A wheel of radius 1 text{ m} rolls forward ha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The horizontal distance covered by the wheel in half a revolution is equal to half of its circumference,which is $\pi R$.The vertical displacement

Vedclass · Accepted Answer

$\sqrt{\pi^2 + 4}$

Vedclass · Answer

(C) The horizontal distance covered by the wheel in half a revolution is equal to half of its circumference,which is $\pi R$.The vertical displacement of the point initially in contact with the ground is equal to the diameter of the wheel,which is $2R$.Let the initial position of the point be $A$ and the final position be $A'$. The displacement is the vector sum of the horizontal and vertical displacements.The magnitude of the displacement is given by $\sqrt{(\pi R)^2 + (2R)^2}$.Given $R = 1 	ext{ m}$,the magnitude of the displacement is $\sqrt{(\pi 	imes 1)^2 + (2 	imes 1)^2} = \sqrt{\pi^2 + 4} 	ext{ m}$.

$A$ wheel of radius $1 \text{ m}$ rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is

Similar Questions

If a solid sphere is rolling without slipping on a horizontal plane,then the ratio of its rotational and total kinetic energies is

The ratio of the total kinetic energy to the rotational kinetic energy for a rolling disc is:

Obtain the necessary condition $v_{cm} = R\omega$ for a body rolling without slipping.

$A$ wheel of radius $1 \ m$ rolls through $180^{\circ}$ over a plane surface. The magnitude of the displacement of the point of the wheel initially in contact with the surface is:

$A$ solid sphere rolls without slipping on a rough surface and the centre of mass has a constant speed $v_0$. If the mass of the sphere is $m$ and its radius is $R$,then find the angular momentum of the sphere about the point of contact $P$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module