A solid uniform sphere having a mass M,radius | vedclass

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

Question

(D) For a sphere in pure rolling,the equations of motion are:$N - Mg \cos 	heta = 0 \quad \dots(i)$$Mg \sin 	heta - f = Ma_{CM} \quad \dots(ii)$$	a

Vedclass · Accepted Answer

the sphere will roll without slipping only,if $	heta \leq 	an^{-1} \frac{7 \mu_s}{2}$

Vedclass · Answer

(D) For a sphere in pure rolling,the equations of motion are:$N - Mg \cos 	heta = 0 \quad \dots(i)$$Mg \sin 	heta - f = Ma_{CM} \quad \dots(ii)$$	au = f 	imes R = I \alpha = I \frac{a_{CM}}{R} \quad \dots(iii)$Since $I = \frac{2}{5} MR^2$,substituting into $(iii)$ gives $f = \frac{2}{5} Ma_{CM}$.Substituting $f$ into $(ii)$:$Mg \sin 	heta - \frac{2}{5} Ma_{CM} = Ma_{CM} \implies Mg \sin 	heta = \frac{7}{5} Ma_{CM} \implies a_{CM} = \frac{5}{7} g \sin 	heta$.Now,substituting $a_{CM}$ back into the expression for $f$:$f = \frac{2}{5} M \left( \frac{5}{7} g \sin 	heta ight) = \frac{2}{7} Mg \sin 	heta$.For pure rolling without slipping,the static friction must satisfy $f \leq \mu_s N$.Substituting $f = \frac{2}{7} Mg \sin 	heta$ and $N = Mg \cos 	heta$:$\frac{2}{7} Mg \sin 	heta \leq \mu_s Mg \cos 	heta$$	an 	heta \leq \frac{7}{2} \mu_s$$	heta \leq 	an^{-1} \left( \frac{7}{2} \mu_s ight)$.

$A$ solid uniform sphere having a mass $M$,radius $R$ and moment of inertia of $\frac{2}{5} M R^2$ rolls down a plane inclined at an angle $\theta$ to the horizontal starting from rest. The coefficient of static friction between the sphere and the plane is $\mu_s$. Then,

Similar Questions

$A$ body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity $V$. If the same body is a ring which rolls down the same inclined plane,then the linear velocity at the bottom of the plane is:

$A$ disc of mass $3 \, kg$ rolls down an inclined plane of height $5 \, m$. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is .......... $J$. (Take $g = 10 \, m/s^2$)

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\,cm$. If the cylinder rolls without slipping,its speed upon reaching the bottom of the inclined plane is $...........\,ms^{-1}$. (Given $g = 10\,ms^{-2}$)

$A$ solid sphere and a solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping throughout the motion. The two climb maximum heights $h_{sph}$ and $h_{cyl}$ on the incline. The ratio $\frac{h_{sph}}{h_{cyl}}$ is given by

$A$ sphere of radius $R$ is rolling down an inclined plane whose angle of inclination is $\theta$. Its acceleration would be

Vedclass Test Series

Exam Paper Generator

Online Exam Module