A solid cylinder of mass m is wrapped with an | vedclass

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

Question

(C) Let the solid cylinder be in equilibrium.The forces acting along the inclined plane are the tension $T$ and the friction $f$. The component of gra

Vedclass · Accepted Answer

$\frac{mg}{5}$

Vedclass · Answer

(C) Let the solid cylinder be in equilibrium.The forces acting along the inclined plane are the tension $T$ and the friction $f$. The component of gravity along the plane is $mg \sin 60^{\circ}$.For translational equilibrium: $T + f = mg \sin 60^{\circ} \quad ......(i)$For rotational equilibrium about the center of the cylinder: $TR - fR = 0 \implies T = f \quad ......(ii)$Substituting $(ii)$ into $(i)$: $2f = mg \sin 60^{\circ} \implies f = \frac{mg \sin 60^{\circ}}{2} = \frac{mg \sqrt{3}}{4} \approx 0.433 mg$.The limiting friction is $f_{L} = \mu_{s} N = \mu_{s} mg \cos 60^{\circ} = 0.4 	imes mg 	imes 0.5 = 0.2 mg$.Since the required friction $(0.433 mg)$ is greater than the limiting friction $(0.2 mg)$,the cylinder will not remain in static equilibrium and will roll down.The friction acting will be kinetic friction: $f_{k} = \mu_{k} N$. Assuming $\mu_{k} = \mu_{s} = 0.4$,we get $f_{k} = 0.4 	imes mg 	imes 0.5 = 0.2 mg = \frac{mg}{5}$.

$A$ solid cylinder of mass $m$ is wrapped with an inextensible light string and is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is: [The coefficient of static friction,$\mu_{s}$,is $0.4$]

Similar Questions

$A$ cylinder of mass $m$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What will be the velocity of its center of mass when the cylinder reaches the bottom?

$A$ solid sphere rolls down from rest on an inclined plane. At the same time,a rectangular block slides down from rest on the same inclined plane. Which of the following is true?

If a sphere is rolling,the ratio of its rotational energy to the total kinetic energy is given by

The ratio of the accelerations for a solid sphere (mass $m$ and radius $R$) rolling down an incline of angle $\theta$ without slipping and slipping down the incline without rolling is

$A$ cylinder rolls up an inclined plane to a certain height and then rolls down. (It does not slip during the entire motion.) What is the direction of the frictional force acting on the cylinder?

Vedclass Test Series

Exam Paper Generator

Online Exam Module