A solid sphere is rolling on a horizontal pla | vedclass

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

Question

(C) For a solid sphere rolling without slipping,the angular momentum $L$ about the axis of rotation is $L = I_{com}\omega = (\frac{2}{5}MR^2)\omega$.S

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) For a solid sphere rolling without slipping,the angular momentum $L$ about the axis of rotation is $L = I_{com}\omega = (\frac{2}{5}MR^2)\omega$.Since $V_{com} = R\omega$,we have $L = \frac{2}{5}MR^2(\frac{V_{com}}{R}) = \frac{2}{5}MRV_{com}$.The total kinetic energy $K$ is $K = \frac{1}{2}I_{com}\omega^2 + \frac{1}{2}MV_{com}^2 = \frac{1}{2}(\frac{2}{5}MR^2)(\frac{V_{com}}{R})^2 + \frac{1}{2}MV_{com}^2 = \frac{1}{5}MV_{com}^2 + \frac{1}{2}MV_{com}^2 = \frac{7}{10}MV_{com}^2$.The ratio $\frac{L}{K} = \frac{\frac{2}{5}MRV_{com}}{\frac{7}{10}MV_{com}^2} = \frac{2}{5} 	imes \frac{10}{7} 	imes \frac{R}{V_{com}} = \frac{4}{7} 	imes \frac{R}{R\omega} = \frac{4}{7\omega}$.Given $\frac{L}{K} = \frac{\pi}{22}$,we have $\frac{4}{7\omega} = \frac{\pi}{22}$. Assuming $\pi \approx \frac{22}{7}$,we get $\frac{4}{7\omega} = \frac{22/7}{22} = \frac{1}{7}$.Therefore,$7\omega = 28$,which gives $\omega = 4\,rad/s$.

$A$ solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about the axis of rotation of the sphere to the total energy of the moving sphere is $\pi: 22$,then the value of its angular speed will be $...........\,rad/s$.

Similar Questions

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

$A$ ring and a disc roll on a horizontal surface without slipping with the same linear velocity. If both have the same mass and the total kinetic energy of the ring is $6 \ J$,then the total kinetic energy of the disc is: (in $/2 \ J$)

$A$ disc of radius $1\,m$ and mass $4\,kg$ rolls on a horizontal plane without slipping in such a way that its centre of mass moves with a speed of $10\,cm/s$. Its rotational kinetic energy is (in $,J$)

$A$ thin uniform circular disc rolls with a constant velocity without slipping on a horizontal surface. Its total kinetic energy is

$A$ solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is $:$

Vedclass Test Series

Exam Paper Generator

Online Exam Module