A solid sphere of mass 2,kg is making pure ro | vedclass

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

Question

(A) The total kinetic energy $(KE)$ of a body in pure rolling motion is the sum of its translational kinetic energy and rotational kinetic energy.$KE

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The total kinetic energy $(KE)$ of a body in pure rolling motion is the sum of its translational kinetic energy and rotational kinetic energy.$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia about its centre of mass is $I = \frac{2}{5}mR^2$.For pure rolling,the relation between linear velocity $(v)$ and angular velocity $(\omega)$ is $v = R\omega$,or $\omega = \frac{v}{R}$.Substituting these into the energy equation:$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$$KE = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$Given $m = 2\,kg$ and $KE = 2240\,J$:$2240 = \frac{7}{10} 	imes 2 	imes v^2$$2240 = \frac{7}{5}v^2$$v^2 = \frac{2240 	imes 5}{7} = 320 	imes 5 = 1600$$v = \sqrt{1600} = 40\,m/s$.

$A$ solid sphere of mass $2\,kg$ is making pure rolling on a horizontal surface with kinetic energy $2240\,J$. The velocity of the centre of mass of the sphere will be $..........\,m/s$.

Similar Questions

$A$ solid sphere of radius $R$ is rolled by a force $F$ acting at the top of the sphere as shown in the figure. The sphere rolls without slipping on a rough stationary surface. Initially,the sphere is at rest. Then:

$A$ uniform solid cylindrical roller of mass $m$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its centre. If the acceleration of the cylinder is $a$ and it is rolling without slipping,then the value of $F$ is:

$A$ ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If the radius of the ball is $R$,then the fraction of total energy associated with its rotational energy is:

$A$ hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$. The value of $x$ is . . . . . . .

$A$ wheel is rolling along the ground with a speed of $2\ m/s$. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module