Question

A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is 8 m/s. The speed of the particle on the rim of the wheel at the same level as the center of the wheel, will be:

• A. \( 4\sqrt{2} \, \text{m/s} \)
• B. 8 m/s
• C. 4 m/s
• D. \( 8\sqrt{2} \, \text{m/s} \)

Hint:

The speed of a point on the rim of a rolling wheel is the sum of the translational speed and the rotational speed. At the highest point, these speeds add up.

Answer 1

The Correct Option is A

In this problem, we need to determine the speed of a particle on the rim of a wheel rolling on a plane surface, at the same level as the center of the wheel. Let's break down the problem step-by-step:

1. The given speed of the particle at the highest point of the rim is 8 m/s. When a wheel is rolling without slipping, the speed at the highest point of the rim is twice the velocity of the center of the wheel. Therefore, let the velocity of the center of the wheel be \( V \). We have:

\[ 2V = 8 \, \text{m/s} \]

\[ V = 4 \, \text{m/s} \]

1. Now, we need to find the speed of a particle on the rim of the wheel at the same level as the center of the wheel. At this point (horizontal from the center, either left or right on the rim), the velocity component due to rotation is perpendicular to the translational velocity of the wheel (i.e., center speed). The speed due to rotation at this point is \( V = 4 \, \text{m/s} \). The wheel moves with uniform translational speed, also \( 4 \, \text{m/s} \).
2. To find the resultant speed, we use the Pythagorean theorem since the translational and rotational velocities are perpendicular to each other:

\[ v_{\text{resultant}} = \sqrt{V^2 + V^2} = \sqrt{4^2 + 4^2} \]

\[ v_{\text{resultant}} = \sqrt{16 + 16} \]

\[ v_{\text{resultant}} = \sqrt{32} \]

\[ v_{\text{resultant}} = 4\sqrt{2} \, \text{m/s} \]

1. Thus, the speed of the particle on the rim of the wheel at the same level as the center of the wheel is \( 4\sqrt{2} \, \text{m/s} \).

Answer 2

Given that the speed of the particle at the highest point of the rim is 8 m/s, and the wheel is rolling without slipping, the speed at any point on the rim is the sum of the velocity of the center of the wheel and the velocity due to the rotational motion. 

Let: 
- \( V_B = 8 \, \text{m/s} \) (speed at the highest point of the rim), 
- \( V = 4 \, \text{m/s} \) (speed of the center of the wheel), 
- \( V_P = \sqrt{2}V \) (velocity at point \( P \)). 

Since the wheel is rolling without slipping, the speed at point \( P \) (which is the same level as the center of the wheel) will be: \[ V_P = \sqrt{2} \times 4 = 4\sqrt{2} \, \text{m/s} \] 
Thus, the correct answer is (1).