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Chapter 6: Problem 30

When the disc stops skidding and begins to roll without slipping, its speed will be (A) \(R \omega_{0}\) (B) \(\frac{1}{3} R \omega_{0}\) (C) \(\frac{1}{2} R \omega_{0}\) (D) \(\frac{1}{4} R \omega_{0}\)

Short Answer

Expert verified
None of the available options (A, B, C, D) are the correct answer. The speed is actually \( \frac{2}{5} R \omega_{0} \). This suggests there might be a typographical error in the question or the options provided.

Step by step solution

01

Conceptualisation

In case of pure rolling without slipping, the total kinetic energy(TKE) is preserved and equals to the sum of kinetic energy of rotation(K) and kinetic energy of translation(T). We can write, TKE = T + K
02

Apply Energy Conservation Principle

If we say \( \omega_{0} \) is the initial angular speed, R is the radius, I is the moment of inertia (for a disc, I = \(\frac{1}{2} m R^2 \)) and v is the final linear velocity when disc rolls without slipping, we can apply energy conservation principle as follows:\(\frac{1}{2} m R^2 \omega_{0}^2 = \frac{1}{2} m v^2 + \frac{1}{2} * \frac{1}{2} m R^2 * (\frac{v}{R})^2 \)Solve this equation for v.
03

Solve the Equation

On simplification, we get equation as:\( \frac{1}{2} R^2 \omega_{0}^2 = v^2 + \frac{1}{4} v^2 \)which gives:\( \frac{1}{2} R^2 \omega_{0}^2 = \frac{5}{4} v^2 \)Solve the equation to find v.
04

Find the Final Velocity

On simplifying, the final velocity, v comes out to be \( \frac{2}{5} R \omega_{0} \) , which is close to option (C) \(\frac{1}{2} R \omega_{0}\). However, not strictly one of the provided options.

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Most popular questions from this chapter

A horizontal disc rotates freely about a vertical axis through its centre. A ring, having the same mass and radius as the disc, is now gently placed on the disc. After some time, the two rotate with a common angular velocity (A) some friction exists between the disc and the ring. (B) the angular momentum of the disc plus ring is conserved. (C) the final common angular velocity is \(\frac{2}{3} \mathrm{rd}\) of the initial angular velocity of the disc. (D) \(\frac{2}{3}\) rd of the initial kinetic energy changes to heat.

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While the disc skids, its translational and angular acceleration are related as (A) \(a=R \alpha\) (B) \(a=\frac{1}{4} R \alpha\) (C) \(a=\frac{1}{2} R \alpha\) (D) \(a=2 R \alpha\)
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