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Chapter 6: Problem 29

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) \(a=2 R \alpha\)

Short Answer

Expert verified
The correct relationship between the translational acceleration (\(a\)) and the angular acceleration (\(\alpha\)) of a skidding disc is given by: \(a = R \times \alpha\). Therefore, the answer is option (A).

Step by step solution

01

Understand the concept of rolling without slipping

When an object undergoes rolling motion, it may be moving translationally and rotating about its axis. For the object to undergo rolling without slipping, the condition to be followed is that the velocity of the contact point with the ground should be zero. Let's consider the point of contact of the disk with the ground (marked O). The no-slip condition can be described as: \[ V_{cm} = R \times \omega \] Where, \(V_{cm}\) is the center of mass' velocity, \(R\) is the disc's radius, \(\omega\) is the angular velocity.
02

Write the relationship for the accelerations

Now, differentiating the equation \(V_{cm} = R \times \omega\) with respect to time, we will get the relationship between the accelerations. \[ a = R \times \alpha \] Where, \(a\) is the translational acceleration, \(\alpha\) is the angular acceleration.
03

Identify the correct relationship

As we derived in step 2, the correct relationship between the translational acceleration and the angular acceleration of a skid is: \[a = R \times \alpha\] Comparing this result with the given choices, we can see that the correct option is: (A) \(a = R \times \alpha\) Hence, the answer is option (A).

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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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