Problem Analysis:
- The question concerns a wheel rolling without slipping on a plane surface.
- The velocity of the wheel's center is given as $v_0$.
- We need to find the speed of a particle on the rim located at the same vertical level as the center.
- The final speed is expressed as $\sqrt{x} v_0$, and we need to find the value of $x$.
Understanding Rolling Motion Velocities
For a wheel rolling without slipping:
- The velocity of the center ($\vec{v}_c$) is related to the angular velocity ($\omega$) and radius ($R$) by $v_0 = \omega R$.
- The velocity of any point on the rim ($\vec{v}_p$) relative to the ground is the vector sum of the center's velocity ($\vec{v}_c$) and the point's velocity relative to the center ($\vec{v}_{rel}$). That is, $\vec{v}_p = \vec{v}_c + \vec{v}_{rel}$.
- The magnitude of the velocity relative to the center is $|\vec{v}_{rel}| = \omega R = v_0$. This velocity is always tangential to the rim.
Determining Particle Velocity
- Define Coordinate System: Let the wheel move along the positive x-axis. The center's velocity is $\vec{v}_c = v_0 \hat{i}$. Assume clockwise rotation for forward motion, meaning the angular velocity vector is $\vec{\omega} = -\omega \hat{k}$.
- Locate the Particle: A particle "at the same level as the centre" means its vertical position is the same as the center's. In a standard coordinate system where the center is at height $R$, this corresponds to points on the rim where the angle $\theta$ (measured from the positive x-axis) satisfies $R \sin\theta = 0$. This occurs when $\theta = 0$ (point horizontally aligned and ahead of the center) or $\theta = \pi$ (point horizontally aligned and behind the center).
- Calculate Relative Velocity ($\vec{v}_{rel}$):
- For $\theta = 0$ (point ahead): The position vector from the center is $\vec{r} = R \hat{i}$. The relative velocity is $\vec{v}_{rel} = \vec{\omega} \times \vec{r} = (-\omega \hat{k}) \times (R \hat{i}) = -\omega R \hat{j} = -v_0 \hat{j}$.
- For $\theta = \pi$ (point behind): The position vector from the center is $\vec{r} = -R \hat{i}$. The relative velocity is $\vec{v}_{rel} = \vec{\omega} \times \vec{r} = (-\omega \hat{k}) \times (-R \hat{i}) = \omega R \hat{j} = v_0 \hat{j}$.
- Calculate Particle Velocity ($\vec{v}_p$): $\vec{v}_p = \vec{v}_c + \vec{v}_{rel}$.
- For $\theta = 0$: $\vec{v}_p = v_0 \hat{i} + (-v_0 \hat{j}) = v_0 \hat{i} - v_0 \hat{j}$.
- For $\theta = \pi$: $\vec{v}_p = v_0 \hat{i} + v_0 \hat{j}$.
- Calculate Speed: The speed is the magnitude of $\vec{v}_p$.
- For $\theta = 0$: Speed $= |\vec{v}_p| = \sqrt{(v_0)^2 + (-v_0)^2} = \sqrt{v_0^2 + v_0^2} = \sqrt{2v_0^2} = \sqrt{2} v_0$.
- For $\theta = \pi$: Speed $= |\vec{v}_p| = \sqrt{(v_0)^2 + (v_0)^2} = \sqrt{v_0^2 + v_0^2} = \sqrt{2v_0^2} = \sqrt{2} v_0$.
Finding the Value of x
The calculated speed is $\sqrt{2} v_0$. We are given that the speed is $\sqrt{x} v_0$. Equating these:
$ \sqrt{x} v_0 = \sqrt{2} v_0 $
Dividing both sides by $v_0$ (assuming $v_0 \neq 0$):
$ \sqrt{x} = \sqrt{2} $
Squaring both sides gives:
$ x = 2 $
Similar Questions
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Important Questions from Mechanics
- A soap bubble of surface tension $0.04 \text{ N/m}$ is blown to a diameter of $7 \text{ cm}$. If $(15000 - x) \text{ }\mu\text{J}$ of work is done in blowing it further to make its diameter $14 \text{ cm}$, then the value of $x$ is _______.
($\pi = 22/7$) - A uniform solid cylinder of length $L$ and radius $R$ has moment of inertia about its axis equal to $I_1$. A small co-centric cylinder of length $L/2$ and radius $R/3$ carved from this cylinder has moment of inertia about its axis equals to $I_2$. The ratio $I_1/I_2$ is _______.
- A fly wheel having mass $3\text{ kg}$ and radius $5\text{ m}$ is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to $3\text{ kg}$ mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by $3\text{ m}$ is _________ $\text{J}$. ($g = 10\text{ m/s}^{2}$)
- A body of density '$\rho$' is dropped slowly on the surface of a lake of depth $d$. If the density of the lake water be '$\rho'$' ($\rho' < \rho$) then the time taken by the body to reach the bottom of the lake is
Three blocks of masses $m_1 = 2\text{ kg}$, $m_2 = 3\text{ kg}$ and $m_3 = 5\text{ kg}$ are placed on a horizontal frictionless surface and a force of 30N pulls the system as shown below. The value of tension in the string between $m_2$ and $m_3$ will be
