The problem involves the concept of rolling motion, which is a combination of translational (linear) and rotational motion.

When the sphere starts rolling purely (without slipping), the condition that must be satisfied is V' = Rω', where V' is the final linear speed, R is the radius of the sphere, and ω' is the final angular speed.

Given that the initial angular speed is V/R, and assuming no external torques act on the sphere (which means angular momentum is conserved), we can write:

I_initial * ω_initial = I_final * ω_final

For a solid sphere, the moment of inertia I = 2/5 * m * R^2, where m is the mass of the sphere. Substituting this in, we get:

(2/5 * m * R^2) * (V/R) = (2/5 * m * R^2) * ω'

Solving for ω', we find that ω' = V/R.

Substituting this back into the condition for pure rolling, V' = Rω', we find that V' = R * (V/R) = V.

So, the linear speed of the sphere when slipping finally ceases and pure rolling starts is the same as the initial linear speed, V.

Question

The problem involves the concept of rolling motion, which is a combination of translational (linear) and rotational motion.

When the sphere starts rolling purely (without slipping), the condition that must be satisfied is V' = Rω', where V' is the final linear speed, R is the radius of the sphere, and ω' is the final angular speed.

Given that the initial angular speed is V/R, and assuming no external torques act on the sphere (which means angular momentum is conserved), we can write:

I_initial * ω_initial = I_final * ω_final

For a solid sphere, the moment of inertia I = 2/5 * m * R^2, where m is the mass of the sphere. Substituting this in, we get:

(2/5 * m * R^2) * (V/R) = (2/5 * m * R^2) * ω'

Solving for ω', we find that ω' = V/R.

Substituting this back into the condition for pure rolling, V' = Rω', we find that V' = R * (V/R) = V.

So, the linear speed of the sphere when slipping finally ceases and pure rolling starts is the same as the initial linear speed, V.

Knowee AI · Accepted Answer

The problem involves the concept of rolling motion, which is a combination of translational (linear) and rotational motion.

When the sphere starts rolling purely (without slipping), the condition that must be satisfied is V' = Rω', where V' is the final linear speed, R is the radius of the sphere, and ω' is the final angular speed.

Given that the initial angular speed is V/R, and assuming no external torques act on the sphere (which means angular momentum is conserved), we can write:

I_initial * ω_initial = I_final * ω_final

For a solid sphere, the moment of inertia I = 2/5 * m * R^2, where m is the mass of the sphere. Substituting this in, we get:

(2/5 * m * R^2) * (V/R) = (2/5 * m * R^2) * ω'

Solving for ω', we find that ω' = V/R.

Substituting this back into the condition for pure rolling, V' = Rω', we find that V' = R * (V/R) = V.

So, the linear speed of the sphere when slipping finally ceases and pure rolling starts is the same as the initial linear speed, V.

A solid sphere is set into motion on a rough horizontal surface with a linear speed V in the forward direction and an angular speed V/R in the anti clock wise direction as shown in figure. Find the linear speed of the sphere when slipping finally ceases and pure rolling starts?

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