ t           -------- (1)

                ω = $w_{0}$ - ∝t

                    = $w_{0}$ - $(\frac{fR}{I})$ t

  ⇒          ω = $w_{0}$ $\frac{fR}{\frac{2}{5} mR^{2} }$ t      -----(2)

  ⇒          v = Rω

  ⇒                $(\frac{f}{m})$ t

                  = Rω - $(\frac{5f}{2m})$ t

  ⇒             $(\frac{f}{m})$ t

               

                  = $(\frac{2}{7})$ R $w_{0}$ = v

Answer 2

Answer:

The linear speed of the sphere is 2/7Rωo .

Explanation:

Rω will be the linear velocity when the body or here the sphere starts to pure roll .

Hence, as the speed from Newtons law of motion v=at or v =(f/m)t, again we know that the ω = (ωo - at) in this  a is the angular acceleration.

Therefore, we will get ω = ωo - (fR/I)t.(I is the moment of inertia).

Since, I is 2/5mR^2 for a sphere of radius R, now ω=ωo - [fR/(2/5mR^2)]t. On multiplying both the sides of the equation with R we will get that                 (f/m)t = 2/7Rωo which will be equal to the linear velocity of the sphere when it starts pure rolling.