A solid sphere of radius R is rolled by a for | vedclass

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

Question

(B) The force $F$ acts at the top of the sphere,which is at a distance $2R$ from the point of contact. When the centre of mass $(CM)$ moves by a dista

Vedclass · Accepted Answer

Speed of the $CM$ when $CM$ moves a distance $S$ is $\sqrt{\frac{20FS}{7M}}$.

Vedclass · Answer

(B) The force $F$ acts at the top of the sphere,which is at a distance $2R$ from the point of contact. When the centre of mass $(CM)$ moves by a distance $S$,the point of application of force moves by $2S$. Thus,the work done by force $F$ is $W_F = F(2S) = 2FS$. Since the sphere rolls without slipping,the work done by static friction is zero. Using the work-energy theorem: $W_F + W_{friction} = \Delta K$. $2FS = \frac{1}{2}Mv^2 + \frac{1}{2}I_{cm}\omega^2$. For a solid sphere,$I_{cm} = \frac{2}{5}MR^2$ and for rolling without slipping,$\omega = \frac{v}{R}$. $2FS = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2$. $2FS = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$. $v^2 = \frac{20FS}{7M} \implies v = \sqrt{\frac{20FS}{7M}}$. Therefore,option $B$ is correct.

$A$ 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:

Similar Questions

For a rolling spherical shell,the ratio of rotational kinetic energy and total kinetic energy is $\frac{x}{5}$. The value of $x$ is ................

$A$ body is rolling on a horizontal surface without slipping. Its rotational kinetic energy is equal to its translational kinetic energy. The body is a:

$A$ solid sphere of mass $2 \ kg$ and radius $0.5 \ m$ is rolling without slipping on a horizontal surface. The ratio of the rotational and translational kinetic energies of the sphere is (in $: 5$)

$A$ solid disc and a ring,both of radius $10\; cm$,are placed on a horizontal table simultaneously,with an initial angular speed equal to $10\pi\; rad\; s^{-1}$. Which of the two will start to roll earlier? The coefficient of kinetic friction is $\mu_k = 0.2$.

$A$ sphere of mass $50 \, g$ and diameter $20 \, cm$ is rolling without slipping with a velocity of $5 \, cm/s$. Its total kinetic energy is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module