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:
Show Hint
- \( 4\sqrt{2} \, \text{m/s} \)
- 8 m/s
- 4 m/s
- \( 8\sqrt{2} \, \text{m/s} \)
The Correct Option is A
Approach Solution - 1
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:
- 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:
\[ 2V = 8 \, \text{m/s} \]
\[ V = 4 \, \text{m/s} \]
- 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 \( 4 \, \text{m/s} \).
- To find the resultant speed, we use the Pythagorean theorem since the translational and rotational velocities are perpendicular to each other:
\[ v_{\text{resultant}} = \sqrt{V^2 + V^2} = \sqrt{4^2 + 4^2} \]
\[ v_{\text{resultant}} = \sqrt{16 + 16} \]
\[ v_{\text{resultant}} = \sqrt{32} \]
\[ v_{\text{resultant}} = 4\sqrt{2} \, \text{m/s} \]
- Thus, the speed of the particle on the rim of the wheel at the same level as the center of the wheel is \( 4\sqrt{2} \, \text{m/s} \).
Approach Solution -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).
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Circular motion Questions
- If the radius of curvature of the path of two particles of the same mass are in the ratio \( 3:4 \), then in order to have constant centripetal force, their velocities will be in the ratio of:
- A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in the string, when the stone is at the lowest point, is (if \( \pi^2 = 9.8 \) and \( g = 9.8 \, \text{m/s}^2 \))
- A bob of mass \( m \) is suspended by a light string of length \( L \). It is imparted a minimum horizontal velocity at the lowest point \( A \) such that it just completes a half circle, reaching the topmost position \( B \). The ratio of kinetic energies \( \frac{(\text{K.E.})_A}{(\text{K.E.})_B} \) is:
- A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at \( t = 0 \) is \( 4 \, \text{m/s} \), the time taken to complete the first revolution will be\[\frac{1}{\alpha} \left[ 1 - e^{2\pi} \right] \, \text{s},\]where \( \alpha = \, \underline{\hspace{2cm}} \).
- A ball of mass 0.5 kg is attached to a string of length 50 cm. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400 N. The maximum possible value of angular velocity of the ball in rad/s is,:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.