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Chapter 6: Problem 9

A ring rolls without slipping on the ground. Its centre \(C\) moves with a constant speed \(u . P\) is any point on the ring. The speed of \(P\) with respect to the ground is \(v\). (A) \(0 \leq v \leq 2 u\) (B) \(v=u\), if \(C P\) is horizontal (C) \(v=u\), if \(C P\) makes an angle of \(30^{\circ}\) with the horizontal and \(P\) is below the horizontal level of \(C\) (D) \(v=\sqrt{2} u\), if \(C P\) is horizontal

Short Answer

Expert verified
The short answer to the question is: For a ring rolling without slipping on the ground with its center $C$ moving at a constant speed $u$, the speed of any point $P$ on the ring with respect to the ground is $v$. The correct options are: (A) \(0 \leq v \leq 2 u\) (C) \(v=u\), if $CP$ makes an angle of \(30^{\circ}\) with the horizontal and $P$ is below the horizontal level of $C$ (D) \(v=\sqrt{2} u\), if $CP$ is horizontal

Step by step solution

01

General velocity relation for point P

Let \(V_C\) be the velocity of the center of the ring, \(C\), and \(V_P\) the velocity of the point \(P\) relative to \(C\). Since ring rolls without slipping, the point P will have a linear velocity \(V'_P\) due to rotation. The velocity of P with respect to the ground is \(V_P = V_C + V'_P\).
02

When CP is horizontal

In this case, the linear velocity of point P due to rotation \(V'_P\) is vertical. Since \(V_C\) is horizontal and of magnitude u, we can use the Pythagorean theorem to find the magnitude of the velocity of point P: \[v = \sqrt{(V_C)^2 + (V'_P)^2} = \sqrt{u^2 + u^2} = u\sqrt{2}\]
03

When CP makes an angle of \(30^\circ\) with the horizontal and P is below C

Let's analyze the components of \(V'_P\). In this case, the horizontal component of \(V'_P\) is \(V'_P\cos\left(30^\circ\right)\), and the vertical component is \(V'_P\sin\left(30^\circ\right)\). Hence, we have the horizontal and vertical components of the velocity of point P: Horizontal component: \(V_C + V'_P\cos\left(30^\circ\right) = u + u\frac{\sqrt{3}}{2} = u\left(1 + \frac{\sqrt{3}}{2}\right)\) Vertical component: \(V'_P\sin\left(30^\circ\right) = u\left(\frac{1}{2}\right)\) Now, using the Pythagorean theorem again to find the magnitude of the velocity of point P: \[v = \sqrt{\left(u\left(1 + \frac{\sqrt{3}}{2}\right)\right)^2 + \left(u\left(\frac{1}{2}\right)\right)^2} = u\]
04

Check against the given options

(A) General velocity relation for P tells us that \(0 \leq v \leq 2u\) is true. (B) Since the velocity \(v\) is equal to \(u\sqrt{2}\) when CP is horizontal, option (B) is false. (C) The velocity \(v\) is equal to \(u\) when CP makes an angle of \(30^\circ\) with the horizontal and P is below C, so option (C) is true. (D) We have found that \(v = u\sqrt{2}\) when CP is horizontal, so option (D) is true. The correct options are (A), (C), and (D).

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