A uniform sphere of mass 500,g rolls without | vedclass

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

Question

(A) The total kinetic energy $(K)$ of a rolling sphere is the sum of its translational kinetic energy and rotational kinetic energy.$K = K_{	ext{tran

Vedclass · Accepted Answer

$1.4 	imes 10^{-4}\,J$

Vedclass · Answer

(A) The total kinetic energy $(K)$ of a rolling sphere is the sum of its translational kinetic energy and rotational kinetic energy.$K = K_{	ext{trans}} + K_{	ext{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$.Since the sphere rolls without slipping,$v = R\omega$,which implies $\omega = \frac{v}{R}$.Substituting these into the energy equation:$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$$K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$Given: $m = 500\,g = 0.5\,kg$ and $v = 0.02\,m/s$.$K = \frac{7}{10} 	imes 0.5 	imes (0.02)^2$$K = 0.7 	imes 0.5 	imes 0.0004 = 0.35 	imes 0.0004 = 0.00014\,J = 1.4 	imes 10^{-4}\,J$.

$A$ uniform sphere of mass $500\,g$ rolls without slipping on a plane surface such that its centre moves at a speed of $0.02\,m/s$. The total kinetic energy of the rolling sphere would be (in $J$):

Similar Questions

$A$ hoop of radius $2 \; m$ weighs $100 \; kg$. It rolls along a horizontal floor so that its centre of mass has a speed of $20 \; cm/s$. How much work has to be done to stop it?

$A$ solid sphere of radius $10 \ cm$ and mass $500 \ g$ is rolling without slipping with a velocity of $20 \ cm/s$. The total kinetic energy of the sphere is: (in $J$)

$A$ 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 . . . . . . .

$A$ uniform disc of radius $R$ is resting on a table on its rim. The coefficient of friction between the disc and the table is $\mu$. Now,the disc is pulled with a force $F$ applied at its center as shown in the figure. What is the maximum value of $F$ for which the disc rolls without slipping?

$A$ wheel of radius $r$ rolls without slipping with a speed $v$ on a horizontal road. When it is at a point $A$ on the road,a small piece of mud separates from the wheel at its highest point $B$ and drops at point $C$ on the road. The distance $AC$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module